Answer:
(a) n₂ = 2
(b) n₂ = 5
(c) n₂ = 4
(d) n₂ = 3
(e) n₂ = 6
Explanation:
The Rydberg equation give us the wavelength of the transition between energy levels according to the formula:
1/λ = Rh x ( 1/n₁² - 1/n₂² )
where n₁ is the final state and n₂ is the initial state.
The strategy here, since we are given the wavelength, is to solve for λ, and then by substituting for n₂ combinations find which ones match our question.
λ = 1 / [ Rh x ( 1/n₁² - 1/n₂² ) ]
Lets express Rh in 1/Angstrom
1.097 x 10 ⁷ / [m x ( 1 m/ 10¹⁰ A) ] = 0.011 / Å
⇒ λ = 1 / [0.011 A x ( 1/n₁² - 1/n₂² )] = 911.6 Å / ( 1/n₁² - 1/n₂² )
For n₁ = 2 n₂ = 3, 4, 5,.......
λ ( n₂ = 3 ) = 911.6 A / ( 1/2² - 1/3² ) = 6563.5 Å
λ ( n₂ = 4 ) = 911.6 A / ( 1/2² - 1/4² ) = 4861.9 Å
λ ( n₂ = 5 ) = 911.6 A / ( 1/2² - 1/5² ) = 4341.0 Å
So we have matched three of the transitions
Now for n₁ = 1 n₂ = 2, 3, 4....
λ ( n₂ = 2 ) = 911.6 A / ( 1/1² - 1/2² ) = 1215.5 Å
For n₁ = 3 n₂ = 4, 5, 6....
λ ( n₂ = 4 ) = 911.6 A / ( 1/3² - 1/4² ) = 18752.9 Å
λ ( n₂ = 5 ) = 911.6 A / ( 1/3² - 1/5² ) = 12819.4 Å
λ ( n₂ = 6 ) = 911.6 A / ( 1/3² - 1/6² ) = 10939.2 Å
a. For [tex]\lambda=1212.7\;\rm \r{A}[/tex], initial transition level is 2.
b. For [tex]\lambda=4340.5\;\rm \r{A}[/tex], initial transition level is 5.
c. For [tex]\lambda=4861.3\;\rm \r{A}[/tex], initial level of transition is 4.
d. For [tex]\lambda=6562.8\;\rm \r{A}[/tex], initial level of transition is 3.
e. For [tex]\lambda=10,938\;\rm \r{A}[/tex], the initial level of transition has been 6.
The wavelength from the transition has been given by the Rydberg equation as:
[tex]\dfrac{1}{\lambda}=\text Rh\dfrac{1}{N_1}-\dfrac{1}{N_2}[/tex]
Where, wavelength of the radiation, [tex]\lambda[/tex]
The initial transition level, [tex]N_1[/tex]
The final transition level, [tex]N_2[/tex]
The constant [tex]Rh=1.097\;\times\;10^7[/tex], or in Armstrong it can be given as, [tex]0.011/\r{\rm A}[/tex]
The wavelength can be given as:
[tex]\lambda=\dfrac{1}{0.011\;\r{\rm A}\;\times\;(\frac{1}{N_1^2} -\frac{1}{N_2^2}) }[/tex]
[tex]\rm \lambda=\dfrac{911.6\;\r{A}}{\frac{1}{N_1^2}-\frac{1}{N_2^2} }[/tex]
a. The initial level ([tex]N_2[/tex]) of transition has been given for [tex]\lambda=1212.7\;\rm \r{A}[/tex], and [tex]N_2=2[/tex] has been given, with substituting [tex]N_1=1[/tex]
[tex]\rm \lambda=\dfrac{911.6\;\r{A}}{\frac{1}{1^2}-\frac{1}{2^2} }\\\lambda=1215.5\;\r{A}[/tex]
Thus, for [tex]\lambda=1212.7\;\rm \r{A}[/tex], initial transition level is 2.
b. For, [tex]\lambda=4340.5\;\rm \r{A}[/tex], and [tex]N_2=2[/tex], [tex]N_1=5[/tex]
[tex]\rm \lambda=\dfrac{911.6\;\r{A}}{\frac{1}{2^2}-\frac{1}{5^2} }\\\lambda=4341.0\;\r{A}[/tex]
Thus, for [tex]\lambda=4340.5\;\rm \r{A}[/tex], initial transition level is 5.
c. For, [tex]\lambda=4861.3\;\rm \r{A}[/tex], the final transition has been, [tex]N_2=2[/tex], the initial level has been substituted as [tex]N_1=4[/tex]:
[tex]\rm \lambda=\dfrac{911.6\;\r{A}}{\frac{1}{2^2}-\frac{1}{4^2} }\\\lambda=4861.3\;\r{A}[/tex]
Thus, for [tex]\lambda=4861.3\;\rm \r{A}[/tex], initial level of transition is 4.
d. For [tex]\lambda=6562.8\;\rm \r{A}[/tex], the final level has been 2, the initial level has been substituted as, [tex]N_1=3[/tex]
[tex]\rm \lambda=\dfrac{911.6\;\r{A}}{\frac{1}{2^2}-\frac{1}{3^2} }\\\lambda=6562.8\;\r{A}[/tex]
Thus, for [tex]\lambda=6562.8\;\rm \r{A}[/tex], initial level of transition is 3.
e. For [tex]\lambda=10,938\;\rm \r{A}[/tex], the initial level has been substituted as, [tex]N_1=6[/tex]
[tex]\rm \lambda=\dfrac{911.6\;\r{A}}{\frac{1}{3^2}-\frac{1}{6^2} }\\\lambda=10,938\;\r{A}[/tex]
Thus, for [tex]\lambda=10,938\;\rm \r{A}[/tex], the initial level of transition has been 6.
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What ions are possible for the two largest stable elements in Group 4A(14)? How does each arise?
Answer:
The elements are Tin (Sn) and Lead (Pb)
Explanation:
They are 4+ and 2+
For the 2+, this happens because of the loss of the outermost two P electrons and for the 4+, because of the loss of two electrons from the P orbital and another two electrons from the s-orbital.
KI KI has a lattice energy of − 649 kJ/mol. −649 kJ/mol. Consider a generic salt, AB AB , where A 2 + A2+ has the same radius as K + , K+, and B 2 − B2− has the same radius as I − . I−. Estimate the lattice energy of the salt AB AB .
Answer:
The answer to this question is
The lattice energy of the salt AB is ≅ -2600 kJ/mol rounded to three significant figures
Explanation:
To solve this we list out the known variables and the required unkown, then we look for the appropriate relation between the known and the unknown thus
Lattice energy of KI = -649 kJ/mol
Latice energy of salt AB = unknown
Charge on A²⁺ =+2
Charge on B²⁻ = -2
radius of A²⁺ = radius of K⁺
radius of B²⁻ = radius of I⁻
the Born-Lande equation for lattice energies is as follows
[tex]E=-\frac{N_{A} Mz^{+}z^{-} e^{2} }{4\pi e_{0} r_{0} } (1-\frac{1}{n})[/tex]
[tex]E=-\frac{N_{A} M e^{2} }{4\pi e_{0} } (1-\frac{1}{n})\frac{|z|^{+}|z|^{-}}{r_{0}}[/tex]
Where to the above question, the following terms are important
z+ = positive ion charge
z− = negative ion charge
r₀ → distance between the positive and the negative ions;
Taking all variables of KI and AB as identical we have.
