Here is the complete question:
Calculate the energy released in the following fusion reaction. The masses of the isotopes are: 14N (14.00307 amu), 32S (31.97207 amu), 12C (12.00000 amu), and 6Li (6.01512 amu).
¹⁴N + ¹²C + ⁶Li ⇒ ³²S
Answer:
68.7372 × 10⁻¹⁶ kJ
Explanation:
Given that the reaction; ¹⁴N + ¹²C + ⁶Li ⇒ ³²S
To calculate for the energy released; we need to determine the mass defect (md) of the reaction and which is given as :
Mass defect (md) = [mass of reactants] -[mass of product]
Mass defect (md) = [ ¹⁴N + ¹²C + ⁶Li ] - [ ³²S ]
Mass defect (md) = [ 14.00307 + 12.00000 + 6.01512 ] amu - [ 31.97207 ] amu
Mass defect (md) = 32.01819 - 31.97207
Mass defect (md) = 0.04612 amu
Having gotten the value of our Mass defect (md); = 0.04612 amu
We know that if 1 amu ⇒ 931.5 Mev of energy
∴ 0.04612 amu = 0.04612 × 931.5 Mev of energy
= 42.96078 Mev of energy
where M = million = 10⁶
1 ev = 1.6 × 10⁻¹⁹ Joules
∴ 42.96078 Mev of energy = 42.96078 × 10⁶ × 1.6 × 10⁻¹⁹ J
= 68.7372 × 10⁻¹³ J
= 68.7372 × 10⁻¹⁶ kJ
Hence; the energy released in the above fusion reaction = 68.7372 × 10⁻¹⁶ kJ.
The Energy released in the fusion reaction is approximately 6.87 × 10⁻¹² J
Step 1: Determine the total mass of the reactants and products
The given reaction is:
[tex]\[ \text{14N} + \text{12C} + \text{6Li} \rightarrow \text{32S} \][/tex]
The masses of the isotopes are:
[tex]\( \text{14N} = 14.00307 \, \text{amu} \)[/tex]
[tex]\( \text{12C} = 12.00000 \, \text{amu} \)[/tex]
[tex]\( \text{6Li} = 6.01512 \, \text{amu} \)[/tex]
[tex]\( \text{32S} = 31.97207 \, \text{amu} \)[/tex]
Total mass of the reactants:
[tex]\[ \text{Mass of reactants} = \text{Mass of 14N} + \text{Mass of 12C} + \text{Mass of 6Li} \][/tex]
[tex]\[ \text{Mass of reactants} = 14.00307 \, \text{amu} + 12.00000 \, \text{amu} + 6.01512 \, \text{amu} \][/tex]
[tex]\[ \text{Mass of reactants} = 32.01819 \, \text{amu} \][/tex]
Mass of the products:
[tex]\[ \text{Mass of products} = \text{Mass of 32S} = 31.97207 \, \text{amu} \][/tex]
Step 2: Calculate the mass defect
The mass defect (\( \Delta m \)) is the difference between the total mass of the reactants and the total mass of the products:
[tex]\[ \Delta m = \text{Mass of reactants} - \text{Mass of products} \][/tex]
[tex]\[ \Delta m = 32.01819 \, \text{amu} - 31.97207 \, \text{amu} \][/tex]
[tex]\[ \Delta m = 0.04612 \, \text{amu} \][/tex]
Step 3: Convert the mass defect into energy
To convert the mass defect into energy, we use Einstein’s equation [tex]\( E = \Delta m c^2 \).[/tex]
First, we need to convert the mass defect from atomic mass units (amu) to kilograms (kg). The conversion factor is:
[tex]\[ 1 \, \text{amu} = 1.66053906660 \times 10^{-27} \, \text{kg} \][/tex]
So,
[tex]\[ \Delta m = 0.04612 \, \text{amu} \times 1.66053906660 \times 10^{-27} \, \text{kg/amu} \][/tex]
[tex]\[ \Delta m \approx 7.656 \times 10^{-29} \, \text{kg} \][/tex]
Now, using [tex]\( c = 3 \times 10^8 \, \text{m/s} \),[/tex] we find the energy released:
[tex]\[ E = \Delta m c^2 \][/tex]
[tex]\[ E = 7.656 \times 10^{-29} \, \text{kg} \times (3 \times 10^8 \, \text{m/s})^2 \][/tex]
[tex]\[ E = 7.656 \times 10^{-29} \, \text{kg} \times 9 \times 10^{16} \, \text{m}^2/\text{s}^2 \][/tex]
[tex]\[ E \approx 6.87 \times 10^{-12} \, \text{J} \][/tex]
Complete question is - Calculate the energy released in the following fusion reaction - [tex]\[ \text{14N} + \text{12C} + \text{6Li} \rightarrow \text{32S} \][/tex]. The masses of the isotopes are: 14N (14.00307 amu), 32S (31.97207 amu), 12C (12.00000 amu), and 6Li (6.01512 amu).
What is the solubility of ethylene (in units of grams per liter) in water at 25 °C, when the C2H4 gas over the solution has a partial pressure of 0.684 atm? kH for C2H4 at 25 °C is 4.78×10-3 mol/L·atm.
Answer: The solubility of ethylene gas in water is [tex]9.16\times 10^{-2}g/L[/tex]
Explanation:
To calculate the molar solubility, we use the equation given by Henry's law, which is:
[tex]C_{C_2H_4}=K_H\times p_{C_2H_4}[/tex]
where,
[tex]K_H[/tex] = Henry's constant = [tex]4.78\times 10^{-3}mol/L.atm[/tex]
[tex]C_{C_2H_4}[/tex] = molar solubility of ethylene gas = ?
