Answer:
[tex]6.022\times 10^{22} [/tex]glucose molecules are included in that 100 mL sample.
Explanation:
Concentration of freshly prepared glucose solution = 1 M = 1 mol/L
1 L = 1000 ml
This means that 1 mole of glucose is present in 1000 mL of water.
If we have 100 mL of solution. then number of moles of glucose will be L;
[tex]\frac{1}{1000}\times 100 mL=0.1 mole[/tex]
1 mole = [tex]N_A=6.022\times 10^{23} [/tex] molecules/atoms
Number of molecules of glucose in 0.1 mole :
= [tex]0.1 mol\times 6.022\times 10^{23} molecules=6.022\times 10^{22} moleules[/tex]
In what ways are microwave and ultraviolet radiation the same? In what ways are they different?
Answer:
Electromagnetic waves are usually defined as those waves that are generated due to the vibrations between an electric as well as a magnetic field. Here, the component comprising the electric field and the component comprising the magnetic field vibrates perpendicular to each other, and both are in phase. Some of the examples of this type of wave are microwaves, infrared, ultra-violet, visible light, X-rays and many more.
The microwave and ultraviolet radiations are two electromagnetic waves that have similar characteristics, travel at a similar speed of about 300,000 km per second.
They differ from one another in many ways. It is because the microwaves have a higher wavelength, low frequency, and low energy. On the other hand, ultraviolet radiations have a low wavelength, high frequency, and high energy.
Household hydrogen peroxide is an aqueous solution containing 3.0% hydrogen peroxide by mass. What is the molarity of this solution? (Assume a density of 1.01 g>mL.)
Answer: Molarity of this solution is 0.88 M
Explanation:
Molarity : It is defined as the number of moles of solute present in one liter of solution.
Formula used :
[tex]Molarity=\frac{n\times 1000}{V_s}[/tex]
where,
Given : 3.0 g of hydrogen peroxide is present in 100 g of solution.
n = moles of solute = [tex]\frac{\text {given mass}}{\text {molar mass}}=\frac{3.0g}{34g/mol}=0.088mol[/tex]
[tex]V_s[/tex] = volume of solution in ml= [tex]\frac{\text {mass of solutuion}}{\text {density of solution}}=\frac{100g}{1.01g/ml}=99.0ml[/tex]
Now put all the given values in the formula of molarity, we get
[tex]Molarity=\frac{0.088moles\times 1000}{99.0ml}=0.88mole/L[/tex]
Thus molarity of this solution is 0.88 M
Which aqueous solution has the lowest boiling point?
a. 1.25 MC6H12O6
b. 1.25 M KNO3
c. 1.25 MCa(NO3)2
d. None of the above.
Answer:
C
Explanation:
The aqueous solution C has three ions where as b 2 ions and A no Ionization
An aqueous solution of 4.57 M H2SO4 has a density of 1.25 g/mL. Calculate the molality of this solution
Answer : The molality of solution is, 5.69 mole/L
Explanation :
The relation between the molarity, molality and the density of the solution is,
[tex]d=M[\frac{1}{m}+\frac{M_b}{1000}][/tex]
where,
d = density of solution = 1.25 g/mL
m = molality of solution = ?
M = molarity of solution = 4.57 M
[tex]M_b =\text{molar mass of solute }(H_2SO_4)[/tex] = 98 g/mole
Now put all the given values in the above formula, we get
[tex]1.25g/ml=4.57M[\frac{1}{m}+\frac{98g/mole}{1000}][/tex]
[tex]m=5.69mol/kg[/tex]
Therefore, the molality of solution is, 5.69 mole/L
Answer:
The molality is 5.7 molal
Explanation:
Step 1: Data given
Molarity of H2SO4 = 4.57 M ( = 4.57 mol/L)
Density of the solution = 1.25 g/mL
Molar mass of H2SO4 = 98.09 g/mol
Step 2: Calculate mass of solution
Suppose we have 1L (= 1000 mL) of solution
Mass of the solution = 1.25 g/mL * 1000 mL
Mass of the solution = 1250 grams
Step 3: Calculate mass H2SO4
In 1L of a 4.57 M solution we have 4.57
Mass H2SO4 = moles H2SO4 * molar mass H2SO4
Mass H2SO4 = 4.57 moles * 98.09 g/mol
Mass H2SO4 = 448.3 grams
Step 4: Calculate mass solvent
Mass solvent = mass solution - mass H2SO4
Mass solvent = 1250 grams - 448.3 grams
Mass solvent = 801.7 grams
Step 5: Calculate molality
Molality = moles H2SO4 / mass solvent
Molality = 4.57 moles / 0.8017 kg
Molality = 5.7 molal
The molality is 5.7 molal
How many grams of methanol (CH 3 OH, FM 32.04) are contained in 0.100 L of 1.71 M aqueous methanol (i.e., 1.71 mol CH 3 OH/L solution)?
Answer:
5.48 g
Explanation:
The concentration in M represents the mol/L, or the number of moles in each L of the solution, and it can be calculated by the number of moles (n) divided by the volume of the solution (V) in L, so:
M = n/V
1.71 = n/0.100
n = 0.171 mol
The molar mass of methanol is 32.04 g/mol, and it is the mass (m) divided by the number of moles:
FM = m/n
32.04 = m/0.171
m = 5.48 g
For which set of crystallographic planes will a first-order diffraction peak occur at a diffraction angle of 44.53° for FCC nickel (Ni) when monochromatic radiation having a wavelength of 0.1542 nm is used? The atomic radius for Ni is 0.1246 nm.
Answer:
The set of planes responsible for this diffraction peak is the [111] set.