[tex]E=-\frac{N_{A} M e^{2} }{4\pi e_{0} } (1-\frac{1}{n})\frac{|z|^{+}|z|^{-}}{r_{0}}[/tex]
[tex]E=CONSTANT\frac{|z|^{+}|z|^{-}}{r_{0}}[/tex]
Where radius of A²⁺ = radius of K⁺ and
radius of B²⁻ = radius of I⁻ = r₀
we have
[tex]E_{AB} = E_{KI} (|z|^{+} |z|^{-} )[/tex]
As for AB z⁺ = +2 and z⁻ = -2 we have
[tex]E_{AB} = E_{KI} (|+2|^{+} |-2|^{-} ) = 4E_{KI}[/tex]
= 4 ×(-649 kJ/mol) = -2596 kJ/mol
[tex]E_{AB}[/tex] ≅ -2600 kJ/mol rounded to three significant figures
The lattice energy of [tex]AB[/tex][tex]=-2596 \ kJ/mol.[/tex]
[tex]\Delta H_{Lattice}=\frac{C\left ( z^{+} \right )\left ( z^{-} \right )}{R_{o}}[/tex]
where,
c=constant
[tex]( z^{+} \right) )\left, ( z^{-} \right )}[/tex]= Charges on ions.
[tex]{R_{o}}[/tex]= Interionic distance (radius of cation+anion).
From the lattice energy formula, we know that lattice energy is directly proportional to the charge of the ions and inversely proportional to the radius of the ions.
In the given question charge of the ions doubled so lattice energy will be increased four-fold.
The lattice energy of AB[tex]=Lattice \ energy \ of \ kl\times4[/tex]
[tex]=-649\times4[/tex]
The lattice energy of [tex]AB[/tex][tex]=-2596 \ kJ/mol.[/tex]
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The angle at which light strikes a surface is the same as the angle at which it is reflected. true or false
Answer: True
Explanation: the angle of incident Ray equals the angle of the reflected Ray.
Cobalt-60 is a radioactive isotope used to treat cancers. A gamma ray emitted by this isotope has an energy of 1.33 MeV (million electron volts; 1 eV = 1.602 x 10¹⁹ J). What is the frequency (in Hz) and the wavelength (in m) of this gamma ray?
Answer:
E = 1.33 MeV = 2.13 x [tex]10^{-13}[/tex] J
v = wavelength = E / h = 2.13 x [tex]10^{-13}[/tex] / 6.626 x [tex]10^{-34}[/tex] = 3.2 x [tex]10^{20}[/tex] m
f = frequency = c / 3.2 x [tex]10^{20}[/tex] m = 3 x [tex]10^{8}[/tex] / 3.2 x [tex]10^{20}[/tex] = 9.375 x [tex]10^{-13}[/tex] Hz
A reaction has a rate constant of 0.0642 sec-1; how long will it take (in minutes) until 0.479 mol/L of the compound is left, if there was 0.502 mol/L at the start? (give answer to 3 decimal places)?
Answer:
[tex]t=0.012\ min[/tex]
Explanation:
Using integrated rate law for first order kinetics as:
[tex][A_t]=[A_0]e^{-kt}[/tex]
Where,
[tex][A_t][/tex] is the concentration at time t
[tex][A_0][/tex] is the initial concentration
Given that:
The rate constant, k = 0.0642 s⁻¹
Initial concentration [tex][A_0][/tex] = 0.502 mol/L
Final concentration [tex][A_t][/tex] = 0.479 mol/L
Time = ?
Applying in the above equation, we get that:-
[tex]0.479=0.502e^{-0.0642\times t}[/tex]
[tex]e^{-0.0642t}=\frac{0.479}{0.502}[/tex]
[tex]t=-\frac{\ln \left(\frac{0.479}{0.502}\right)}{0.0642}[/tex]
[tex]t=0.731\ sec[/tex]
Also, 1 sec = 1/60 min
So, [tex]t=0.012\ min[/tex]
Final answer:
Using the first-order rate law equation and the provided rate constant, it will take approximately 0.012 minutes (rounded to three decimal places) until the concentration of the compound decreases from 0.502 mol/L to 0.479 mol/L.
Explanation:
To find out how long it will take until 0.479 mol/L of the compound is left, given that the initial concentration is 0.502 mol/L and the rate constant is 0.0642 sec⁻¹, we need to use the first-order rate law equation, which is given by:
ln([A]t/[A]0) = -kt
Where: t is the time
Plugging in the values we have:
ln(0.479/0.502) = -0.0642 * t
The calculation would proceed as follows:
ln(0.479/0.502) = -0.0642 * t
ln(0.9545) = -0.0642 * t
-0.0465 = -0.0642 * t
t = 0.0465 / 0.0642
t ≈ 0.724 sec
To convert seconds to minutes, divide by 60:
t ≈ 0.724 sec / 60
t ≈ 0.012 min
Rounding up to three decimal places, the time t will be 0.012 min.
Suppose you have 75 gas-phase molecules of methanol (CH3OH) at T = 470 K. These molecules are contained in a spherical container of volume 0.500 liters.At this temperature, the root mean square speed of methanol molecules is 605 m/sec.What is the average pressure in the container due to these 75 molecules?(The molar mass of methanol is 32.0 g/mol.)
Answer:
The average pressure in the container due to these 75 gas molecules is [tex]P=9.72 \times 10^{-16} Pa[/tex]
Explanation:
Here Pressure in a container is given as
[tex]P=\frac{1}{3} \rho <u^2>[/tex]
Here
P is the pressure which is to be calculatedρ is the density of the gas which is to be calculated as below[tex]\rho =\frac{mass}{Volume of container}[/tex]
Here
mass is to be calculated for 75 gas phase molecules as
[tex]m=n_{molecules} \times \frac{1 mol}{6.022 \times 10^{23} molecules} \times \frac{32 g/mol}{1 mol}\\m=75 \times \frac{1 mol}{6.022 \times 10^{23} molecules} \times \frac{32 g/mol}{1 mol}\\m=3.98 \times 10^{-21} g[/tex]
Volume of container is 0.5 lts
So density is given as
[tex]\rho =\frac{mass}{Volume of container}\\\rho =\frac{3.98 \times 10^{-21} \times 10^{-3} kg}{0.5 \times 10^{-3} m^3}\\\rho =7.97 \times 10^{-21}\, kg/m^3[/tex]
[tex]<u^2>[/tex] is the mean squared velocity which is given as[tex]<u^2>=RMS^2[/tex]
Here RMS is the Root Mean Square speed given as 605 m/s so
[tex]<u^2>=RMS^2\\<u^2>=(605)^2\\<u^2>=366025[/tex]
Substituting the values in the equation and solving
[tex]P=\frac{1}{3} \rho <u^2>\\P=\frac{1}{3} \times 7.97 \times 10^{-21} \times 366025\\P=9.72 \times 10^{-16} Pa[/tex]
So the average pressure in the container due to these 75 gas molecules is [tex]P=9.72 \times 10^{-16} Pa[/tex]
Complete each of the definitions with the appropriate phrase. Precision means that measurements are close to Accuracy means that measurements are close to Anawer Bank each other an accepted value |individual several
Answer: Precision means that measurements are close to each other . Accuracy means that measurements are close to accepted value
Explanation:
Precision refers to the closeness of two or more measurements to each other.