[tex]p_{C_2H_4}[/tex] = partial pressure of ethylene gas = 0.684 atm
Putting values in above equation, we get:
[tex]C_{C_2H_4}=4.78\times 10^{-3}mol/L.atm\times 0.684atm\\\\C_{C_2H_4}=3.27\times 10^{-3}mol/L[/tex]
Converting this into grams per liter, by multiplying with the molar mass of ethylene:
Molar mass of ethylene gas = 28 g/mol
So, [tex]C_{C_2H_6}=3.27\times 10^{-3}mol/L\times 28g/mol=9.16\times 10^{-2}g/L[/tex]
Hence, the solubility of ethylene gas in water is [tex]9.16\times 10^{-2}g/L[/tex]
The solubility of ethylene in water at 25 °C with a partial pressure of 0.684 atm is approximately 0.0919 grams per liter.
Explanation:To calculate the solubility of ethylene in water, we can use Henry's Law. Henry's Law states that the solubility of a gas in a liquid is directly proportional to the partial pressure of the gas above the liquid. The equation is given by:
S = kH * P
Where:
First, we can calculate the solubility of ethylene in units of moles per liter:
S = (4.78×10-3 mol/L·atm) * (0.684 atm) = 0.00327 mol/L
Then, we can convert the solubility to grams per liter using the molar mass of ethylene:
Molar mass of C2H4 = 2(12.01 g/mol) + 4(1.01 g/mol) = 28.05 g/mol
Grams/Liter = (0.00327 mol/L) * (28.05 g/mol) = 0.0919 g/L
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In a reaction involving iron, Fe, and oxygen, O. it was determined that 4.166 grams of iron reacted with 1.803 grams of oxygen. From this information, determine the empirical formula of the compound that resulted.a. FEO2b. FeO3c. Fe2Od. Fe2O3
Answer: The empirical formula for the given compound is [tex]Fe_2O_3[/tex]
Explanation:
We are given:
Mass of Fe = 4.166 g
Mass of O = 1.803 g
To formulate the empirical formula, we need to follow some steps:
Step 1: Converting the given masses into moles.Moles of Iron =[tex]\frac{\text{Given mass of Iron}}{\text{Molar mass of Iron}}=\frac{4.166g}{55.85g/mole}=0.0746moles[/tex]
Moles of Oxygen = [tex]\frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{1.803g}{16g/mole}=0.113moles[/tex]
Step 2: Calculating the mole ratio of the given elements.For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 0.0746 moles.
For Iron = [tex]\frac{0.0746}{0.0746}=1[/tex]
For Oxygen = [tex]\frac{0.113}{0.0746}=1.5[/tex]
Converting the mole ratio into whole number by multiplying with '2'
Mole ratio of Fe = (1 × 2) = 2
Mole ratio of O = (1.5 × 2) = 3
Step 3: Taking the mole ratio as their subscripts.The ratio of Fe : O = 2 : 3
Hence, the empirical formula for the given compound is [tex]Fe_2O_3[/tex]
Draw the structure of the aromatic compound para-aminochlorobenzene (para-chloroaniline). Draw the molecule on the canvas by choosing buttons from the Tools
Answer:
Explanation:
In the picture you have the answer.
Now, let's analize the structure, so you can know why the structure in the picture is the correct structure.
The aniline is the name that receives the benzene with a NH2 group as one of it's substituent. Now, This group is a really strong activating group and in the nomenclature priority, it has more order priority than any halide.
Now, it says that the chloro it's on the para position. The "para" position in a aromatic ring, in this case, the benzene, refers to the position of this substituent to the first substitued position. In this case, the NH2 it's on the position 1 or carbon 1, the para position, means that it's on position 4 of the ring. The ortho position is carbon 2, and meta position is carbon 3 of the benzene. So, according to this, the p-chloroaniline it's on picture attached.
State Hund’s rule in your own words, and show its application in the orbital diagram of the nitrogen atom.
Answer:
.
Explanation:
Answer:
Hund's rule: states that electrons always enter an empty orbital before they pair up.
In this exercise, we notice that orbital p has three suborbitals, then,
we must start filling each suborbitals with one electrons and after that we start to pairing them up.
The result must be,
1st suborbital with 2 electrons
2nd suborbital with 1 electron
3 rd suborbital with 1 electron
A chemist determines by measurements that moles of fluorine gas participate in a chemical reaction. Calculate the mass of fluorine gas that participates. Be sure your answer has the correct number of significant digits.
The question is incomplete, here is the complete question:
A chemist determines by measurements that 0.0850 moles of fluorine gas participate in a chemical reaction. Calculate the mass of fluorine gas that participates. Be sure your answer has the correct number of significant digits.
Answer: The mass of fluorine gas that is precipitated is 3.23 grams
Explanation:
To calculate the number of moles, we use the equation:
[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex]
We are given:
Moles of fluorine gas = 0.0850 moles
Molar mass of fluorine gas = 38.0 g/mol
Putting values in above equation, we get:
[tex]0.0850mol=\frac{\text{Mass of fluorine gas}}{38.0g/mol}\\\\\text{Mass of fluorine gas}=(0.0850mol\times 38.0g/mol)=3.23g[/tex]
Hence, the mass of fluorine gas that is precipitated is 3.23 grams
To calculate the mass of fluorine gas that participates in a chemical reaction, multiply the number of moles by the molar mass of fluorine. The mass is approximately 76 grams with 2 significant digits.
Explanation:To calculate the mass of fluorine gas that participates in a chemical reaction, you need to know the number of moles of fluorine gas involved and the molar mass of fluorine (F2). The molar mass of fluorine is approximately 38 grams per mole. Multiply the number of moles by the molar mass to obtain the mass of fluorine gas participating in the reaction.
Example:
If the chemist determines that 2 moles of fluorine gas participate in the reaction, the mass of fluorine gas can be calculated as follows:
Mass = number of moles x molar mass
Mass = 2 moles x 37.996 grams/mole
Mass = 75.992 grams
Therefore, the mass of fluorine gas that participates in the reaction is approximately 76 grams . Remember to use the correct number of significant digits in your final answer, which in this case is 2 significant digits.
