Answer:
23.38 mmHg is the vapor pressure at 25 °C of an aqueous solution that is 5.50% NaCl by mass.
Explanation:
Vapor pressure of water at 25 °C ,[tex]p^o= 23.8 mmHg[/tex]
Vapor pressure of the solution = [tex]p_s[/tex]
Number moles of water in 5.50% NaCl solution.In 100 gram of solution, 94.5 g of water is present.
[tex]n_1=\frac{94.5 g}{18 g/mol}=5.25 mol[/tex]
Number moles of NaCl in 5.50% NaCl solution.5.50 g of NaCl in 100 grams of solution.
[tex]n_2=\frac{5.50 g}{58.5 g/mol}=0.09401 mol [/tex]
Mole fraction of the solute = [tex]\chi_2=\frac{n_2}{n_1+n_2}[/tex]
[tex]\chi_2=\frac{0.09401 mol}{0.09401 mol+5.25 mol}=0.01759[/tex]
The relative lowering in vapor pressure of the solution with non volatile solute is equal to the mole fraction of solute in the solution:
[tex]\frac{p^o-p_s}{p^o}=\chi_2=\frac{n_2}{n_1+n_2}[/tex]
[tex]\frac{23.8 mmHg - p_s}{23.8 mmHg}=0.01759[/tex]
[tex]p_s=23.38 mmHg[/tex]
23.38 mmHg is the vapor pressure at 25 °C of an aqueous solution that is 5.50% NaCl by mass.
23.38 mmHg is the vapour pressure.
The pressure enforced by the vapours when in the thermodynamical equilibrium state on the system is called the vapour pressure.
How to calculate the vapour pressure?Vapour pressure [tex](p^{\circ})[/tex] of water at [tex]25 ^{\circ} \rm C[/tex] = 23.8 mmHgVapour pressure of aqueous solution = [tex](p_{s})[/tex]Step 1: Moles of water in 5.50% NaCl solution when 100 grams of solution = 94.5 g of water
[tex]\begin{aligned}\rm n &= \dfrac{94.5 \;\rm g}{18\;\rm g/mol}\\\\\\&= 5.25\;\rm mol\end{aligned}[/tex]
Step 2: Moles of NaCl in 5.50% solution when 5.50 g NaCl in 100 grams solution then,
[tex]\begin{aligned}\rm n &= \dfrac{5.50\;\rm g}{58.5\;\rm g/mol}\\\\\\&= 0.09401\;\rm mol\end{aligned}[/tex]
Step 3: Calculate the mole fraction of the solute:
[tex]\begin{aligned}\rm X_{2} &= \dfrac{\rm n_{2}}{\rm n_{1}+n_{2}}\\\\\rm X_{2} &= \dfrac{0.09401\;\rm mol}{0.09401 + 5.25\;\rm mol}\\\\&= 0.0175\end{aligned}[/tex]
Step 4: Calculate the vapour pressure
[tex]\begin{aligned}\rm X_{2} &= \dfrac{\rm n_{2}}{\rm n_{1}+n_{2}} = \dfrac{p^{\circ}-p_{s}}{p^{\circ}}\\\\0.0175 &= \dfrac{23.38 \;\text{mmHg} -p_{s}}{23.38 \;\rm mmHg}\\\\&= 23.38\;\rm mmHg\end{aligned}[/tex]
Therefore, 23.38 mmHg is the vapour pressure.
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Write a chemical equation for the reaction that occurs in the following cell: Cu|Cu2+(aq)||Ag+(aq)|Ag Express your answer as a balanced net ionic equation. Identify all of the phases in your answer.
Answer:
Explanation:
The cell reaction properly written is shown below:
Cu|Cu²⁺[tex]_{aq}[/tex] || Ag⁺[tex]_{aq}[/tex] | Ag
From this cell reaction, to get the net ionic equation, we have to split the reaction into their proper oxidation and reduction halves. This way, we can know that is happening at the electrodes and derive the overall net equation.
Oxidation half:
Cu[tex]_{s}[/tex] ⇄ Cu²⁺[tex]_{aq}[/tex] + 2e⁻
At the anode, oxidation occurs.
Reduction half:
Ag⁺[tex]_{aq}[/tex] + 2e⁻ ⇄ Ag[tex]_{s}[/tex]
At the cathode, reduction occurs.
To derive the overall reaction, we must balance the atoms and charges:
Cu[tex]_{s}[/tex] ⇄ Cu²⁺[tex]_{aq}[/tex] + 2e⁻
Ag⁺[tex]_{aq}[/tex] + e⁻ ⇄ Ag[tex]_{s}[/tex]
we multiply the second reaction by 2 to balance up:
2Ag⁺[tex]_{aq}[/tex] + 2e⁻ ⇄ 2Ag[tex]_{s}[/tex]
The net reaction equation:
Cu[tex]_{s}[/tex] + 2Ag⁺[tex]_{aq}[/tex] + 2e⁻⇄ Cu²⁺[tex]_{aq}[/tex] + 2e⁻ + 2Ag[tex]_{s}[/tex]
We then cancel out the electrons from both sides since they appear on both the reactant and product side:
Cu[tex]_{s}[/tex] + 2Ag⁺[tex]_{aq}[/tex] ⇄ Cu²⁺[tex]_{aq}[/tex] + 2Ag[tex]_{s}[/tex]
The net ionic equation for the cell - Cu|Cu2+(aq)||Ag+(aq)|Ag is Cu(s) + 2Ag+(aq) --> Cu2+(aq) + 2Ag(s). This reaction involves the transfer of electrons from copper to silver ions, converting copper metal to copper ions and silver ions to silver metal.