For Example: If we weigh a given substance five times and you get 5.0 kg each time. Then the measurement is very precise.
Accuracy refers to the closeness of a measured value to a standard or known value.
For Example: If the mass of a substance is 5.0 kg and one person weighed 4.9 kg and another person weighed 3.9 kg. Then, the weight measured by first person is more accurate.
Thus Precision means that measurements are close to each other . Accuracy means that measurements are close to accepted value.
Precision in measurements means the values are close to each other, indicating consistency or reproducibility. Accuracy describes how close a measurement is to the correct or true value. For the best scientific outcomes, both high precision and high accuracy are desired.
Explanation:When discussing measurements in scientific terms, precision and accuracy are two critical concepts that often get confused. Precision means that measurements are close to each other. It refers to the consistency of repeated measurements. On the other hand, accuracy means that measurements are close to an accepted value, which is the correct or true value of the quantity being measured.
For example, in archery, if several arrows hit close to each other but not near the target's center, the shots are precise but not accurate. Conversely, if an arrow hits the center of the target, it's considered accurate, especially if repeated shots hit the same spot, which would show both accuracy and precision.
The aim in scientific measurement is to achieve both high precision and high accuracy, thereby ensuring that measurements not only agree with each other but also with the true or accepted value. However, a measurement instrument can sometimes have one without the other. It's possible to have a set of measurements that are very close to each other (precise) but far from the correct value (inaccurate), or to have a measurement that is very close to the correct value (accurate) but when repeated fluctuates significantly (imprecise).
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Classify each of the following chemical reaction (check all that apply) :
combination, precipitation, single replacement, combustion, double replacement, acid-base, decomposition
(1) HI(aq) + NaOH(aq) → Nal(aq) + H20(l)
(2) CuSO4(aq) + ZnCrO4(aq) → ZnSO4(aq) + CuCrO4(s)
(3) CH3CH2OH(l) + 302(g) → 2CO2(g) + 3H2O(g)
(4) NaCl(aq) + AgNO3(aq) → NaNO3(aq) + AgCl(s)
Reactions (1), (2), (3) and (4) are classified as acid-base, double replacement, combustion, and both double replacement and precipitation reactions respectively.
Explanation:The reactions can be classified as follows:
(1) HI(aq) + NaOH(aq) → NaI(aq) + H₂O(l) is an acid-base reaction. In this reaction, an acid (HI) and a base (NaOH) react to form a salt (NaI) and water.(2) CuSO₄(aq) + ZnCrO₄(aq) → ZnSO₄(aq) + CuCrO₄(s) is a double replacement reaction. Double replacement reactions involve the exchange of ions between two compounds.(3) CH₃CH₂OH(l) + 3O₂(g) → 2CO₂(g) + 3H₂O(g) is a combustion reaction. Combustion reactions involve a compound reacting with oxygen to produce carbon dioxide and water.(4) NaCl(aq) + AgNO₃(aq) → NaNO₃aq) + AgCl(s) is both a double replacement and a precipitation reaction. AgCl forms a precipitate during this reaction as it is insoluble in water.Learn more about Classifying Chemical Reactions here:https://brainly.com/question/32766737
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You need to purchase an expensive chemical for a series of enzyme assays. Sigma Chemical Company sells the chemical in 1g ($30), 5g ($120), 10g ($200), and 25g ($400) vials. You will perform 40 assays. Each assay requires 3 ml of a 250mM solution of the chemical (M.W. = 156 g/mole). Which vial(s) is the best choice for your limited budget?
Answer:
5g ($120)
Explanation:
The amount of chemical needed for assay is 3 ml of 250 mM solution of a chemical
Molarity (M) = number of moles / volume in L
number of mole = M × volume in L
M of the chemical = 250 mM = 250 / 1000 = 0.25 M
number of moles = 0.25 × ( 3 / 1000) in L = 0.00075 m
mass of the chemical needed = 0.00075 × molar mass = 0.00075 × 156 = 0.117 g for each assay.
mass needed for 40 assay = 40 × 0.117 = 4.68 g
It is therefore wise for him to buy the 5 g ( $ 120), though the 10 g and 25 g yields better prices per gram, they much more than what he needed for the assay.
A West Virginia coal is burned at a rate of 8.02 kg/s. The coal has a sulfur content of 4.40 % and the bottom ash contains 2.80 % of the input sulfur. What is the annual rate (in kg/yr) of stack emissions of SO2
Answer: The annual emission rate of SO2 is 1.08 × [tex]10^{7}[/tex] kg/yr
Explanation:
The rate r at which the coal is been burnt is 8.02 kg/s.Amount of sulphur in the burning coal is given as 4.40 %i.e., 4.4/100 × 8.02 = 0.353 kg/s. Which is equivalent to the rate at which the sulphur is been burnt.
Since the burning of sulphur oxidizes it to produce SO2, it follows that the non-oxidized portion of the sulphur will go with the bottom ash.The bottom ash is said to contain 2.80 % of the input sulphur.Hence the portion of the SO2 produced is 100 — 2.80 = 97.20 %.The rate of the SO2 produced is percentage of SO2 × rate at high sulphur is been burnt.= 97.20/100 × 0.353 kg/s.
= 0.343 kg/s.
To get the annual emission rate of SO2, we convert the kg/s into kg/yr.
1 kg/s = 1 kg/s × (60 × 60 × 24 × 365) s/yr
1 kg/s = 31536000 kg/yr
Therefore, 0.343 kg/s = 0.343 × 31536000 kg/yr
= 10816848 kg/yr
= 1.08 × 10^7 kg/yryr.
What is the pH of an aqueous solution with the hydronium ion concentration : [H3O+] = 6.3 x 10-6 M ?
Answer:
pH = 5.2
Explanation:
pH of a solution is defined as the concentration of hydrogen ion in a solution. It is also the negative logarithmic value of the concentration of hydrogen ion. It determines the acidity or alkalinity of a solution.
HA(aq) + H2O(l) --> A-(aq) + H3O+(aq)
pH = -log[H3O+]
= - log [6.3 x 10-6]
= 5.200
In a game of "Clue," Ms. White is killed in the conservatory. You have a device in each room to help you find the murderer— a spectrometer that emits the entire visible spectrum to indicate who is in that room. For example, if someone wearing yellow is in a room, light at 580 nm is reflected. The suspects are Col. Mustard, Prof. Plum, Mr. Green, Ms. Peacock (blue), and Ms. Scarlet. At the time of the murder, the spectrometer in the dining room recorded a reflection at 520 nm, those in the lounge and study recorded reflections of lower frequencies, and the one in the library recorded a reflection of the shortest possible wavelength. Who killed Ms. White? Explain.
Answer: Ms. Scarlet is the murderer.
Explanation: Each color has a wavelength band, some visible to the human eye (visible spectrum), some for the other animals. In this game of "Clue", the color are yellow (Col. Mustard), violet (Prof. Plum), green (Mr. Green), blue (Ms. Peacock) and red (Ms. Scarlet). Using the device and knowing each band, it's determined that at the time of murder, Mr. Green was in the dining room, because green light has the wavelength between 495 to 570nm; Prof. Plum was in the library, due to violet's light has the shortest wavelength; Col. Mustard and Ms. Peacock were either in the lounge or in the study, because of their color's light wavelength measurement. So, the person responsible for the murderer was Ms. Scarlet, as the other devices in the other places didn't register a higher wavelength, which is compatible to the red light.