Explanation:
We need to calculate the interplanar spacing, dₕₖₗ for nickel.
dₕₖₗ = nλ/2 sin θ
where θ = half of the diffraction angle = 44.53°/2 = 22.265°, n is the order of reflection = 1 and λ is the wavelength of the monochromatic radiation = 0.1542 nm.
dₕₖₗ = 1×0.1542/2sin 22.265°
dₕₖₗ = 0.2035 nm
But, interplanar spacing, dₕₖₗ is related to the plane (hkl), by the relation
√(h² + k² + l²) = a/dₕₖₗ
a is the lattice parameter.
Since Nickel has an FCC structure, a = 2R√2, R is given as 0.1242 nm
√(h² + k² + l²) = 2R√2/dₕₖₗ = (2×0.1246√2)/0.2035 = 1.732
(h² + k² + l²) = 1.732² = 3
The only 3 integers the values of h, k and l which fit properly into the equation is [111]
Therefore, the answer is [111].
Hope this Helps!!
√(h² + k² + l²) = a/dₕₖₗ
lattice parameterize is When the Nickel has an FCC structure, so that = 2R√2, R is given as 0.1242 nm= √(h² + k² + l²) = 2R√2/dₕₖₗ = (2×0.1246√2)/0.2035 = 1.732
= (h² + k² + l²) = 1.732² = 3
Then The only 3 integers are the values of h, k, and which is fit properly into the equation is 1Learn more about:
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Gastric juice is made up of substances secreted from parietal cells, chief cells, and mucous‑secreting cells. The cells secrete HCl , proteolytic enzyme zymogens, mucin, and intrinsic factor. The p H of gastric juice is acidic, between 1–3. If the p H of gastric juice is 2.2 , what is the amount of energy ( Δ G ) required for the transport of hydrogen ions from a cell (internal p H of 7.4) into the stomach lumen? Assume that the potential difference across the membrane separating the cell and the interior of the stomach is − 60.0 mV (inside of the cell is negative relative to the lumen of the stomach). Assume that the temperature is 37 °C. The Faraday constant is 96.5 kJ / ( V ⋅ mol ) and the gas constant is 8.314 × 10 − 3 kJ / ( mol ⋅ K ) . Express your answer in kilojoules per mole.
Answer:
what is the amount of energy ( Δ G ) required = 36.65KJ
Explanation:
The concept of VANT HOFF ISOTHERM EQUATION was used, as well as the relationship between the gibb's free energy and the electromotive force, all the steps and appropriate calculations is as shown in the attached file in order to get the amount of energy required for the transport of the hydrogen ions.
The energy required for the transport of hydrogen ions from a cell to the stomach lumen can be calculated using the Nernst equation and the equation for Gibbs free energy.
The amount of energy required for the transport of hydrogen ions from a cell into the stomach lumen can be calculated using the Nernst equation and the equation for Gibbs free energy. The Nernst equation relates the equilibrium potential for an ion to the concentrations of the ion on either side of the membrane, while the equation for Gibbs free energy relates the change in free energy to the change in potential across the membrane and the Faraday constant. By substituting the given values into these equations, the amount of energy required can be calculated.
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Within its genome, an organism must encode all of the genes necessary for maintaining life and for multiplying. As organisms increase in complexity, the size of their genomes and the total number of genes tend to increase. While there are exceptions to this trend, organism complexity can still be useful for the estimation of genome size.
Final answer:
The genome of an organism contains all its genetic material necessary for life and reproduction. Genome size varies widely across species and can inform us about genetic diversity and evolutionary history. The Human Genome Project and similar research have deepened our understanding of genomes and their role in growth, development, and evolution.
Explanation:
The genome of an organism includes all of its genetic material, which contains the necessary information for sustaining life and reproduction. As the complexity of organisms increases, generally, so does the size of their genomes. However, genome size doesn't correlate directly to the number of genes an organism has.
For instance, humans have about 3.5 pg of DNA in their genome, which translates to roughly 3.4 billion base pairs, despite not having the largest genome when compared to certain plants or even other animals. Intriguingly, a significant portion of the genome consists of non-coding DNA that doesn't seem to have a direct function in gene encoding.
Genome size differs widely across species, with eukaryotes often having multiple chromosomes and bacteria generally having smaller genomes. The genome size can provide information on an organism's genetic diversity and evolutionary history. The role of the genome is vital in the growth and development of organisms, dictating the specific instructions for these processes.
Additionally, the sequencing of genomes, as in the Human Genome Project, has provided valuable insights into genetic variations between and within species, fostering a greater understanding of evolutionary biology and potential medical advancements.
Based on the molecular formula, determine whether each compound is an alkane, alkene, or alkyne. (Assume that the hydrocarbons are noncyclical and there is no more than one multiple bond.)
a. C6H10
b. C4H10
c. C8H14
d. C8H16
Answer:
a. alkyne
b. alkane
c. alkyne
d. alkene
Explanation:
The general formula for each class of compound is given below
Alkane: [tex]C_nH_{2n+2}[/tex]
Alkene: [tex]C_nH_{2n}[/tex]
Alkyne: [tex]C_nH_{2n-2}[/tex] (assuming single multiple bonds)
Now let us classify according to the above formulas:
a. It has two hydrogen atoms less than the two times of carbon atoms hence, it's alkyne
b. It has two hydrogen atoms more than the two times of carbon atoms hence, it's alkane
c. It has two hydrogen atoms less than the two times of carbon atoms hence, it's alkyne
d. It has hydrogen atoms two times of carbon atoms hence, it's alkene
Answer:Based on the molecular formula, the following compounds belongs to the following groups:
a. C6H10 ( alkyne)
b. C4H10 ( alkane)
c. C8H14 (alkyne)
d. C8H16 (alkene)
Explanation: Hydrocarbon is an organic chemical compound that contains Hydrogen and carbons. There are three main types of Hydrocarbons which includes:
- Saturated Hydrocarbons.( They are composed entirely of single bonds and are saturated with hydrogen.). They are the alkane.