Explanation:The chemical equation for the reaction in the given cell, Cu|Cu2+(aq)||Ag+(aq)|Ag, can be expressed as follows:
Cu(s) --> Cu2+(aq) + 2e−
2Ag+(aq) + 2e− --> 2Ag(s)
Combining these two half-reactions gives the net ionic equation:
Cu(s) + 2Ag+(aq) --> Cu2+(aq) + 2Ag(s)
In this reaction, solid copper (Cu) reacts with silver ions (Ag+) in the aqueous solution to produce copper ions (Cu2+) and solid silver (Ag). All the phases for each species are defined in the equation. The copper metal is solid, represented by (s), both ionic species are in aqueous solution denoted by (aq).
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1. Define the term disproportionation reaction. 2. Write the balanced equation for the conversion of 4-chlorobezaldehye into 4-chlorobenzyl alcohol and 4-chlorobenzoic acid. 3. What is the oxidation number of the aldehyde carbon in each product of the Cannizzaro reaction? Show all work for credit.
Answer:
3 is the answer
Explanation:
An ideal gas is compressed isothermally and reversibly at 400K from 1 m3 to 0.5 m3 9200 J heat is evolved during compression. What is the work done and how many moles of gas were compressed during this process?
Answer : The work done on the system is 9200 J and the number of moles of gas is 3.99 moles.
Explanation : Given,
Initial volume of gas = [tex]1m^3[/tex]
Final volume of the gas = [tex]0.5m^3[/tex]
Temperature of the gas = 400 K
Heat evolved = -9200 J
According to the question, this is the case of isothermal reversible compression of gas.
As per first law of thermodynamic,
[tex]\Delta U=q+w[/tex]
where,
[tex]\Delta U[/tex] = internal energy
q = heat
w = work done
As we know that, the term internal energy is the depend on the temperature and the process is isothermal that means at constant temperature.
So, at constant temperature the internal energy is equal to zero.
[tex]q=-w[/tex]
Thus, the work done on the system = -q = -(-9200J) = 9200 J
The expression used for work done will be,
[tex]w=-nRT\ln (\frac{V_2}{V_1})[/tex]
where,
w = work done on the system = 9200 J
n = number of moles of gas = ?
R = gas constant = 8.314 J/mole K
T = temperature of gas = 400 K
[tex]V_1[/tex] = initial volume of gas = [tex]1m^3[/tex]
[tex]V_2[/tex] = final volume of gas = [tex]0.5m^3[/tex]
Now put all the given values in the above formula, we get the number of moles of gas.
[tex]9200J=-n\times 8.314J/moleK\times 400K\times \ln (\frac{0.5m^3}{1m^3})[/tex]
[tex]n=3.99mole[/tex]
Therefore, the work done on the system is 9200 J and the number of moles of gas is 3.99 moles.
A student placed 13.0 g of glucose (C6H12O6) in a volumetric flask, added enough water to dissolve the glucose by swirling, then carefully added additional water until the 100.-mL mark on the neck of the flask was reached. The flask was then shaken until the solution was uniform. A 60.0-mL sample of this glucose solution was diluted to 0.500 L. How many grams of glucose are in 100. mL of the final solution?
Answer:
There are 1.5597 grams of glucose in 100 mL of 0.08658 M glucose solution.
Explanation:
13.0 grams of glucose was added to volumetric flask and solution was made up to 100 mL.
Moles of glucose = [tex]\frac{13.0 g}{180.156 g/mol}=0.07215 mol[/tex]
Volume of the solution = 100 mL = 0.1 L
Molarity of the solution = [tex]\frac{Moles}{\text{Volume of solution}}[/tex]
[tex]M=\frac{0.07215 mol}{0.1 L}=0.7215 M[/tex]
60 mL of 0.7215 M was diluted to 0.500 L, the molarity of the solution after dilution be [tex]M_2[/tex]
[tex]M_1=0.7215 M,V_2=60 mL=0.060 L[/tex]
[tex]M_2=?,V_2=0.500 L[/tex]
[tex]M_1V_1=M_2V_2[/tex] (Dilution)
[tex]M_2=\frac{0.7215 M\times 0.060 L}{0.500 L}[/tex]
[tex]M_2=0.08658 M[/tex]
Mass of glucose in 0.08658 M glucose solution:
In 1 L of solution = 0.08658 moles
In 1000 mL of solution = 0.08658 moles
Then in 100 mL of solution :
[tex]\frac{0.08658}{1000}\times 100 mol=0.008658 mol[/tex] of glucose
Mass of 0.008658 moles of glucose:
[tex]0.008658 mol\times 180.156 g/mol=1.5597 g[/tex]
There are 1.5597 grams of glucose in 100 mL of 0.08658 M glucose solution.