According to the spectrometer readings, Ms. Peacock is the murderer of Ms. White in the conservatory, as the reflected colors in the other rooms correspond to the other characters' colors and leave only Ms. Peacock unaccounted for at the crime scene.
To determine who killed Ms. White in the conservatory in a game of "Clue" using a spectrometer, we need to match the color associated with the wavelength of light reflected in each room with the corresponding suspect's color. The dining room spectrometer recorded a reflection at 520 nm, which corresponds to green.
Since Mr. Green reflects green and that is his character color, he is in the dining room. The lounge and study recorded reflections of lower frequencies i.e., longer wavelengths than green, which suggest colors towards the red end of the spectrum, likely placing Ms. Scarlet (red) and Prof. Plum (purple) in those rooms.
The library recorded a reflection of the shortest possible wavelength, meaning violet, which is typically associated with Prof. Plum; however, since he is presumed to be in a room with a longer wavelength, that leaves Col. Mustard (who wears yellow) as the suspect in the library because yellow requires the absorption of wavelengths at the violet end of the spectrum.
Therefore, Ms. Peacock, whose color is blue and wasn't indicated in the other rooms, must have been in the conservatory and is determined to be the murderer according to the spectrometer readings.
Solution of the Schrödinger wave equation for the hydrogen atom results in a set of functions (orbitals) that describe the behavior of the electron. Each function is characterized by 3 quantum numbers: n, l, and ml ... n is known as the quantum number. ... l is known as the quantum number. ... ml is known as the quantum number. ... n specifies l specifies ml specifies ... A.The orbital orientation. B.The energy and average distance from the nucleus. C.The subshell - orbital shape.
Answer :
'n' specifies → (B) The energy and average distance from the nucleus.
'l' specifies → (C) The subshell orbital shape.
'ml' specifies → (A) The orbital orientation.
Explanation :
Principle Quantum Numbers : It describes the size of the orbital. It is represented by n. n = 1,2,3,4....
Azimuthal Quantum Number : It describes the shape of the orbital. It is represented as 'l'. The value of l ranges from 0 to (n-1). For l = 0,1,2,3... the orbitals are s, p, d, f...
Magnetic Quantum Number : It describes the orientation of the orbitals. It is represented as [tex]m_l[/tex]. The value of this quantum number ranges from [tex](-l\text{ to }+l)[/tex]. When l = 2, the value of
Spin Quantum number : It describes the direction of electron spin. This is represented as [tex]m_s[/tex]. The value of this is [tex]+\frac{1}{2}[/tex] for upward spin and [tex]-\frac{1}{2}[/tex] for downward spin.
As per question we conclude that,
'n' specifies → The energy and average distance from the nucleus.
'l' specifies → The subshell orbital shape.
'ml' specifies → The orbital orientation.
The Schrödinger wave equation for a hydrogen atom uses three quantum numbers to characterize wave functions, or orbitals, of an electron. The principal quantum number 'n' indicates the energy and average distance from the nucleus, the azimuthal or orbital quantum 'l' represents the shape of the orbital, and the magnetic quantum number 'ml' describes the orientation of the orbital.
Explanation:The solution of the Schrödinger wave equation for the hydrogen atom leads to the development of wave functions, also known as orbitals, that specify the behavior of the electron. Each of the wave functions or orbitals are defined by three quantum numbers: n, l, and ml.
The principal quantum number, n, specifies the energy and average distance of the electron from the nucleus. The azimuthal or orbital quantum number, l, determines the shape of the orbital, essentially the subshell. The magnetic quantum number, ml, gives the orientation of these orbitals in space.
In summary, the quantum numbers describe the discrete states that an electron can occupy in a hydrogen atom and provide the physical significance for the electron's behavior within the atom based on the Schrödinger equation.
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How many inner, outer, and valence electrons are present in an atom of each of the following elements?
(a) O (b) Sn (c) Ca (d) Fe (e) Se
Oxygen has 2 inner electrons, 6 outer electrons, and 6 valence electrons. Tin has 40 inner electrons, 10 outer electrons, and 10 valence electrons. Calcium has 18 inner electrons, 2 outer electrons, and 2 valence electrons. Iron has 18 inner electrons, 8 outer electrons, and 8 valence electrons. Selenium has 26 inner electrons, 6 outer electrons, and 6 valence electrons.
Explanation:Let's determine the number of inner, outer, and valence electrons for each element:
Oxygen (O): Atomic number 8 means it has 8 electrons. The distribution is 2 electrons in the first energy level, and the remaining 6 in the second energy level. Outer electrons: 6, Inner electrons: 2, Valence electrons: 6.Tin (Sn): Atomic number 50 means it has 50 electrons. The distribution is 2 electrons in the first energy level, 8 electrons in the second energy level, 18 electrons in the third energy level, and the remaining 22 in the fourth energy level. Outer electrons: 10, Inner electrons: 40, Valence electrons: 10.Calcium (Ca): Atomic number 20 means it has 20 electrons. The distribution is 2 electrons in the first energy level, 8 electrons in the second energy level, and the remaining 10 in the third energy level. Outer electrons: 2, Inner electrons: 18, Valence electrons: 2.Iron (Fe): Atomic number 26 means it has 26 electrons. The distribution is 2 electrons in the first energy level, 8 electrons in the second energy level, 14 electrons in the third energy level, and the remaining 2 in the fourth energy level. Outer electrons: 8, Inner electrons: 18, Valence electrons: 8.Selenium (Se): Atomic number 34 means it has 34 electrons. The distribution is 2 electrons in the first energy level, 8 electrons in the second energy level, 16 electrons in the third energy level, and the remaining 8 in the fourth energy level. Outer electrons: 6, Inner electrons: 26, Valence electrons: 6.Learn more about Electron distribution here:https://brainly.com/question/37566080
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The number of electrons in an atom can be determined by the electron configuration. For example, Oxygen has 2 inner, 6 outer, and 6 valence electrons while Tin has 50 inner, 14 outer, and 4 valence electrons.
Explanation:The electron configuration can be used to determine the number of inner, outer, and valence electrons in an atom.
O (Oxygen) has 2 inner electrons (in the 1s orbital), 6 outer electrons (in the 2s and 2p orbitals), and 6 valence electrons (also in the 2s and 2p orbitals). Sn (Tin) has 50 inner electrons, 14 outer electrons, and 4 valence electrons. Ca (Calcium) has 18 inner electrons, 2 outer electrons, and 2 valence electrons. Fe (Iron) has 18 inner electrons, 8 outer electrons, and 2 valence electrons. Se (Selenium) has 28 inner electrons, 6 outer electrons, and 6 valence electrons. Learn more about Electron Configuration here:
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The key to separating liquids with similar boiling points is to maximize the number of ____________.
Group of answer choices:
A. component plates
B. intermolecular forces
C. vaporization plates
D. theoretical plates
Answer:
D
Explanation: In order to separate liquids with similar boiling points,theoretical plates are the ones that must be maximized because two liquids would be at equilibrium with all properties being the same
To separate liquids with similar boiling points effectively, one needs to maximize the number of theoretical plates in the distillation process. Each theoretical plate represents a vaporization-condensation cycle, refining the separation and increasing the purity of the distillate.