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Some versions of the periodic table show hydrogen at the top of Group 1A(1) and at the top of Group 7A(17). What properties of hydrogen justify each of these placements?
Answer:Hydrogen is placed such because it exhibits some similar characteristics of both group1 and group VII elements.
Explanation:
The reason why hydrogen is similar to group 1 metals:
#It has same valence electron and inorder achieve octet state it can lose that electron and forms H+ ion
#It acts as a good reducing agent similar to group1 metals
#It can also halides
Similarity to halogens:
#hydrogen can also gain one electron to gain noble gas configuration. It can combine with other non metals to form molecules with covalent bonding.
#It exists as diatomin molecule,H2
#Have the same electronegativity nature
#its reaction with other metal
A standard sheet of paper is 21.59cm x 27.94cm. There are two ways to bend it into a cylinder, a thin tall cylinder or a wide-short cylinder. Describe each method and calculate the volume of each.
Explanation:
Case 1. When circumference of cylinder is 27.94 cm and its height is 21.59 cm. Therefore, we will calculate the radius of cylinder as follows.
Circumference = [tex]2 \pi r[/tex]
27.94 cm = [tex]2 \times 3.14 \times r[/tex]
r = 4.45 cm
Now, we will calculate the volume of cylinder as follows.
Volume = [tex]\pi r^{2}h[/tex]
= [tex]3.14 \times (4.45 cm)^{2} \times 21.59 cm[/tex]
= [tex]1342.46 cm^{3}[/tex]
Case 2. When circumference of cylinder is 21.59 cm and its height is 27.94 cm. Therefore, we will calculate the radius of cylinder as follows.
Circumference = [tex]2 \pi r[/tex]
21.59 cm = [tex]2 \times 3.14 \times r[/tex]
r = 3.44 cm
Now, we will calculate the volume of cylinder as follows.
Volume = [tex]\pi r^{2}h[/tex]
= [tex]3.14 \times (3.44 cm)^{2} \times 27.94 cm[/tex]
= [tex]1038.18 cm^{3}[/tex]
An unknown compound, X, is thought to have a carboxyl group with a pKa of 2.0 and another ionizable group with a pKa between 5 and 8. When 75 ml of 0.10 M NaOH is added to 100 ml of a 0.10 M solution of X at pH 2.0, the pH increases to 6.72. Calculate the pKa of the second ionizable group.
Answer:
pKa = 7.2
Explanation:
When pH=pKa=2.0, there are equal amounts of the X carboxyl group and its ionized form.
100 mL * 0.1 M = 10 mmol compound X5 mmol ionized carboxyl X-COO⁻ & 5 mmol unionized carboxyl X-COOHThen, (75 mL * 0.10 M) 7.5 mmol of OH⁻ are added, so all 5 mmol of X-COOH converts into X-COO⁻. Then the remaining (7.5 - 5) 2.5 mmol of OH⁻ react with the second ionizable group of X
Number of X-COO⁻ = 5mmol (from the beginning) + 5mmol (from X-COOH that reacted) - 2.5 mmol (from the OH⁻ remaining) = 7.5 mmol
Because the total number of X compound moles did not change, we have (10 - 7.5) 2.5 mmol of the conjugate base of X-COO⁻.
Now we have all required data to solve this problem using Henderson-Hasselbach's equation:
pH = pKa + log [A⁻]/[HA]
6.72 = pKa + log (2.5/7.5)
pKa = 7.2
Answer:
[tex]pK_a[/tex] = 7.20
Explanation:
Given that our Molarity of X(unknown compound)= 0.10 M
Volume of X = 100 ml = 0.1 L
Molarity = [tex]\frac{number of moles of X}{Volume of X}[/tex]
Number of moles of X = Molarity × Volume of X
= 0.1 M × 0.1 L
= 0.01 mol
[tex]pK_a[/tex] = [tex]pH[/tex]
∴[tex][H^+][/tex] =[tex][HA][/tex] (this literaly implies and point out that the beginning the amount of carboxyl group and the second ionizable group must be equal.)
Having said that,
Number of moles of carboxyl group in X = [tex]\frac{0.01mol}{2}[/tex]
= 0.005 mol
When 75 ml of 0.10 M NaOH is added to 100 ml of a 0.10 M solution of X;
we have the number of moles of NaOH that is being added as:
Molarity × volume of NaOH
= 0.1 M × 0.075 L
= 0.0075 mol
From the question, if NaOH molecules thoroughly dissociate the carboxyl group of X.
The excess NaOH can be calculated as:
Excess of (NaOH) = Number of moles of NaOH added - Number of moles of carboxyl group in X
Excess of (NaOH) = 0.0075 -0.005 = 0.0025 mol
∴ Applying Henderson-Hesselbalch equation; it will be easier to determine the [tex]pK_a[/tex] for second group:
[tex]pK_a[/tex] = [tex]pH[/tex] [tex]-\frac{log[A]}{HA}[/tex]
= 6.72 - log[tex](\frac{0.0025}{0.0075})[/tex]
= 6.72 - log (0.333)
= 6.72 + 0.477
[tex]pK_a[/tex] = 7.197
≅ 7.20
The rate constant is 0.556 L mol-1 s-1 at some temperature. If the initial concentration of NOBr in the container is 0.32 M, how long will it take for the concentration to decrease to 0.039 M
Answer:
12.96 seconds
Explanation:
Assuming the reaction follows a first order
Rate = K[NOBr] = change in concentration of NOBr/time
K = 0.556 L mol^-1 s^-1
Change in concentration of NOBr = 0.32M - 0.039M = 0.281M
0.281/t = 0.556×0.039
t = 0.281/0.021684 = 12.96 seconds
It will take approximately 40.48 seconds for the concentration of NOBr to decrease from 0.32 M to 0.039 M.