- Unsaturated Hydrocarbons: They are composed of either a double or triple bonds. The double bonds are the alkene and the triple bonds are the alkynes.
- Aromatic Hydrocarbons: they contain an aromatic ring. Example is the Benzene.
I hope this helps. Thanks.
What is the pH of a solution that contains three parts of acetic acid and one part sodium acetate? The p K for acetic acid is 4.76.
Answer:
The pH of the solution is 4.28
Explanation:
The dissolution reaction as below
CH₃COOH ⇔ CH₃COO⁻ + H⁺
[tex]K_{a} = \frac{[CH_{3}COO^{-}][H^{+}]}{[CH_{3}COOH} = 10^{-4.76}[/tex]
Assume the concentration of the ion, [H⁺] = a,
so [CH₃COO⁻] = a and [CH₃COOH] = 3a
Then use the formula of Ka, we get
Ka = a * a / 3a = 10^-4.76 ⇔ a = 3 x 10^-4.76 = 5.21 x 10^-5
Hence pH = -log(a) = - log(5.21 x 10^-5) = 4.28
When molten sulfur reacts with chlorine gas, a vile-smelling orange liquid forms that has an empirical formula of SCl. The structure of this compound has a formal charge of zero on all elements in the compound. Draw the Lewis structure for the vile-smelling orange liquid.
Answer:
The structure is shown below.
Explanation:
The formal charge (FC) is the charge that is more close to the actual charge in the real molecules and ions. It can be calculated based on the number of valence electrons (V), the shared electrons (S) and the electrons in the lone pairs (L) by the equation:
FC = V - (L + S/2)
Sulfur is in group 16 of the periodic table, so it has 6 valence electrons, and chlorine is from group 17 of the periodic table, and so it has 7 valence electrons. Chlorine can share only one electron, so it is stable. Sulfur can expand its octet (because it's from the third period) and can have more than 8 electrons when stable.
The possible formulas, from the empiric one, are:
SCl, S₂Cl₂, and S₃Cl₃.
To have FC = 0, chlorine must done only one bond, because S = 2, and L = 6, so:
FC = 7 - (6 + 2/2) = 0
So, it can not be the central atom of a structure. In the SCl, it will hav only a simple bond, so for sulfur, S = 2, and L = 4 (only the lone pairs are counted)
FC = 6 - (4+ 2/2) = +1
For S₂Cl₂, the two sulfurs must be bonded to a simple bond, and each one to one chlorine, thus, for both od them S = 4, and L = 4. so
FC = 6 - (4 + 4/2) = 0
So, it is the correct structure. The lewis structure represents the bonds by lines and the lone pairs of electrons by dots, and it is shown below.
Final answer:
The molecular formula for the compound formed when molten sulfur reacts with chlorine gas is [tex]S_2Cl_2[/tex], known as sulfur monochloride. Its Lewis structure shows two sulfur atoms, each bonded to a chlorine atom and to each other, with no formal charges.
Explanation:
When molten sulfur reacts with chlorine gas, the resulting compound with an empirical formula of SCl is sulfur monochloride. However, the molecular formula for sulfur monochloride is [tex]S_2Cl_2[/tex], since each molecule contains two sulfur atoms and two chlorine atoms. To draw its Lewis structure, we place the two sulfur atoms at the center, bonded to each other, and each sulfur atom has a single bond to a chlorine atom. Each sulfur atom needs to complete its octet by sharing one pair of electrons with chlorine and two pairs with another sulfur atom, resulting in a S-S single bond and S-Cl single bond for each sulfur. Since every atom needs to have a formal charge of zero, no charges are present in the Lewis structure. The Lewis structure will look like this:
Cl-S-S-Cl
Observe the reactions given below. Classify the reactants based on whether they are oxidized or reduced. Zn(s)+Cu2+(aq)→Zn2+(aq)+Cu(s) 3Fe(s)+4H2O(g)→Fe2O3(s)+4H2(g) Drag the appropriate items to their respective bins. View Available Hint(s)
Answer:
1. Zn is oxidize.
Cu is reduced.
2. Fe is oxidize.
H2 is reduced.
Explanation:
1. Zn(s)+Cu2+(aq)→Zn2+(aq)+Cu(s)
From the equation,
Zn changes oxidation number from 0 to +2. Therefore Zn is oxidized.
Cu2+ changes oxidation number from +2 to 0. Therefore, Cu2+ is reduced
2. 3Fe(s)+4H2O(g)→Fe2O3(s)+4H2(g)
From the equation,
Fe changes oxidation number from 0 to +3. Therefore, Fe is oxidized.
H changes oxidation number from +1 to 0. Therefore, H is reduced
Which one of the following has the largest acid equilibrium constant, Ka? Multiple Choice CH3CO2H CH2ClCO2H CHCl2CO2H CCl3CO2H
Answer:CCl3CO2H
Explanation:
CCl3CO2H is the strongest among the list because it ionize to give all its hydrogen ion
CCl3CO2H <==> H+ + CCL3COO-
CCl3CO2H has the largest acid equilibrium constant, Ka, because it contains the most Cl atoms per molecule, which enhances its ability to donate protons.