Explanation:The key to separating liquids with similar boiling points is to maximize the number of theoretical plates. Theoretical plates are a concept used to describe the efficiency of the distillation process. Each theoretical plate represents a single vaporization-condensation cycle, where the composition of the vapor becomes richer in the more volatile component. For compounds with boiling points close to each other, multiple theoretical plates are needed to achieve high purity in the distillation product. The greater the number of theoretical plates in a distillation column, the more refined the separation process can be, allowing for a more effective separation of components with similar boiling points.
For example, if we distill a mixture with components having a small difference in boiling points, with enough theoretical plates, the more volatile component vaporizes first and, upon condensation, can be collected separately from the less volatile component. Using Raoult's law, we can predict the composition of the vapor and liquid at each stage and design the distillation process accordingly to optimize the number of theoretical plates and, in turn, the purity of the distillate.
Write the ground-state electron configurations of the following transition metal ions. (a) Sc3 (b) Cr3 (c) Cu (d) Au
The ground-state electron configurations for Sc3+, Cr3+, Cu, and Au are [Ar], [Ar]18 3d3, [Ar]18 3d10 4s1, and [Xe]6s1 4f14 5d10 respectively. Electrons are removed from the 4s orbital first and then from 3d in transition metals.
Explanation:The ground-state electron configurations of the transition metal ions are expressed as follows:
Sc3+: [Ar] (Scandium loses three electrons from 4s and 3d orbitals.)Cr3+: [Ar]18 3d3 (Chromium ion loses one electron from 4s and two from 3d orbital.)Cu: [Ar]18 3d10 4s1 (Copper has one electron in 4s and 10 in 3d in its ground state.) Au: [Xe]6s1 4f14 5d10 (Gold has one electron in 6s, ten in 5d, and 14 in 4f in its ground state.)It's essential to remember that for transition metals, electrons are removed from the 4s orbital first and then from the 3d orbital.
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The ground-state electron configurations for Sc3+, Cr3+, Cu, and Au ions are [Ar], [Ar] 3d3, [Ar] 3d10, and [Xe] 4f14 5d9, respectively, with the higher energy electrons being removed first to form the ions.
To write the ground-state electron configurations of the given transition metal ions, we must understand how electrons are removed when forming an ion. For transition metals, electrons are removed from the s orbital before the d orbital. Now let's determine the electron configurations:
Sc3+: The atomic number of Scandium (Sc) is 21. The electron configuration for Sc is [Ar] 3d1 4s2. When forming Sc3+, we remove three electrons, two from the 4s orbital and one from the 3d orbital, resulting in [Ar] as the electron configuration for Sc3+.Cr3+: Chromium (Cr) has an atomic number of 24. Its electron configuration is [Ar] 3d5 4s1, but upon forming Cr3+, three electrons are removed (one from the 4s orbital and two from the 3d orbital), leading to [Ar] 3d3 as the electron configuration for Cr3+.Cu: Copper (Cu) has an atomic number of 29. The electron configuration of Cu is [Ar] 3d10 4s1. In its monatomic ion form (Cu+), only one electron (from the 4s orbital) is removed, resulting in [Ar] 3d10 as the electron configuration for Cu+.Au: Gold (Au) has an atomic number of 79. Its electron configuration is [Xe] 4f14 5d10 6s1. The most common monatomic ion of gold is Au3+, which means three electrons are removed (two from the 6s orbital and one from the 5d orbital), giving [Xe] 4f14 5d9 as the electron configuration for Au3+.List three ways in which metals and nonmetals differ.
Explanation:
The point of difference between metals and non metals are as follows -
Metals -
Metals are malleable in nature , so can be beaten into thin sheets .
Non - metals -
Non - metals are brittle in nature , so when beaten , it can be powdered .
Metals -
Metals have one to three electrons in the very outer most shell .
Non - metals -
Non - metals have four to eight electrons in the very outer most shell .
Metals -
Metals are lustrous in appearance .
Non - metals -
Non - metals re not lustrous in appearance .
Draw the Lewis structure (including all lone pair electrons) for one isomer of the molecular formula C3H8O.
Answer:
Here's how I would do it.
Explanation:
A Lewis structure shows the valence electrons surrounding the atoms.
1. Identify the central atoms
The central atoms are the least electronegative atoms (C).
I chose as my central atoms a chain of four C atoms: C-C-C-C
I stuck the most electronegative atom (O) at the end of the chain:
C-C-C-C-O
2. Complete the structure
The hydrogen atoms must be terminal. Add enough to give every atom an octet.
You should get a structure like the one below
The Lewis structure for an isomer of C3H8O (like propanol) represents the atomic structure of the compound, including each atom and its valence electrons. Bonds are formed from the valence electrons and arranged around the carbon atoms enabling one to visualize the molecular structure.
Explanation:The Lewis structure for the isomer of C3H8O (propanol as an example) is a visual representation of the compound's atomic structure. It entails representing each atom and its valence electrons. By arranging the atoms in a molecule, we can then place the bonding and lone pair electrons. In propanol, the three carbon (C) atoms are aligned with one end bonded to a hydrogen (H) atom and the other end to an oxygen (O) atom. The oxygen atom is then also bonded to a hydrogen atom.
C-C-C-O-H
The remaining hydrogen atoms are evenly distributed around the carbon atoms and are bound to the carbon atoms. As a result, each carbon atom forms four bonds, oxygen forms two bonds and bears two lone pairs, and each hydrogen atom forms a single bond.
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To determine the molar mass of a protein, a 0.891 g sample of it is added to 5.00 g of water and the osmotic pressure is measured as 0.179 atm at 22.0°C, what is the molar mas of the protein? (The protein is non-ionizing)
Answer:
MW = 24,097 g/mol
Explanation:
The osmotic pressure of a solution is given by the equation:
πV = nRT
where π is the osmotic pressure, V is the volume, R is the gas constant 0.08205 Latm/kmol , and T is the temperature.
n , the number of moles is equal to m/MW, substituting into the equation:
πV = ( mass/MW ) RT
MW = mass x R x T / ( πV )
V is given by the density of solution assumed to be that of water:
d = m/v ⇒ v= m/d = 5.00 g / 1 g/mL = 5.00 mL
The volume we need to convert to liters for units consistency in the metric system:
5.00 mL x 1 L / 1000 mL = 5 x 10⁻³ L
solving for
MW = 0.891 g x 0.08205 Latm/Kmol x ( 22 + 273 )K / (0.179 atm x 5 x 10⁻³ L )
= 24,097 g/mol
What is the wavelength (in nm) of the least energetic spectral line in the visible series of the H atom?
Answer:
Brackett Series (n = 4)
Explanation:
The least energetic line of Hydrogen atom lies in Brackett Series when n = 4 because these are least energetic, have longer wavelengths and lies in Infrared region of spectrum. No traces of Pfund series are formed by H=atoms.
The following questions deal with safety issues in the Solutions and Spectroscopy experiment. (Select all that apply.) (a) Which hazards are associated with copper(II) sulfate? carcinogenic irritant toxic corrosive flammable (b) What actions should you take if you spill copper(II) sulfate solution on yourself? Dry the area with paper towels, but avoid getting it wet. Neutralize it with a solution of sodium bicarbonate. Neutralize it with a solution of sodium bisulfite. Flush the affected area with water. (c) What clothing is appropriate for the laboratory? shorts tank tops without sleeves shirt and pants that overlap 6" sandals pants that extend to the ankles shoes that cover heels and toes a shirt that covers the shoulders (has sleeves)
Answer:
(a) Which hazards are associated with copper(II) sulfate?
irritant toxic(b) What actions should you take if you spill copper(II) sulfate solution on yourself?