To determine how long it will take for the concentration of NOBr to decrease from 0.32 M to 0.039 M, we need to use the integrated rate law for a second-order reaction:
For a second-order reaction:
1 / [A]₁ = kt + 1 / [A]₀
Where:
k = 0.556 L mol⁻¹ s⁻¹[A]₀ = 0.32 M (initial concentration)[A]₁ = 0.039 M (final concentration)Now, let's rearrange and solve for time (t):
(1 / [A]₁) - (1 / [A]₀) = kt
[1 / 0.039 M] - [1 / 0.32 M] = (0.556 L mol⁻¹ s⁻¹) * t
25.64 - 3.125 = 0.556t
22.515 = 0.556t
t = 22.515 / 0.556
t ≈ 40.48 seconds
How many unpaired electrons are present in the ground state of an atom from each of the following groups?
(a) 2A(2) (b) 5A(15) (c) 8A(18) (d) 3A(13)
Answer:
a. Zero unpaired electron
b. 3 unpaired electrons
c. Zero unpaired electron
d. 1 unpaired electron
Explanation:
a. 2A(2) has configuration => 1s2. Since the s-orbital is completely filled, Therefore it has zero unpaired electrons
b. 5A(15) has configuration =>
1s2 2s2 2p6 3s2 3p3
Since the p-orbital is not completely filled, It has 3 unpaired electrons
c. 8A(18) has configuration =>
1s2 2s2 2p6 3s2 3p6
Since the p-orbital is completely filled, therefore it has zero unpaired electrons
d. 3A(13) has configuration =>
1s2 2s2 2p6 3s2 2p1
Since the p-orbital is not completely filled, therefore it has 1 unpaired
Answer:
(a) 2A(2) - it has 2 valence electrons
(b) 5A(15) -
Explanation:
A)To determine the number of unpaired electrons for atoms in group 2A (2)
Using beryllium (it belongs to group 2A) as an example
The atomic number of Be is 4
The electronic configuration is 1s²2s²
The highest principal quantum number is 2, therefore all electrons with n=2 are valence/unpaired electron
Beryllium has 2 valence/unpaired electrons, this applies to all other elements in this group
Therefore group 2A atoms have 2 unpaired electrons
B) To determine the number of unpaired electrons for atoms in group 5A(15)
Using Nitrogen as an example
The atomic number of Nitrogen is 7
The electronic configuration is 1s²2s²2p³
The highest principal quantum number for nitrogen is 2, therefore all electrons with n=2 are valence/unpaired electrons
Nitrogen has 2+3= 5 valence/unpaired electrons, this applies to all other elements in this group
Therefore, group 5A atoms have 5 unpaired electrons
C) To determine the number of unpaired electrons for atoms in group 8A(18)
Using Neon as an example
The atomic number of Neon is 2
The electronic configuration of Neon is 1s²2s²2p⁶3s²3p⁶
The highest principal quantum number for 3, therefore all electrons with n=3 are valence/unpaired electrons
Neon has 2+6 = 8 valence/unpaired electrons,this applies to all other elements in this group except Helium whose number of unpaired electrons is 2
Therefore, group 8A atoms have 8 unpaired electrons
D) To determine the number of unpaired electrons for atoms in group 3A(13)
Using Aluminium as an example
The atomic number of Aluminium is 13
The electronic configuration of Neon is 1s²2s²2p⁶3s²3p¹
The highest principal quantum number for 3, therefore all electrons with n=3 are valence/unpaired electrons
Neon has 2+1 = 3 valence/unpaired electrons,this applies to all other elements in this group
Therefore, group 3A atoms have 3 unpaired electrons
Note: The number of valence electrons of atoms in a group is the same as the group number that the atom belongs to
An unknown compound has the following chemical formula:
P2Ox
where x stands for a whole number. Measurements also show that a certain sample of the unknown compound contains 8.20 mol of oxygen and 3.3 mol of calcium. Write the complete chemical formula for the unknown compound.
The question provides us with the chemical formula for an unknown compound, and values for moles of oxygen and calcium. However, to solve this problem, we need to have the number of moles for phosphorus, not calcium. Without this, it's impossible to proceed further and derive a concrete answer.
Explanation:The task requires us to find the accurate value of x in the chemical formula P2Ox. To do this, we consider that the unknown compound is composed entirely of phosphorus (P) and oxygen (O), but not calcium (Ca). Calcium seems to be irrelevant for this particular problem.
In the formula P2Ox, '2' is the stoichiometric index for phosphorus and 'x' for oxygen. However, we are not given the moles of phosphorus, which makes the problem unsolvable with the information provided.
Ideally, we would divide the number of moles of oxygen by the number of moles of phosphorus to find the value of 'x', but since the moles of phosphorus are not provided, we are unable to proceed further with the provided information.
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A couple purchases a house for $400,000.00. They pay 20% down at closing, and take out a mortgage of $320,000.00. The mortgage company offers them a 4.80% annual rate with monthly compounding. The mortgage will require monthly payments for the next 30 years.
A couple purchases a house for $400,000.00. They pay 20% down at closing, and take out a mortgage of $320,000.00. The mortgage company offers them a 4.80% annual rate with monthly compounding. The mortgage will require monthly payments for the next 30 years.
What will be the monthly payment on this mortgage?
Answer:
$ 1678.91
Explanation:
Given that;
Cost of purchasing a house = $400,000.00
Down payment =20%
Mortgage Value (MV) = $320,000.00
Annual rate offered by the mortgage company = 4.80% yearly i.e 0.4 per month
Duration of the Mortgage Loan (n) = 30 years which is equivalent to 360 months.
if we represent the monthly repayment with MR ,To calculate the monthly repayment MR;we have;
MV = MR × [tex](\frac{1}{i})*[1-(\frac{1}{(1+i)^{n}} )}][/tex]
where i = 0.004 (4.8 % annually expressed as 0.48, divided by 12 monthly payments per year)
∴
[tex]320,000.00[/tex] = [tex]MR[/tex] [tex]*(\frac{1}{0.004})*[1-(\frac{1}{1+0.004)^{360}}})][/tex]
320,000.00 = MR × 190.60
Monthly repayment (MR) = $ 1678.90870933
Monthly repayment (MR) ≅ $ 1678.91
When 10.1 g of an unknown, non-volatile, non-electrolyte, X was dissolved in 100. g of benzene, the vapor pressure of the solvent decreased from 100 torr to 87.7 torr at 299 K. Calculate the molar mass of the solute, X.