Explanation:The equilibrium constant Ka in acid-base chemistry is the acid dissociation constant. It represents the extent to which a compound donates protons (H+) in a solution. A larger Ka value indicates a stronger acid. When we compare CH3CO2H, CH2ClCO2H, CHCl2CO2H, and CCl3CO2H, the acid with the most Cl atoms has the largest Ka because Cl is a highly electronegative atom. This electronegativity pulls electron density away from the acidic H atom, making it easier to lose, which means the acid is stronger. So, CCl3CO2H has the largest acid equilibrium constant, Ka.
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The properties of several unknown solids were measured. Solid Melting point Other properties A >1000 °C does not conduct electricity B 850 °C conducts electricity in the liquid state, but not in the solid state C 750 °C conducts electricity in the solid state D 150 °C does not conduct electricity Classify the solids as ionic, molecular, metallic, or covalent. Note that covalent compounds are also known as covalent network solids or macromolecular solids. Ionic Molecular Metallic Covalent
Answer:
Explanation:
A >1000 °C does not conduct electricity : covalent ( usually do not conduct electricity, they are formed by the sharing of electrons)
B 850 °C conducts electricity in the liquid state, but not in the solid state: Ionic ( ionic or electrovalent compounds are formed between atoms where one loses an electron while the other gains e.g NaCl, they conduct electricity when dissolved in a polar solvent because they dissociate into ions and have high melting and boiling points)
750 °C conducts electricity in the solid state : Metallic ( metals generally have delocalized electrons that enables them to conduct electron since they are no associated with bond and are therefore free to move)
D 150 °C does not conduct electricity : Molecular ( consist mainly of molecules; they do not have charge)
Barium sulfate, BaSO4, is a white crystalline solid that is insoluble in water. It is used by doctors to diagnose problems with the digestive system. Barium hydroxide, Ba(OH)2, is also a white crystalline solid and is used in waste water treatment. How many more oxygen atoms are represented in the formula for barium sulfate than in the formula for barium hydroxide?
Answer:2
Explanation:
Ba(OH)2 contains two oxygen atoms
BaSO4 contains four oxygen atoms.
This means that barium sulphate contains two more oxygen atoms than barium hydroxide in its formula. This is clearly seen from the two formulae shown above.
The number of oxygen atom represented in the formula for barium sulfate than in the formula for barium hydroxide is 2 atoms
The molecular formula of barium sulfate = BaSO₄
The molecular formula of barium hydroxide = Ba(OH)₂
The number of oxygen atoms in BaSO₄ = 4 atoms
The number of oxygen atoms in Ba(OH)₂ = 2 atoms
Differences in atoms of oxygen = 4 – 2 = 2 atoms
From the above illustration, we can see that there are 2 more atoms of oxygen in barium sulfate, BaSO₄ than in barium hydroxide, Ba(OH)₂
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A company issued 6%, 15-year bonds with a face amount of $67 million. The market yield for bonds of similar risk and maturity is 6%. Interest is paid semiannually. At what price did the bonds sell? (FV of $1, PV of $1, FVA of $1, PVA of $1, FVAD of $1 and PVAD of $1)
To calculate the price at which the bonds sold, we use the present value formula for bonds.
Explanation:To calculate the price at which the bonds sold, we need to use the formula for present value of a bond. The formula is:
PV = (C * (1 - (1 + r)^-n) / r) + (FV / (1 + r)^n)
Where PV is the present value, C is the coupon payment, r is the market yield, n is the number of periods, and FV is the face value.
In this case, the coupon payment is $2,010,000 (6% of $67 million). The market yield is 6%, the number of periods is 30 (15 years * 2 semiannual payments per year), and the face value is $67 million. Plugging these values into the formula, we get:
PV = (2,010,000 * (1 - (1 + 0.06)^-30) / 0.06) + (67,000,000 / (1 + 0.06)^30)
Calculating this expression will give us the price at which the bonds sold.
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Tryptophan is an essential amino acid important in the synthesis of the neurotransmitter serotonin in the body. What are the hybridization state, molecular geometry and approximate bond angle at the indicated carbon and nitrogen atoms?
Answer:
The molecules are hybridized at different angles as shown in the explanation.
Explanation:
The hybridization and the geometries of the carbon atoms are as follows:
C1 → sp², trigonal planar, angle = 120⁰
C (II)→ sp³, tetrahedral, angle = 109.5⁰
C(III)→ sp², trigonal planar, 120⁰
C(IV)→ sp³, trigonal pyramidal, 109, 5⁰
A hydrate of beryllium nitrate has the following formula: Be(NO3)2⋅xH2O . The water in a 3.41-g sample of the hydrate was driven off by heating. The remaining sample had a mass of 2.43 g .Find the number of waters of hydration (x) in the hydrate.
Answer:
The formula is Be(NO3)2*3H2O. The number of waters in the hydrate is 3
Explanation:
Step 1: Data given
Mass of the hydrate sample = 3.41 grams
Mass after heating = 2.43 grams
Step 2: Calculate mass of water
After heating, all the water is gone. So the mass of water can be calculated by
Mass of hydrate before heating - mass after heating
Mass of water = 3.41 -2.43 = 0.98 grams
Step 2 : Calculate moles H2O
Moles H2O = mass H2O / molar mass H2O
Moles H2O = 0.98 grams / 18.02 g/mol
Moles H2O = 0.054 moles
Step 3: Calculate moles Be(NO3)2
Moles Be(NO3)2 = 2.43 grams / 133.02
Moles Be(NO3)2 = 0.0183 moles
Step 4: Calculate molecules water
Molecules H2O = 0.054 moles / 0.0183 moles
Molecules H2O = 3.0
The formula is Be(NO3)2*3H2O. The number of waters in the hydrate is 3
The number of waters of hydration in the hydrate is 3.