Flush the affected area with water.(c) What clothing is appropriate for the laboratory?
shirt and pants that overlap 6" pants that extend to the ankles shoes that cover heels and toes a shirt that covers the shoulders (has sleeves)Explanation:
Copper sulfate (ii) is a toxic compound by ingestion. Irritant in prolonged skin contact.
In case of skin contact, contaminated clothing and shoes should be removed, wash the affected areas with soap and large amounts of water. Seek medical assistance.
The shoe is recommended that this be completely closed, since if any substance falls do not burn the feet, and comfortable to perform the practice in the best way.
The clothes to be used in a laboratory should cover as much skin as possible, in addition it is necessary to use a cotton lab coat, long sleeve and that reaches the knees.
Copper II sulfate is a toxic irritant, if ever your skin comes in contact with it flush the affected area with water.
How to avoid accidents in the laboratory?In labs, there are many chemicals that can harm the student while haphazardly handling them.
For example - Copper II sulfate is a toxic irritant, if ever your skin comes in contact with it flush the affected area with water.
The students should wear full sleeve shirts, overlapping with pants and shoes that cover the ankle and toe.
Therefore, Copper II sulfate is a toxic irritant, if ever your skin comes in contact with it flush the affected area with water.
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Calcium nitrate and ammonium fluoride react to form calcium fluoride, dinitrogen monoxide, and water vapor. How many grams of dinitrogen monoxide are present after 31.3 g of calcium nitrate and 38.7 g of ammonium fluoride react completely? Hint: Write a balanced chemical reaction first, then find which one is the limiting reactant.
Answer:
16.79 grams of dinitrogen monoxide are present after 31.3 g of calcium nitrate and 38.7 g of ammonium fluoride react completely. And calcium nitrate is a limiting reactant. Explanation:
[tex]Ca(NO_3)_2(aq)+2NH_4F(aq)\rightarrow CaF_2(aq)+2N_2O(g)+4H_2O(g)[/tex]
Moles of calcium nitrate = [tex]\frac{31.3 g}{164 g/mol}=0.1908 mol[/tex]
Moles of ammonium fluoride = [tex]\frac{38.7 g}{37 g/mol}=1.046 mol[/tex]
According to reaction , 2 moles of ammonium fluoride reacts with 1 mole of calcium nitrate.
Then 1.046 moles of ammonium fluoride will react with :
[tex]\frac{1}{2}\times 1.046 mol=0.523 mol[/tex] calcium nitarte .
This means that ammonium fluoride is in excess amount and calcium nitrate is in limiting amount.
Hence, calcium nitrate is a limiting reactant.
So, amount of dinitrogen monoxide will depend upon moles of calcium nitrate.
So, according to reaction , 1 mole of calcium nitarte gives 2 moles of dinitrogen monoxide gas .
Then 0.1908 moles of calcium nitrate will give:
[tex]\frac{2}{1}\times 0.1908 mol=0.3816 mol[/tex]of dinitrogen monoxide gas.
Mass of 0.03816 moles of dinitrogen monoxide gas:
0.03816 mol × 44 g/mol = 16.79 g
16.79 grams of dinitrogen monoxide are present after 31.3 g of calcium nitrate and 38.7 g of ammonium fluoride react completely. And calcium nitrate is a limiting reactant.
After the reaction is complete the mass of dinitrogen monoxide N2O 8.39 grams.
let's follow these steps.
Step 1. Write the balanced chemical equation.
The reaction given is between calcium nitrate [tex]\( \text{Ca(NO}_3\text{)}_2 \)[/tex] and ammonium fluoride [tex]\( \text{NH}_4\text{F} \)[/tex], producing calcium fluoride [tex]\( \text{CaF}_2 \),[/tex] dinitrogen monoxide nitrous oxide, [tex]\ \text{N}_2\text{O} \)[/tex], and water vapor [tex]\( \text{H}_2\text{O} \)[/tex]
The balanced chemical equation is.
[tex]\[ \text{Ca(NO}_3\text{)}_2 + 2 \text{NH}_4\text{F} \[/tex] right arrow [tex]\text{CaF}_2 + \text{N}_2\text{O} + 4 \text{H}_2\text{O} \][/tex]
Step 2. Calculate the molar masses.
- Molar mass of. [tex]\( \text{Ca(NO}_3\text{)}_2 \)[/tex]
[tex]\[ \text{Ca}[/tex] = [tex]40.08 \, \text{g/mol} \][/tex]
[tex]\[ \text{N} = 14.01 \, \text{g/mol} \][/tex]
[tex]\[ \text{O} = 16.00 \, \text{g/mol} \][/tex]
[tex]\[ \text{Molar mass of } \text{Ca(NO}_3\text{)}_2 = 40.08 + 2 \cdot (14.01 + 3 \cdot 16.00) = 164.10 \, \text{g/mol} \][/tex]
- Molar mass of [tex]\( \text{NH}_4\text{F} \):[/tex]
[tex]\[ \text{N} = 14.01 \, \text{g/mol} \][/tex]
[tex]\[ \text{H} = 1.01 \, \text{g/mol} \][/tex]
[tex]\[ \text{F} = 19.00 \, \text{g/mol} \][/tex]
[tex]\[ \text{Molar mass of } \text{NH}_4\text{F} = 14.01 + 4 \cdot 1.01 + 19.00 = 37.05 \, \text{g/mol} \][/tex]
- Molar mass of [tex]\( \text{N}_2\text{O} \)[/tex].
[tex]\[ \text{N} = 14.01 \, \text{g/mol} \][/tex]
[tex]\[ \text{O} = 16.00 \, \text{g/mol} \][/tex]
[tex]\[ \text{Molar mass of } \text{N}_2\text{O} = 2 \cdot 14.01 + 16.00 = 44.02 \, \text{g/mol} \][/tex]
Step 3. Determine the limiting reactant.
Calculate the number of moles for each reactant.
- Moles of[tex]\( \text{Ca(NO}_3\text{)}_2 \).[/tex]
[tex]\[ \text{Moles} = \frac{31.3 \, \text{g}}{164.10 \, \text{g/mol}} \approx 0.1908 \, \text{mol} \][/tex]
- Moles of[tex]\( \text{NH}_4\text{F} \):[/tex]
[tex]\[ \text{Moles} = \frac{38.7 \, \text{g}}{37.05 \, \text{g/mol}} \approx 1.0451 \, \text{mol} \][/tex]
According to the balanced equation,[tex]\( \text{Ca(NO}_3\text{)}_2 \) and \( \text{NH}_4\text{F} \)[/tex] react in a 1:2 molar ratio. Therefore, [tex]\( \text{Ca(NO}_3\text{)}_2 \)[/tex] is the limiting reactant because it produces fewer moles of products compared to [tex]\( \text{NH}_4\text{F} \).[/tex]
Step 4. Calculate the mass of [tex]\( \text{N}_2\text{O} \)[/tex] produced.
From the balanced equation,[tex]\( 1 \) mole of \( \text{Ca(NO}_3\text{)}_2 \) produces \( 1 \) mole of \( \text{N}_2\text{O} \).[/tex]
- Moles of [tex]\( \text{N}_2\text{O} \)[/tex] produced.