Answer:
56.06 g/mol is the molar mass
Explanation:
Vapor pressure lowering → P° - P' = P° . Xm
Where P° is vapor pressure of pure solvent
P' is vapor pressure of solution
Xm is the mole fraction (moles of solute / total moles)
Total moles = moles of solute + moles of solvent
Let's replace the data.
100 Torr - 87.7 Torr = 100 Torr . Xm
12.3 Torr = 100 Torr . Xm
0.123 = Xm
We know the moles of solvent because we know the molar mass from benzene and its mass in the solution. (mass / molar mass)
100 g / 78 g/mol = 1.28 moles
Let's build the equation where the unknown is the moles of solute
0.123 = moles of solute / moles of solute + 1.28 moles
0.123 (moles of solute + 1.28 moles) = moles of solute
0.123 moles of solute + 0.158 moles = moles of solute
0.158 = 1moles of solute - 0.123moles of solute
0.158 moles = 0.877 moles of solute
0.158 / 0.877 = moles of solute → 0.180
These moles corresponds to 10.1 g of the unknown, non volatile and non electrolyte X compound so:
molar mass (g/mol) → 10.1 g / 0.180 mol = 56.06 g/mol
To calculate the molar mass of the solute, we can use the formula: molar mass = (mass of solute / moles of solute). First, find the moles of solute using the relationship between the freezing point depression and the moles of solute. Next, calculate the molality of the solution using the given freezing point depression constant and mass of the solute, and use that to calculate the moles of solute. Finally, divide the mass of solute by the moles of solute to find the molar mass.
Explanation:To calculate the molar mass of the solute, we can use the formula:
molar mass = (mass of solute / moles of solute)
We first need to find the moles of solute using the relationship between the freezing point depression and the moles of solute:
ΔTf = Kf * m
where ΔTf is the freezing point depression, Kf is the freezing point depression constant, and m is the molality of the solution.
In this case, we are given that the freezing point depression is 0.40°C, the freezing point depression constant of benzene is 5.12 K kg/mol, and the mass of the solute is 2 grams. We can use these values to calculate the molality of the solution:
m = (ΔTf / Kf)
m = (0.40°C / 5.12 K kg/mol)
m = (0.078125 mol/kg)
Now we can calculate the moles of solute:
moles of solute = (m * mass of solvent)
moles of solute = (0.078125 mol/kg * 0.1 kg)
moles of solute = 0.0078125 mol
Finally, we can calculate the molar mass:
molar mass = (mass of solute / moles of solute)
molar mass = (2 grams / 0.0078125 mol)
molar mass ≈ 256 g/mol
Therefore, the molar mass of the solute, X, is approximately 256 g/mol.
When formic acid is heated, it decomposes to hydrogen and carbon dioxide in a first-order decay: HCOOH(g) →CO2(g) + H2 (g) The rate of reaction is monitored by measuring the total pressure in the reaction container.Time (s) . . . P (torr)0 . . . . . . . . . 22050 . . . . . . . . 324100 . . . . . . . 379150 . . . . . . . 408200 . . . . . . . 423250 . . . . . . . 431300 . . . . . . . 435At the start of the reaction (time = 0), only formic acid is present.What is the formic acid pressure (in torr) when the total pressure is 319?
Answer : The formic acid pressure is, 99 torr
Explanation :
The given chemical reaction is:
[tex]HCOOH(g)\rightarrow CO_2(g)+H_2(g)[/tex]
Initial pressure a 0 0
At time 't' (a-x) x x
According to the Dalton's law,
[tex]P_{Total}=P_{HCOOH}+P_{CO_2}+P_{H_2}[/tex]
[tex]P_{Total}=(a-x)+x+x=a+x[/tex] .........(1)
As we are given that:
Initial pressure = a = 220 torr
[tex]P_{Total}=319torr[/tex]
Now put the value of 'a' in equation 1, we get:
[tex]P_{Total}=a+x[/tex]
[tex]319torr=220torr+x[/tex]
[tex]x=99torr[/tex]
Thus, the formic acid pressure is, 99 torr
Find the percentage composition of a compound that contains 1.94 g of carbon, 0.48 g of hydrogen, and 2.58 g of sulfur.
Answer : The percentage composition of carbon, hydrogen and sulfur in a compound is, 38.8 %, 9.6 % and 51.6 % respectively.
Explanation :
To calculate the percentage composition of element in sample, we use the equation:
[tex]\%\text{ composition of element}=\frac{\text{Mass of element}}{\text{Mass of sample}}\times 100[/tex]
Given:
Mass of carbon = 1.94 g
Mass of hydrogen = 0.48 g
Mass of sulfur = 2.58 g
First we have to calculate the mass of sample.
Mass of sample = Mass of carbon + Mass of hydrogen + Mass of sulfur
Mass of sample = 1.94 + 0.48 + 2.58 = 5.0 g
Now we have to calculate the percentage composition of a compound.
[tex]\%\text{ composition of carbon}=\frac{1.94g}{5.0g}\times 100=38.8\%[/tex]
[tex]\%\text{ composition of hydrogen}=\frac{0.48g}{5.0g}\times 100=9.6\%[/tex]
[tex]\%\text{ composition of sulfur}=\frac{2.58g}{5.0g}\times 100=51.6\%[/tex]
Hence, the percentage composition of carbon, hydrogen and sulfur in a compound is, 38.8 %, 9.6 % and 51.6 % respectively.