Based on the given information,
• The formula of the beryllium nitrate hydrate is Be(NO3)2⋅xH2O.
• The weight of the sample is 3.41 grams.
• The weight of the sample after heating of the sample is 2.43 grams {weight of Be(NO3)2}.
Now, the mass of xH2O will be,
= 3.41 g - 2.43 g
= 0.98 grams
The moles of Be(NO3)2 will be calculated as,
Mass of Be(NO3)2/Molecular weight of Be(NO3)2 = 2.43 g/133 = 0.01827 moles
The moles of xH2O = 0.98/18 = 0.05444 moles
Now the value of x will be,
= 0.05444/0.01827
= 3
Thus, the number of waters of hydration in the hydrate is 3.
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Benzoyl peroxide, the substance most widely used against acne, has a half-life of 9.8 × 103 days when refrigerated. How long will it take to lose 5% of its potency (95% remaining)? Assume that this is a first-order reaction. Give your answer in scientific notation.
Answer:
[tex]7.3\times 10^2\ days[/tex]
Explanation:
Given that:
Half life = [tex]9.8\times 10^3[/tex] days
[tex]t_{1/2}=\frac{\ln2}{k}[/tex]
Where, k is rate constant
So,
[tex]k=\frac{\ln2}{t_{1/2}}[/tex]
[tex]k=\frac{\ln2}{9.8\times 10^3}\ days^{-1}[/tex]
The rate constant, k = 0.00007 days⁻¹
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:
5 % is lost which means that 0.05 of [tex][A_0][/tex] is decomposed. So,
[tex]\frac {[A_t]}{[A_0]}[/tex] = 1 - 0.05 = 0.95
t = ?
[tex]\frac {[A_t]}{[A_0]}=e^{-k\times t}[/tex]
[tex]0.95=e^{-0.00007\times t}[/tex]
t = 732.76 days = [tex]7.3\times 10^2\ days[/tex]
It will take approximately [tex]\( 7.25 \times 10^2 \)[/tex] days for benzoyl peroxide to lose 5% of its potency when refrigerated.
To determine the time, it takes for benzoyl peroxide to lose 5% of its potency, we can use the first-order reaction kinetics formula:
[tex]\[ N(t) = N_0 \times e^{-kt} \][/tex]
where:
[tex]\( N(t) \)[/tex] is the amount of substance remaining after time [tex]\( t \)[/tex],
[tex]\( N_0 \)[/tex] is the initial amount of substance,
[tex]\( k \)[/tex] is the rate constant,
[tex]\( t \)[/tex] is the time.
The half-life [tex]\( t_{1/2} \)[/tex] is related to the rate constant [tex]\( k \)[/tex] by the equation:
[tex]\[ t_{1/2} = \frac{\ln(2)}{k} \][/tex]
Given the half-life [tex]\( t_{1/2} = 9.8 \times 10^3 \)[/tex] days, we can solve for [tex]\( k \)[/tex]:
[tex]\[ k = \frac{\ln(2)}{t_{1/2}} = \frac{\ln(2)}{9.8 \times 10^3} \][/tex]
Now, we want to find the time [tex]\( t \)[/tex] when 95% of the substance remains, so [tex]\( N(t) = 0.95N_0 \)[/tex]. Plugging this into the first-order reaction formula:
[tex]\[ 0.95N_0 = N_0 \times e^{-kt} \][/tex]
[tex]\[ 0.95 = e^{-kt} \][/tex]
Taking the natural logarithm of both sides:
[tex]\[ \ln(0.95) = -kt \][/tex]
Solving for [tex]\( t \)[/tex]:
[tex]\[ t = -\frac{\ln(0.95)}{k} \][/tex]
Substituting [tex]\( k \)[/tex] with the expression involving the half-life:
[tex]\[ t = -\frac{\ln(0.95)}{\ln(2)/t_{1/2}} \][/tex]
[tex]\[ t = -\frac{\ln(0.95) \cdot t_{1/2}}{\ln(2)} \][/tex]
[tex]\[ t = -\frac{\ln(0.95) \cdot 9.8 \times 10^3}{\ln(2)} \][/tex]
[tex]\[ t \approx -\frac{\ln(0.95) \cdot 9.8 \times 10^3}{0.693} \][/tex]
[tex]\[ t \approx -\frac{-0.051293 \cdot 9.8 \times 10^3}{0.693} \][/tex]
[tex]\[ t \approx \frac{0.5027 \times 10^3}{0.693} \][/tex]
[tex]\[ t \approx 7.25 \times 10^2 \][/tex]
Calcium salts give bone its a. tensile strength. b. torsional strength. c. flexibility. d. compressional strength.
Calcium salts in the bone provide it with compressional strength, which is the ability to withstand loads that might reduce size. This is crucial for bones to perform tasks such as standing, sitting, or lifting heavy objects.
Explanation:Calcium salts play a vital role in the structure and mechanical properties of bones. Particularly, they give bone its d. compressional strength. Compressional strength refers to the capacity of a material or structure to withstand loads tending to reduce size. This property is crucial to bones as they need to resist compression from the body's weight and movements. For instance, when we stand up, sit down or lift heavy objects, our leg bones or vertebrae need to bear the compressional force.
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Calcium salts provide bones with their compressional strength by crystallizing on a matrix of collagen fibers, providing hardness and rigidity. The collagen fibers add flexibility to prevent the bones from becoming too brittle.