[tex]\[ \text{Moles} = 0.1908 \, \text{mol} \][/tex]
- Mass of [tex]\( \text{N}_2\text{O} \).[/tex]
[tex]\[ \text{Mass} = \text{Moles} \times \text{Molar mass of } \text{N}_2\text{O} \][/tex]
[tex]\[ \text{Mass} = 0.1908 \, \text{mol} \times 44.02 \, \text{g/mol} \][/tex]
[tex]\[ \text{Mass} = 8.39 \, \text{g} \][/tex]
Beryllium-11 is a radioactive isotope of the alkaline metal Beryllium. It decays at a rate of 4.9% every second. Assuming you started with 100%, how much would be left after 45 seconds?\
Answer:
11.02 % of an isotope will be left after 45 seconds.
Explanation:
[tex]N=N_o\times e^{-\lambda t}\\\\\lambda =\frac{0.693}{t_{\frac{1}{2}}}[/tex]
where,
[tex]N_o[/tex] = initial mass of isotope
N = mass of the parent isotope left after the time, (t)
[tex]t_{\frac{1}{2}}[/tex] = half life of the isotope
[tex]\lambda[/tex] = rate constant
We have :
Mass of Beryllium-11 radioactive isotope= [tex]N_o=100[/tex]
Mass of Beryllium-11 radioactive isotope after 45 seconds = [tex]N=?[/tex]
t = 45 s
[tex]\lambda[/tex] = rate constant = [tex]4.9 \% s^{-1}=0.049 s^{-1}[/tex]
[tex]N=N_o\times e^{-(\lambda \times t}[/tex]
Now put all the given values in this formula, we get
[tex]N=100\times e^{-0.049 s^{-1}\ties 45 s}[/tex]
[tex]N=11.02 g[/tex]
Percentage of isotope left :
[tex]\frac{N}{N_o}\times 100=\frac{11.02 g}{100 g}\times 100=11.02\%[/tex]
11.02 % of an isotope will be left after 45 seconds.
Final answer:
To calculate the remaining amount of Beryllium-11 after 45 seconds based on a 4.9% decay rate per second, the exponential decay formula is used. The remaining percentage is found by the expression 100 × (1 - 0.049) ^ 45.
Explanation:
The question asks how much of the radioactive isotope Beryllium-11 would be left after 45 seconds if it decays at a rate of 4.9% per second starting from 100%. To find the remaining amount, we can use the exponential decay formula, which is:
Remaining amount = Initial amount × (1 - decay rate) ^ time
In this case, the initial amount is 100%, decay rate is 0.049 (4.9% as a decimal), and time is 45 seconds. Plugging the values into the formula gives us:
Remaining amount = 100 × (1 - 0.049) ^ 45
Calculating this, we get:
Remaining amount = 100 × (0.951) ^ 45
Utilizing a calculator for the exponential part, we'll find the percentage of Beryllium-11 left after 45 seconds.
For each of the following, give the sublevel designation, the allowable ml values, and the number of orbitals:
(a) n = 4, l = 2
(b) n = 5, l = 1
(c) n = 6, l = 3
Answer :
(a) n = 4, l = 2
At l = 2, [tex]m_l=+2,+1,0,-1,-2[/tex]
Number of orbitals = 5
(b) n = 5, l = 1
At l = 1, [tex]m_l=+1,0,-1[/tex]
Number of orbitals = 3
(c) n = 6, l = 3
At l = 3, [tex]m_l=+3,+2,+1,0,-1,-2,-3[/tex]
Number of orbitals = 7
Explanation:
There are 4 quantum numbers :
Principle Quantum Numbers : It describes the size of the orbital. It is represented by n. n = 1,2,3,4....
Azimuthal Quantum Number : It describes the shape of the orbital. It is represented as 'l'. The value of l ranges from 0 to (n-1). For l = 0,1,2,3... the orbitals are s, p, d, f...
Magnetic Quantum Number : It describes the orientation of the orbitals. It is represented as m_l. The value of this quantum number ranges from [tex](-l\text{ to }+l)[/tex]. When l = 2, the value of [tex]m_l[/tex] will be -2, -1, 0, +1, +2.
Spin Quantum number : It describes the direction of electron spin. This is represented as [tex]m_s[/tex]The value of this is [tex]+\frac{1}{2}[/tex] for upward spin and [tex]-\frac{1}{2}[/tex] for downward spin.
(a) n = 4, l = 2
At l = 2, [tex]m_l=+2,+1,0,-1,-2[/tex]
Number of orbitals = 5
(b) n = 5, l = 1
At l = 1, [tex]m_l=+1,0,-1[/tex]
Number of orbitals = 3
(c) n = 6, l = 3
At l = 3, [tex]m_l=+3,+2,+1,0,-1,-2,-3[/tex]
Number of orbitals = 7
A reaction at 12.0 °C evolves 358, mmol of sulfur hexafluoride gas. Calculate the volume of sulfur hexafluoride gas that is collected. You can assume the pressure in the room is exactly 1 atm. Round your answer to 3 significant digits. volume: LL
Answer:
The volume of the gas is 8.37 L
Explanation:
Step 1: Data given
Temperature = 12.0 °C
Moles sulfur hexafluoride gas = 358 mmol = 0.358 mol
Pressure = 1 atm
Step 2: Calculate volume
p*V = n*R*T
⇒ with p = the pressure = 1 atm
⇒ with V = The volume = TO BE DETERMINED
⇒ with n = moles of sulfur hexafluoride gas
⇒ with R = the gas constanr = 0.08206 L*atm/mol*K
⇒ with T = The temperature = 12.0 °C = 285 K
V = (n*R*T)/p
V = ( 0.358 * 0.08206 * 285) / 1
V = 8.37 L
The volume of the gas is 8.37 L
Final answer:
Using the ideal gas law, we calculate the volume of sulfur hexafluoride gas at 12.0 °C and 1 atm pressure as 8.378 L, after converting the given moles and temperature to the appropriate units.
Explanation:
To calculate the volume of sulfur hexafluoride gas evolved at 12.0 °C and 1 atm pressure, we can use the ideal gas law, which relates the number of moles (n) of a gas to its volume (V) at a specific temperature (T) and pressure (P): PV = nRT, where R is the ideal gas constant. Here, we know n = 358 mmol, which we convert to 0.358 mol. The temperature must be in Kelvin; therefore, we add 273.15 to the Celsius temperature: 12.0 + 273.15 = 285.15 K. Using R = 0.0821 L·atm/mol·K, we can solve for V.
Calculation: V = nRT/P
V = (0.358 mol)(0.0821 L·atm/mol·K)(285.15 K) / (1 atm)
V = 8.378 L (rounded to three significant digits)
The volume of sulfur hexafluoride gas collected at these conditions is 8.378 L.
substance formed of crystals of equal numbers of cations and anions held together by ionic bonds is called a(n) _____.
Answer:Salt
Explanation:
In chemistry a salt is produced from a neutralization reaction, when an acid react with a base.
HCl(aq) + NaOH (aq) ----> NaCl (aq) + H2O(l)
A salt consists of the positive ion (cation) of an acid and the negative ion (anion) of a base.
H+(aq) + Cl-(aq) + Na+(aq) + OH-(aq) -------> Na+Cl-(aq) + H2O(l)
When the water is evaporated, the negatively charged chlorine ions combine with the positively charged sodium ions to form a solid salt.