Menthol (FW = 156.3 g/mol), the strong-smelling substance in many cough drops, is a compound of carbon, hydrogen, and oxygen. When 0.1595 g of menthol was burned in a combustion apparatus, 0.449 g of CO2 and 0.184 g of H2O formed. What is menthol's molecular formula? enter as C#H#O#
Answer: the molecular formula is C10H20O
Explanation:Please see attachment for explanation
For each of the following, give the sublevel designation, the allowable ml values, and the number of orbitals:
(a) n = 2, l = 0
(b) n = 3, l = 2
(c) n = 5, l = 1
Answer:
(a) n = 2, l = 0 ⇒ sublevel s, ⇒ ml = 0, number of orbitals = 1
(b) n = 3, l = 2 ⇒ sublevel d, ⇒ ml = 0, ±1, ±2, number of orbitals = 5
(c) n = 5, l = 1 ⇒ sublevel p, ⇒ ml = 0, ±1, number of orbitals = 3
Explanation:
The rules for electron quantum numbers are:
1. Shell number, 1 ≤ n,
2. Subshell number, 0 ≤ l ≤ n − 1, from s, p, d, f, g, h...
3. Orbital energy shift, -l ≤ ml ≤ l
4. Spin, either -1/2 or +1/2
In our case
(a) n = 2, l = 0 ⇒ sublevel s
-l ≤ ml ≤ l ⇒ ml = 0, number of orbitals = 1
(b) n = 3, l = 2 ⇒ sublevel d
-l ≤ ml ≤ l ⇒ ml = 0, ±1, ±2, number of orbitals = 5
(c) n = 5, l = 1 ⇒ sublevel p
-l ≤ ml ≤ l ⇒ ml = 0, ±1, number of orbitals = 3
The first pair of quantum numbers (n, l) represents the 2s sublevel with 1 orbital. The second pair represents the 3d sublevel with 5 orbitals. The third pair represents the 5p sublevel with 3 orbitals.
Explanation:The information given refers to quantum numbers in the quantum mechanical model of the atom, a fundamental concept in high school physics and chemistry. This model explains the behavior of electrons in atoms.
(a) For n = 2 and l = 0, the sublevel designation is 2s. The permissible ml value is 0 and there is 1 orbital.
(b) For n = 3 and l = 2, the sublevel designation is 3d. The permissible ml values range from -2, -1, 0, 1, 2 and there are 5 orbitals.
(c) For n = 5 and l = 1, the sublevel designation is 5p. The permissible ml values are -1, 0, 1 and there are 3 orbitals.
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What is the Gibbs energy, LaTeX: \Delta GΔ G, when the very first crystal of potassium nitrate forms in solution while cooling, given that 19.1 grams were dissolved in 192 milliliters of water?
Answer:
74.344 kJ.
Explanation:
Below is an attachment containing the solution.
Police often monitor traffic with "K-band" radar guns, which operate in the microwave region at 22.235 GHz (1 GHz = 10⁹ Hz). Find the wavelength (in nm and Å) of this radiation.
Answer:
Wavelength = 13492242 nm
Wavelength = 134922420 Å
Explanation:
The relation between frequency and wavelength is shown below as:
[tex]c=frequency\times Wavelength [/tex]
c is the speed of light having value [tex]3\times 10^8\ m/s[/tex]
Given, Frequency = [tex]22.235\ GHz=22.235\times 10^{9}\ Hz[/tex]
Thus, Wavelength is:
[tex]Wavelength=\frac{c}{Frequency}[/tex]
[tex]Wavelength=\frac{3\times 10^8}{22.235\times 10^{9}}\ m[/tex]
[tex]Wavelength=0.013492242\ m=13492242\times 10^{-9}\ m[/tex]
Also, 1 m = [tex]10^{-9}[/tex] nm
So,
Wavelength = 13492242 nm
Also, 1 m = [tex]10^{-10}[/tex] Å
Wavelength = 134922420 Å
Calculate the number of grams of sodium chloride in the solution. (Hint: Remember that sodium chloride is a strong electrolyte.)
The number of grams of sodium chloride in the solution is 3.43 g.
Explanation:The number of grams of sodium chloride in the solution can be calculated using the molar mass and molarity.
First, we need to calculate the number of moles of NaCl in the solution. Using the given molarity (0.470 M) and volume (125.0 mL) of the solution, we can use the formula:
moles = molarity × volume (in L)
Therefore, moles of NaCl = 0.470 mol/L × 0.125 L = 0.05875 mol NaCl
Next, we can use the formula mass of NaCl (58.44 g/mol) to calculate the mass:
mass = moles × formula mass
Therefore, mass of NaCl = 0.05875 mol × 58.44 g/mol = 3.43 g NaCl
How many electrons in an atom can have each of the following quantum number or sublevel designations?
(a) n = 2, l = 1, ml = 0
(b) 5p
(c) n = 4, l = 3
Answer: (a) 2 (b) 6 (c) 14
Explanation:
In the Azimuthal quantum number(l) electrons in a particular subshell (such as s, p, d, or f) are defined by values of l (0, 1, 2, or 3).
s is l=0, p is l=1, d is l=2, f is l=3.
The magnetic quantum number (ml) The value of ml can range from -l to +l, including zero. Thus the s, p, d, and f subshells contain 1, 3, 5, and 7 orbitals each, with values of m within the ranges 0, ±1, ±2, ±3 respectively. Each shell can have 2 x l + 1 sublevels, and each of these sublevel can accommodate up to two electrons.
(a) n=2, l=1, ml=0. If l=1 then 2 x 1+ 1=3 sublevels, 3*2=6 electrons. When l=1, ml =-1,0,+1, ml=0 accommodate two(2)electrons
(b) 5p. p is l=1 If l=1 then 2 x 1+ 1=3 sublevels, 3*2= electrons. This means in the 5 shell, the p orbital has 3 subshell and accommodate 6 electrons.