Explanation:Calcium salts in the bone provide its d. compressional strength. This occurs when calcium phosphate and calcium carbonate combine to create hydroxyapatite, a mineral that provides bones with hardness and strength. These salts crystallize on a matrix of collagen fibers within the bone. The role of the collagen fibers is to provide bones with flexibility, preventing them from becoming too brittle.
However, it's important to note that both calcium salts and collagen fibers play crucial roles in the overall health and functionality of bones. The calcium salts allow the bones to withstand a great deal of pressure without deforming (compressional strength), while the collagen fibers give bones their resilience and flexibility. The combination of these two components allows bones to handle a variety of physical stresses.
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In the manufacture of nitric acid by the oxidation of ammonia, the first product is nitric oxide. The nitric oxide is then oxidized to nitrogen dioxide:
2 NO(g) + O2(g) → 2 NO2(g)
Calculate the standard reaction enthalpy for the reaction above (as written) using the following data:
N2(g) + O2(g) → 2 NO(g) ∆H ◦ = 180.5 kJ
N2(g) + 2 O2(g) → 2 NO2(g) ∆H◦ = 66.4 kJ
(A) −128.2 kJ/mol rxn
(B) −100.3 kJ/mol rxn
(C) −520.2 kJ/mol rxn
(D) −252.4 kJ/mol rxn
(E) −690.72 kJ/mol rxn
(F) −114.1 kJ/mol rxn
Answer:
F
Explanation:
simple subtration
In the manufacture of nitric acid by the oxidation of ammonia, the first product is nitric oxide. The nitric oxide is then oxidized to nitrogen dioxide. The standard reaction enthalpy for the reaction is −114.1 kJ/mol rxn.
What is an enthalpy ?A thermodynamic system's enthalpy is the sum of its internal energy and the product of its pressure and volume. It is a constant-pressure state function that is used in many measurements in chemical, biological, and physical systems.
Enthalpy is significant because it tells us how much heat is present in a system (energy). Heat is important because it allows us to generate valuable work. An enthalpy shift indicates how much enthalpy was lost or gained during a chemical reaction, with enthalpy referring to the heat energy of the system.
During chemical reactions, atom bonds can dissolve, reform, or both in order to absorb or release energy. The heat absorbed or emitted by a device under constant pressure is referred to as enthalpy.
Thus, option F is correct.
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Which isomer was produced from the bromination of trans-stilbene and cis-stilbene? Draw the structures of the products (use solid and dashed wedges).
The bromination of trans-stilbene results in meso-1,2-dibromo-1,2-diphenylethane, an achiral molecule. In contrast, the bromination of cis-stilbene yields enantiomeric products, (1R,2S)-1,2-dibromo-1,2-diphenylethane and (1S,2R)-1,2-dibromo-1,2-diphenylethane.
The bromination of trans-stilbene and cis-stilbene leads to different products due to the arrangement of the double bonds in the starting materials.
1. **Trans-Stilbene:**
- In trans-stilbene, the two phenyl rings are on opposite sides of the double bond. Bromination in the presence of bromine ([tex]\(Br_2\)[/tex]) or other brominating agents typically occurs with syn-addition across the double bond.
- The product is meso-1,2-dibromo-1,2-diphenylethane. The meso compound has a plane of symmetry, resulting in an achiral molecule.
2. **Cis-Stilbene:**
- In cis-stilbene, the two phenyl rings are on the same side of the double bond. Bromination of cis-stilbene can yield two enantiomeric products due to anti-addition across the double bond.
- The products are (1R,2S)-1,2-dibromo-1,2-diphenylethane and (1S,2R)-1,2-dibromo-1,2-diphenylethane.
The EA₂ of an oxygen atom is positive, even though its EA₁ is negative. Why does this change of sign occur? Which other elements exhibit a positive EA₂? Explain.
Explanation:
Electron affinity of 7th group elements whose electron affinity of their anionic forms ( EA2) are higher than the electron affinity of their neutral form(EA1). This is because their anionic forms are more stable than their neutral form. If the reaction is endothermic ,change in the energy is negative this means that electron affinity is positive. And vice versa for negative electron affinity.
The EA2 of oxygen is positive because energy is required to add an electron to a small space that is already dense with negative charge. The EA2 of sulfur is also positive.
The electron affinity refers to the energy evolved when one mole of electrons is added to an atom. We know that the energy required to add the second electron to oxygen is positive rather than negative implying that the process is endothermic.
This is because, energy is required to add an electron to a small space that is already dense with negative charge. The EA2 of sulfur is also positive.
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(3) 1. HCl NaOH: In Part I: An aqueous solution of sodium hydroxide reacts with an aqueous solution of hydrochloric acid, yielding water. Write the balanced molecular, ionic, and net ionic equations
Answer:HCl(aq) + NaOH (aq) -> NaCl (aq) + H2O(l)
Explanation:
The reaction of HCl and NaOH is a neutralization reaction. When an acid and a base react salt and water is produced.
HCl(aq) + NaOH (aq) ----> NaCl (aq) + H2O(l)
Because the reactants and products are ionic compounds, they exist as ions in a solution.
H+(aq) + Cl-(aq) + Na+(aq) + OH-(aq) -------> Na+(aq) + Cl-(aq) + H2O(l)
The aim of a neutralization reaction is the formation of water. Na+(aq) + Cl-(aq) remain the same on both the reactants side and the products side, the net ionic reaction will be
H+(aq) + OH-(aq) -------> H2O(l)
The reaction of HCl and NaOH forms water in a neutralization reaction. The net ionic equation, H⁺ + OH⁻ → H₂O, emphasizes the essential combination of ions to produce water.