An Ionic Compound is a substance formed of crystals of equal numbers of cations and anions held together by ionic bonds. They are typically created when metals transfer electrons to non-metals. Sodium Chloride is a common example.
Explanation:A substance formed of crystals of equal numbers of cations and anions held together by ionic bonds is called an Ionic Compound. These are usually formed when metals transfer electrons to non-metals, resulting in the creation of ions which are attracted to each other due to their opposite charges. They are known for high melting and boiling points as well as their ability to conduct electricity when dissolved in water or melted. Sodium Chloride, or table salt, is a common example of an ionic compound.
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Write the full ground-state electron configuration for each:
(a) Rb (b) Ge (c) Ar
Answer:
(a) Rb
[tex]1s^22s^22p^63s^23p^63d^{10}4s^24p^65s^1[/tex]
(b) Ge
[tex]1s^22s^22p^63s^23p^63d^{10}4s^24p^2[/tex]
(c) Ar
[tex]1s^22s^22p^63s^23p^6[/tex]
Explanation:
(a) Rb
Atomic number = 37
The electronic configuration is -
[tex]1s^22s^22p^63s^23p^63d^{10}4s^24p^65s^1[/tex]
(b) Ge
Atomic number = 32
The electronic configuration is -
[tex]1s^22s^22p^63s^23p^63d^{10}4s^24p^2[/tex]
(c) Ar
Atomic number- 18
The electronic configuration is -
[tex]1s^22s^22p^63s^23p^6[/tex]
For the equilibrium Br2(g) + Cl2(g) ⇌ 2 BrCl(g) the equilibrium constant Kp is 7.0 at 400 K. If a cylinder is charged with BrCl(g) at an initial pressure of 1.00 atm and the system is allowed to come to equilibrium what is the final (equilibrium) pressure of BrCl? For the equilibrium the equilibrium constant is 7.0 at 400 . If a cylinder is charged with at an initial pressure of 1.00 and the system is allowed to come to equilibrium what is the final (equilibrium) pressure of BrCl?
a. 0.31 atm b. 0.15 atmc. 0.57 atmd. 0.22 atme. 0.45 atm
Answer:
c. 0.57 atm
Explanation:
For the equilibrium:
Br₂(g) + Cl₂(g) ⇄ 2BrCl (g)
Let construct the ICE table for this reaction.
Br₂(g) + Cl₂(g) ⇄ 2BrCl (g)
Initial 0 0 1
Change +x +x - 2x
Final x x 1 - 2x
[tex]K_p = 7[/tex]
[tex]K_p= \frac{[BrCl_2]^2}{[Br_2][Cl_2]}[/tex]
[tex]7 = \frac{[1-2x]^2}{[x][x]}[/tex]
[tex]7 = \frac{[1-2x]^2}{[x]^2}[/tex]
[tex]7x^2=(1-2x)(1-2x)[/tex]
[tex]7x^2=1-2x-2x+4x^2[/tex]
[tex]7x^2=1-4x+4x^2[/tex]
[tex]7x^2-4x^2=1-4x[/tex]
[tex]3x^2=1-4x[/tex]
[tex]3x^2+4x-1 = 0[/tex] ----------(quadratic equation)
using the formula;
[tex]\frac{-b+/-\sqrt{b^2-4ac} }{2a}[/tex]
where ± represent +/-
a =3 ; b = 4 ; c = -1
[tex]\frac{-4+/-\sqrt{4^2-4(3)(1)} }{2*3}[/tex]
[tex]\frac{-4+/-\sqrt{16+12} }{6}[/tex]
[tex]\frac{-4+/-\sqrt{28} }{6}[/tex]
[tex]\frac{-4+\sqrt{28} }{6}[/tex] or [tex]\frac{-4-\sqrt{28} }{6}[/tex]
[tex]\frac{-4+5.292}{6}[/tex] or [tex]\frac{-4-5.292}{6}[/tex]
x = 0.215 or -1.549
So, we chose the one with a positive integer which is = 0.215
BrCl = 1 - 2(x)
BrCl = 1 - 2(0.215)
BrCl = 1 - 0.43
BrCl = 0.57 atm
∴ the final (equilibrium) pressure of BrCl = 0.57 atm
The final equilibrium pressure of the reaction is 0.57 atm
Data;
Kp = 7.0T = 400KP = 1.00atmEquilibrium Constant for Pressure(Kp)Let's write the equation of reaction for this
Br2(g) + Cl2(g) ⇌ 2 BrCl(g)
initial 1
final x x (1 - 2x)
The equilibrium pressure (Kp) is given as
[tex]kp = \frac{[BrCl]^2}{Br_2][Cl_2]} \\7 = \frac{(1-2x)^2}{x^2} \\7x^2 = 1 + 4x^2 - 4x\\7x^2 - 4x^2 + 4x - 1 = 0\\3x^2 + 4x - 1 = 0[/tex]
We can solve the above quadratic equation to find the value of x
x = 0.215
Let's substitute this into the concentration of BrCl
[tex][BrCl] = 1-2x = 1 - 2(0.215)= 1 - 0.43 = 0.57[/tex]
The final equilibrium pressure of the reaction is 0.57 atm
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As an athlete exercises, sweat is produced and evaporated to help maintain a proper body temperature. On average, an athlete loses approximately 443 g of sweat during an hour of exercise. How much energy is needed to evaporate the sweat that is produced? The heat of vaporization for water is 2257 J/g.
Answer: 999851 Joules of energy is needed to evaporate the sweat that is produced
Explanation:
Latent heat of vaporization is the amount of heat required to convert 1 gram of liquid water to gaseous form at its boiling point.
Given: The heat of vaporization for water is 2257 J/g.
Thus if for 1 g , the heat required is = 2257 J
For 443 g , heat required is =[tex]\frac{2257}{1}\times 443=999851J[/tex]
Thus 999851 Joules of energy is needed to evaporate the sweat that is produced
To calculate the energy required to evaporate the sweat produced during exercise, we multiply the mass of the sweat (443 g) by the heat of vaporization for water (2257 J/g). The result is approximately 999,791 Joules.
Explanation:To calculate the amount of energy required to evaporate the sweat that an athlete produces during exercise, we can use the formula energy = mass × heat of vaporization. Given that the heat of vaporization for water is 2257 J/g and the average sweat loss is approximately 443 g, we can substitute these values into the formula: energy = 443 g × 2257 J/g = 999791 J. So, on average, an athlete needs about 999,791 Joules of energy to evaporate the sweat produced in an hour of exercise.
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A certain molecular compound has a solubility in hexane of at . Calculate the mass of that's dissolved in of a saturated solution of in hexane at this temperature. Be sure your answer has the correct unit symbol and significant digits.
The question is incomplete. The complete question is:
A certain molecular compound M has a solubility in acetone of 0.737g/mL at 10.°C .Calculate the mass of M that's dissolved in 9.0 L of a saturated solution of M in acetone at this temperature.
Be sure your answer has the correct unit symbol and number of significant digits.
Answer: 6633 g of M is dissolved in 9.0 L of saturated solution in hexane at this temperature.
Explanation:
Given : 0.737g of solute is present in 1ml of solution
i.e 1ml of solution contains = 0.737 g of solute
Thus 9L or 9000ml of solution contains=[tex]\frac{0.737g}{1ml}\times 9000ml=6633g[/tex]
Thus 6633 g of M is dissolved in 9.0 L of saturated solution in hexane at this temperature.