(c) n = 4, l = 3 if l=3 then 2 x 3+ 1=7 sublevel 7*2=14 electrons. This means the in the 4 shell, the f orbital has 7 subshell and accomdate 14 elections.
(a) The maximum number of electrons an atom can have with the sublevel designation is 2 electrons.
(b) The maximum number of electrons for p-orbital is 6 electrons.
(c) The maximum number of electrons for f-orbital is 14 electrons.
The energy level of each of an atom with the given sublevel designations determines the maximum number of electrons the atom can occupy.
(a) For n = 2, I = 1, ml = 0,
The energy level is calculated as;
Energy level = 2(1) + 1 = 3 sub-orbtials
= -1 0 1 : ( 2 electrons each)
Thus, the maximum number of electrons an atom can have with the sublevel designation is 2 electrons.
(b) 5p
p-orbital has 3 sub-orbitals and maximum of 6 electrons
(c) n= 4, I = 3
energy level = 2(3) + 1 = 7 sub-orbitals
7 sub-orbitals corresponds f-orbital and it has maximum 14 electrons.
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Select the sentence that accurately describes a pure substance. Please choose the correct answer from the following choices, and then select the submit answer button
a.A pure substance is made up of only 1 type of particle.
b.A pure substance is made up of more than 1 type of particle.
c.Only compounds can be considered pure substances.
d.Only elements can be considered pure substances.
Answer:
a.A pure substance is made up of only 1 type of particle
Explanation:
When a substance is pure, it has only one type of particle. These particles maybe molecules, ions or atoms linked in a definite way throughout the substance. If a substance contains different particles, it cannot be regarded as a pure substance because its properties will be observed as a compromise of the individual properties of its different components.
The statement that accurately describes a pure substance is that it is made up of only one type of particle. Both elements and compounds can be considered as pure substances.
Explanation:The correct statement to describe a pure substance is:
a. A pure substance is made up of only 1 type of particle.
This means that a pure substance is only made up of identical atoms if it is an element, or identical molecules if it is a compound. Therefore, both elements and compounds can be considered as pure substances, contradicting options c and d. It also contradicts option b because a pure substance is not made up of more than one type of particle.
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A 232-lb fullback runs the 40-yd dash at a speed of 19.8 ± 0.1 mi/h.
(a) What is his de Broglie wavelength (in meters)?
(b) What is the uncertainty in his position?
Answer:
(a) 7.11 x 10⁻³⁷ m
(b) 1.11 x 10⁻³⁵ m
Explanation:
(a) The de Broglie wavelength is given by the expression:
λ = h/p = h/mv
where h is plancks constant, p is momentum which is equal to mass times velocity.
We have all the data required to calculate the wavelength, but first we will have to convert the velocity to m/s, and the mass to kilograms to work in metric system.
v = 19.8 mi/h x ( 1609.34 m/s ) x ( 1 h / 3600 s ) = 8.85 m/s
m = 232 lb x ( 0.454 kg/ lb ) = 105.33 kg
λ = h/ mv = 6.626 x 10⁻³⁴ J·s / ( 105.33 kg x 8.85 m/s ) = 7.11 x 10⁻³⁷ m
(b) For this part we have to use the uncertainty principle associated with wave-matter:
ΔpΔx > = h/4π
mΔvΔx > = h/4π
Δx = h/ (4π m Δv )
Again to utilize this equation we will have to convert the uncertainty in velocity to m/s for unit consistency.
Δv = 0.1 mi/h x ( 1609.34 m/mi ) x ( 1 h/ 3600 s )
= 0.045 m/s
Δx = h/ (4π m Δv ) = 6.626 x 10⁻³⁴ J·s / (4π x 105.33 kg x 0.045 m/s )
= 1.11 x 10⁻³⁵ m
This calculation shows us why we should not be talking of wavelengths associatiated with everyday macroscopic objects for we are obtaining an uncertainty of 1.11 x 10⁻³⁵ m for the position of the fullback.
This question discusses the fundamentals quantum mechanics. It asks about the de Broglie wavelength and uncertainty principle of a football player treating as a particle. The de Broglie wavelength can be calculated using Planck's constant and momentum, and the uncertainty can be determine using the uncertainty principal formula.
Explanation:In this question, we're asked about two core principles in quantum mechanics: the de Broglie wavelength and the uncertainty principle. As such, we're essentially treating the football player as both a particle and a wave, which is a fundamental concept of quantum mechanics.
Let's address the points one by one.
(a) The de Broglie wavelength of a particle is given by λ=h/p. Where h is Planck's constant (6.62607015 × 10-34 m2 kg / s) and p is momentum. The momentum of a player can be calculated as mass x velocity. Convert the player's mass from lbs to kg and speed from mi/h to m/s. Substitute these values into the equation to get the de Broglie wavelength.
(b) The uncertainty principle states that the more precisely the momentum of a particle is known, the less precisely its position can be known, and vice versa. In this case, we're given the uncertainty in the player's speed (which contributes to an uncertainty in his momentum, and thus in his position). By using the uncertainty principal formula Δx=Δp*h/4π, where Δx represents the position uncertainty and Δp is the momentum uncertainty. Using the figures provided in the question, you can calculate the position uncertainty.
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State the periodic law, and explain its relation to electron configuration. (Use Na and K in your explanation.)
Answer:
Explanation:
The period law state that when elements are listed in order of their atomic numbers, the elements fall into recurring groups, so that there is a recurrence of similar properties at regular intervals.
Na and K in the periodic table fall into the same group, this is because they both have one electrons in their outermost shell.
Na 11 -1s2 2s2 2p6 3s1
K 19 - 1s2 2s2 2p6 3s2 3p6 4s1
They share similar chemical and physical properties. Na and K are very reactive metals, they can loose/donate their outermost electron to non metals in other to attain stable octet state.
The form ionic compound when they react with non metals.