The reaction between hydrochloric acid (HCl) and sodium hydroxide (NaOH) is a neutralization reaction, resulting in the formation of water and salt. In molecular form, the equation is written as:
HCl(aq) + NaOH(aq) → NaCl(aq) + H₂O(l)
Considering the ionic nature of these compounds in solution, they exist as ions. The complete ionic equation includes the dissociation of each compound into its constituent ions:
H⁺(aq) + Cl⁻(aq) + Na⁺(aq) + OH⁻(aq) → Na⁺(aq) + Cl⁻(aq) + H₂O(l)
Sodium cations (Na⁺) and chloride anions (Cl⁻) remain unchanged on both sides, acting as spectator ions. Eliminating these spectator ions yields the net ionic equation:
H⁺(aq) + OH⁻(aq) → H₂O(l)
This simplified representation emphasizes the essential components of the neutralization reaction, focusing on the combination of hydrogen ions (H⁺) from the acid and hydroxide ions (OH⁻) from the base to form water. The net ionic equation succinctly captures the core chemical transformation occurring in the neutralization process.
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What is the molecular weight of a gas that diffuses through a porous membrane 1.86 times faster than XeXe?
Answer:
Explanation:
Graham's law of diffusion states that the rate at which effusion occurs in two gases is equal to the square inverse of each of their molar masses.
Final answer:
Using Graham's law of effusion, we can calculate the molecular weight of the gas that diffuses 1.86 times faster than XeXe. The molecular weight of the unknown gas is approximately 37.98 g/mol.
Explanation:
To find the molecular weight of the gas that diffuses through a porous membrane 1.86 times faster than Xe, we need to use Graham's law of effusion. Graham's law states that the rate of effusion of a gas is inversely proportional to the square root of its molar mass.
Let the molecular weight of Xe be 'a' and the molecular weight of the unknown gas be 'b'.
According to Graham's law, (Rate of effusion of unknown gas)/(Rate of effusion of Xe) = sqrt((molar mass of Xe)/(molar mass of unknown gas)).
Squaring both sides of the equation, we get, (Rate of effusion of unknown gas)^2 = (Rate of effusion of Xe)^2 * (molar mass of Xe)/(molar mass of unknown gas).
Since the rate of diffusion of the unknown gas is 1.86 times faster than Xe, we can substitute this value into the equation to get (1.86)^2 = 1.86^2 * (molar mass of Xe)/(molar mass of unknown gas).
Simplifying the equation, (molar mass of unknown gas) = (molar mass of Xe) / (1.86^2).
Substituting the molar mass of Xe, which is 131.293 g/mol, into the equation, we get (molar mass of unknown gas) = 131.293 / (1.86^2).
Calculating the value, (molar mass of unknown gas) = 131.293 / 3.4596 = 37.98 g/mol.
Arrange the following H atom electron transitions in order of decreasing wavelength of the photon absorbed or emitted:
(a) n = 2 to n = [infinity]
(b) n = 4 to n = 20
(c) n = 3 to n = 10
(d) n = 2 to n = 1
Answer:
(b)>(c)>(a) >(d)
Explanation:
We know that from Rydberg´s equation:
1/ λ = Rh x (1/n₁² - 1/n₂² )
where n₁ and n₂ are the principal quantum numbers involved in the transition, and n₁ < n₂.
Therefore the wavelength will be given by taking the reciprocal of this equation:
λ = 1 / [Rh x (1/n₁² - 1/n₂² ) ]
So lets calculate the wavelengths for the transitions in this question expressed in terms of the constant Rh
(a ) λ = 1 / [Rh x (1/2² ) ] = 4/ Rh
(b ) λ = 1 / [Rh x (1/4² - 1/20² ) ] = 16.7 / Rh
(c) λ = 1 / [Rh x (1/3² - 1/10² ) ] = 9.9 / Rh
(d) λ = 1 / [Rh x (1/1² - 1/2² ) ] = 1.33 / Rh
Therefore in decreasing wavelength is (b)>(c)>(a) >(d)
as shown in these calculations, be careful with this type of question, since one might erroneously think that the transition for example as in (a) will have a shorter wavelength than (d) which is not the case as shown here. One must use Rydbergs formula.
Arranging the H atom electron transitions in decreasing order ;
B ---> C ----> A -----> D
Determine the wavelength of the photon absorbed by the H atom electron
we will apply Rydberg's equation
Rydberg equation = 1/ λ = Rh * (1/n₁² - 1/n₂² ) --- ( 1 )
n = quantum numbers
λ = wavelength
∴ λ = 1 / [Rh * (1/n₁² - 1/n₂² ) ] ---- ( 2 )
a) For n = 2 to n = [infinity]
λ = 1 / [ Rh * ( 1 / 2² - ∞ ) = 4 / Rh
b) For n = 4 to n= 20
λ = 1 / [ Rh * ( 1 / 4² - 1 / 20² ) = 16.7 Rh ( highest wavelength )
c) For n = 3 to n = 10
λ = 1 / [ Rh * ( 1 / 3² - 1 / 10² ) = 9.9 Rh
d) For n = 2 to n = 1
λ = 1 / [ Rh * ( 1 / 2² - 1 / 1² ) = 1.33 Rh ( Lowest wavelength )
Therefore arranging the H atom electron transitions in order of decreasing wavelength will be; B ---> C ----> A -----> D.