More polar solvents (eluents) move molecules more rapidly than less polar solvents. If you used a 1:1 hexanes:methanol mixture as solvent, would you expect the products to elute faster or slower? Based on your experiment, would a 1:1 hexanes:methanol mixture be a good choice as eluent? Explain why/why not
Answer:
For the first question the products elute slower. The answer to the second question is it would not be a good option as an eluent
Explanation:
Solvents are classified into polar and nonpolar. While in polar solvents, the distribution of the electronic cloud is asymmetric; therefore, the molecule has a positive and a negative pole. Low molecular weight alcohols such as methanol belong to this type.
While in apolar solvents, the distribution of the electronic cloud is symmetric; Therefore, these substances lack a positive and negative pole in their molecules. Some solvents such as hexane.
The miscibility of methanol (polar solvent) in hexane (apolar solvent) is low, miscibility is the main reason for not using this mixture as eluent.
The frequency factors for these two reactions are very close to each other in value. Assuming that they are the same, compute the ratio of the reaction rate constants for these two reactions at 25 ∘C. k1k2 =
Final answer:
The rate law and the Arrhenius equation help calculate the ratio of rate constants for reactions with similar frequency factors. In this case, with identical frequency factors, the ratio of the rate constants is 1.
Explanation:
The rate law for a reaction:
rate = k[A] [B]
The Arrhenius equation:
k = Ae(-Ea/RT)
For the given question:
Given k1 and k2 are the same, k1/k2 = 1.
The ratio of rate constants depends on the exponential term in the Arrhenius equation, specifically the difference in activation energies. If the activation energies are very close, the ratio [tex]\( \frac{k_1}{k_2} \)[/tex] will be close to 1, indicating similar rate constants for the two reactions at 25 °C.
The ratio of reaction rate constants [tex](\(k_1/k_2\))[/tex] for two reactions with similar frequency factors can be expressed using the Arrhenius equation, which relates the rate constant (k) to temperature. The Arrhenius equation is given by:
[tex]\[ k = A \cdot e^{-\frac{E_a}{RT}} \][/tex]
Where:
- ( k ) is the rate constant,
- ( A) is the frequency factor (pre-exponential factor),
- ( E_a ) is the activation energy,
- ( R ) is the gas constant (8.314 J/(mol·K)),
- ( T ) is the temperature in Kelvin.
Assuming the frequency factors[tex](\( A_1 \) and \( A_2 \))[/tex] are the same, the ratio of rate constants [tex](\( \frac{k_1}{k_2} \))[/tex] can be simplified to the ratio of the exponential term:
[tex]\[ \frac{k_1}{k_2} = e^{-\frac{E_{a1} - E_{a2}}{RT}} \][/tex]
Given that the reactions are at 25 °C (298 K), the ratio [tex]\( \frac{k_1}{k_2} \)[/tex]will depend on the difference in activation energies [tex](\( E_{a1} - E_{a2} \)).[/tex]
Rank the ions in each set in order of increasing size, and explain your ranking:
(a) Li⁺, K⁺, Na⁺ (b) Se²⁻, Rb⁺, Br (c) O²⁻, F⁻, N³⁻
Explanation:
Lithium, sodium and potassium are all group 1A elements and when we move down a group then there occurs an increase in atomic size of the elements. As lithium is the smallest and potassium being the largest so, when each of them will lose an electron and obtain a positive charge then size of lithium will further decrease.
Therefore, ions are ranked according to their increase in size as follows.
[tex]Li^{+} < Na^{+} < K^{+}[/tex]
When an atom tends to gain electrons then it acquires a negative charge. This means that size of the atom increases.
So, more is the negative charge present on an atom more will be its atomic size. Therefore, correct order of increasing size for [tex]Se^{2-}, Rb^{+}, Br[/tex] is as follows.
[tex]Br < Rb^{+} < Se^{2-}[/tex]
Similarly, order of increasing size of [tex]O^{2-}, F^{-}, N^{3-}[/tex] is as follows.
[tex]F^{-} < O^{2-} < N^{3-}[/tex]
Draw an orbital diagram showing valence electrons, and write the condensed ground-state electron configuration for each:
(a) Mn (b) P (c) Fe
Answer:
The complete answer is in the diagram.
Diatomic oxygen has a molar mass 16 times that of diatomic hydrogen. The root-mean-square speed vrms for diatomic oxygen at 50∘C is: Choose the correct value of vrms. View Available Hint(s) Choose the correct value of . (16)(2000m/s)=32000m/s (4)(2000m/s)=8000m/s 2000m/s (14)(2000m/s)=500m/s (116)(2000m/s)=125m/s none of the above
The root-mean-square speed for diatomic oxygen at 50°C can be calculated using the formula Urms =
oot(3RT/M), where R is the gas constant, T is the temperature in Kelvin, and M is the molar mass of oxygen. Multiplication options presented in the question are incorrect and should not be used.
To calculate the root-mean-square speed (Urms) for diatomic oxygen at a given temperature, we use the formula derived from the kinetic theory of gases:
Urms =
oot(3RT/M)
Where:
R is the universal gas constant (8.314 J/mol·K)T is the absolute temperature in Kelvins (K)M is the molar mass of the gas in kilograms per mole (kg/mol)The student provides information that diatomic hydrogen has a particular Urms value at 50°C (which needs to be converted to Kelvin by adding 273.15, resulting in 323.15K). However, to find the Urms for oxygen directly, we will use the molar mass of oxygen (32.00 g/mol or 0.032 kg/mol) and the same temperature in Kelvins:
Urms =
oot(3 imes 8.314 imes 323.15 / 0.032)
Calculating the above will give us the correct Urms for oxygen, the presented multiplication options (16)(2000 m/s), etc., are misleading and should not be used for this calculation.
Consider the following molecule. In common nomenclature, what Greek letter would be assigned to the carbon indicated by an asterisk?
alpha
beta
gamma
epsilon
Answer: beta
Explanation: the beta carbon is the second after the alpha carbon( carbon bonding to the functional group)