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A B C D 1) NaNH2 2) MeI 3) 9-BBN 4) H2O2, NaOH 1) Br2 2) Excess NaNH2 3) H2O 1) Br2 2) Excess NaNH2 3) H2O 4) H2SO4, H2O, HgSO4 1) NaNH2 2) EtI 3) Na, NH3 (l) E F G H 1) 9-BBN 2) H2O2, NaOH 1) Excess NaNH2 2) H2O 3) Br2 (1 eq), CCl4 1) Excess NaNH2 2) H2O 3) NaNH2 4) MeI 5) Na, NH3 (l) 1) NaNH2 2) 3) H2, Pt
Answer:
The question has some part missing which I have added in the attachment.
Explanation:
The reactions given occur under certain reagents, which we are told to pick for each of the reaction synthesis. Conventionally, halogens have the ability to undergo addition reactions with hydrocarbons by breaking down the double or triple bond in them to a single bond, this usually occur by electron donation and electron acceptor.
The attachment shows the reactions and the necessary reagents required for each
A tree was cut down and used to make a statue 2300 years ago. What fraction of 14C that was present originally remains today? t½ of 14C is 5730 yr
Answer:
0.76
Explanation:
N = No(0.5)^t/t1/2
N/No = (0.5)^t/t1/2
t = 2300 years, t1/2 = 5730 years
N/No = (0.5)^2300/5730 = 0.5^0.401 = 0.76
Calcium chloride contains calcium and chloride ions. Write the ground-state electron configuration for the calcium ion. You may write either the full or condensed electron configuration.
Answer:
Explanation:
Calcium is the element of the group 2 and period 4 which means that the valence electronic configuration is [tex]1s^22s^22p^63s^23p^64s^2[/tex] or [tex][Ar]4s^2[/tex].
Chlorine is the element of the group 17 and period 3 which means that the valence electronic configuration is [tex]1s^22s^22p^63s^23p^5[/tex] or [tex][Ne]3s^23p^5[/tex].
Thus, calcium losses 2 electrons to 2 atoms of chlorine and these 2 atoms of chlorine accepts each electron to form ionic bond. This is done in order that the octet of the atoms are complete and they become stable.
Thus, the formula of calcium chloride is [tex]CaCl_2[/tex].
Hence, in [tex]CaCl_2[/tex], Calcium exits in [tex]Ca^{2+}[/tex] form which has electronic configuration of [tex]1s^22s^22p^63s^23p^6[/tex] or [tex][Ar][/tex]
The ground-state electron configuration for a calcium ion (Ca²+) is 1s²2s²2p63s²3p6, the same configuration as the noble gas Argon (Ar). This is because a calcium ion loses its two 4s valence electrons when it forms a cation.
Explanation:The calcium ion, denoted as Ca²+, has lost its valence electrons to form a cation. In an uncharged, ground state calcium atom, it contains 20 electrons with configuration as 1s²2s²2p63s²3p64s². When it loses its two 4s valence electrons to become a cation, its electron configuration changes to 1s²2s²2p63s²3p6 which is similar to the noble gas Argon (Ar).
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In each reaction, identify the Brønsted–Lowry acid, the Brønsted–Lowry base, the conjugate acid, and the conjugate base. a. H2CO3(aq) + H2O(l) ∆ H3O+(aq) + HCO3 -(aq) b. NH3(aq) + H2O(l) ∆ NH4 +(aq) + OH-(aq)
In an acid-base reaction, the Brønsted–Lowry acid donates a proton and becomes a conjugate base while the Brønsted–Lowry base accepts a proton and becomes a conjugate acid. The reactions given as examples illustrate these changes with H2CO3 and H2O acting as acids in the two reactions while H2O and NH3 act as bases respectively.
Explanation:In the reaction, Brønsted–Lowry acid is the species that donates a proton (H+), and the Brønsted–Lowry base is the species that accepts a proton. After the acid has lost a proton, it becomes the conjugate base. Conversely, after the base gains a proton, it becomes the conjugate acid.
For the reaction H2CO3(aq) + H2O(l) ∆ H3O+(aq) + HCO3 -(aq), H2CO3 is the Brønsted-Lowry acid and H2O is the Brønsted-Lowry base. Upon losing a proton, H2CO3 becomes HCO3- which is the conjugate base and upon gaining a proton, H2O becomes H3O+ which is the conjugate acid.
For the reaction NH3(aq) + H2O(l) ∆ NH4 +(aq) + OH-(aq), NH3 is the Brønsted-Lowry base and H2O is the Brønsted-Lowry acid. Upon gaining a proton, NH3 becomes NH4+ which is the conjugate acid and upon losing a proton, H2O becomes OH- which is the conjugate base.
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In a Brønsted-Lowry reaction, the acid donates a proton and the base accepts the proton. The species formed when an acid donates a proton becomes a conjugate base, and the species formed when a base accepts a proton becomes a conjugate acid. In the given reactions, H2CO3 and H2O function as acids, and H2O and NH3 function as bases.
Explanation:In a Brønsted-Lowry acid-base reaction, an acid is a substance that donates a proton (H+) and a base is a substance that accepts a proton. A conjugate acid is the species formed when a Brønsted-Lowry base gains a proton, and a conjugate base is the species remaining after a Brønsted-Lowry acid has lost a proton.
For the reaction H2CO3(aq) + H2O(l) ∆ H3O+(aq) + HCO3 -(aq), the Brønsted-Lowry acid is H2CO3, which donates a proton to become the conjugate base, HCO3-. The Brønsted-Lowry base is H2O, which accepts a proton to become the conjugate acid, H3O+.
For the reaction NH3(aq) + H2O(l) ∆ NH4 +(aq) + OH-(aq), the Brønsted-Lowry base is NH3, which accepts a proton to become the conjugate acid, NH4+. The Brønsted-Lowry acid is H2O, which donates a proton to become the conjugate base, OH-.
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