Answer:
m_{C_6H_{12}O_6}=1.56g
Explanation:
Hello,
After the dilution, the concentration of glucose is:
[tex]c=\frac{13.0g}{100mL}=0.13g/mL[/tex]
Now, we apply the dilution equation to compute the concentration after the dilution to 0.500L (500mL):
[tex]c_1*V_1=c_2*V_2\\c_2=\frac{c_1*V_1}{V_2}=\frac{0.13g/mL*60.0mL}{500mL} \\c_2=0.0156g/mL[/tex]
Finally, we compute the present grams in 100mL of the 0.0156g/mL glucose solution as shown below:
[tex]m_{C_6H_{12}O_6} =100mL*0.0156g/mL\\m_{C_6H_{12}O_6}=1.56g[/tex]
Best regards.
Sulfur dioxide and oxygen react to form sulfur trioxide during one of the key steps in sulfuric acid synthesis. An industrial chemist studying this reaction fills a 25.0L tank with 4.5 mol of sulfur dioxide gas and 4.5 mol of oxygen gas at 30.°C. He then raises the temperature, and when the mixture has come to equilibrium measures the amount of sulfur trioxide gas to be 1.4 mol. Calculate the concentration equilibrium constant for the reaction of sulfur dioxide and oxygen at the final temperature of the mixture. Round your answer to 2 significant digits.
Answer:
1.3 is the concentration equilibrium constant for the reaction of sulfur dioxide and oxygen at the final temperature of the mixture.
Explanation:
Initial Concentration of sulfur dioxide = [tex][SO_2]=\frac{4.5 mol}{25 L}=0.18 M[/tex]
Initial Concentration of oxygen= [tex][O_2]=\frac{4.5 mol}{25 L}=0.18 M[/tex]
[tex]2SO_2+O_2\rightleftharpoons 2SO_3[/tex]
Initially (0.18 M) (0.18 M) 0
Eq'm (0.18 -2x) (0.18 -x) 2x
Equilibrium concentration of sulfur trioxide =[tex][SO_3]=2x=\frac{1.4 mol}{25 L}=0.056 M[/tex]
x = 0.028 M
Equilibrium concentration of sulfur dioxide =[tex][SO_2]'=(0.18 -2x)=0.18 - 0.056 =0.124 M[/tex]
Equilibrium concentration of oxygen=[tex][O_2]'=(0.18 -x)=0.18 - 0.028 =0.152 M[/tex]
The expression for an equilibrium constant will be :
[tex]K_c=\frac{[SO_3]^2}{[SO_2]'^2[O_2]'}[/tex]
[tex]K_c=\frac{(0.056 M)^2}{(0.124 M)^2(0.152 M)}=1.3418\approx 1.3[/tex]
1.3 is the concentration equilibrium constant for the reaction of sulfur dioxide and oxygen at the final temperature of the mixture.
Final answer:
To calculate the concentration equilibrium constant (Kc) for the reaction of SO2 and O2 to form SO3, find the equilibrium concentrations from the initial amounts and amount at equilibrium, then apply the equilibrium expression. The resulting Kc for the reaction at the final temperature is 0.020 when rounded to two significant digits.
Explanation:
The calculation of the concentration equilibrium constant for the reaction between sulfur dioxide and oxygen to form sulfur trioxide at a certain temperature involves using the equilibrium concentrations of reactants and products. The balanced chemical equation for the reaction is:
2 SO2(g) + O2(g) = 2 SO3(g)
Given 4.5 mol of SO2 and 4.5 mol of O2 initially in a 25.0 L tank, and 1.4 mol of SO3 at equilibrium, we can calculate the change in moles during the reaction (δ) and thus the equilibrium concentrations ([SO2], [O2], and [SO3]). The concentration equilibrium constant (Kc) is then found using the expression:
Kc = ([SO3]2)/([SO2]2 × [O2])
Through stoichiometry and equilibrium concentration calculations, the concentrations are:
[SO3] = 1.4 mol / 25.0 L = 0.056 M
[SO2] = (4.5 mol - 1.4 mol) / 25.0 L = 0.124 M (since 1 mol of SO2 is consumed for every mol of SO3 produced)
[O2] = (4.5 mol - 0.7 mol) / 25.0 L = 0.152 M (since 0.5 mol of O2 is consumed for every mol of SO3 produced)
Plugging these values into the Kc expression gives:
Kc = (0.0562) / (0.1242 × 0.152) = 0.0197
The calculated Kc at the final temperature is 0.020 (rounded to two significant digits).
Equal volumes of hydrogen and helium gas are at the same pressure. The atomic mass of helium is four times that of hydrogen. If the total mass of both gases is the same, the ratio of the temperature of helium (He) to that of hydrogen (H2) is:
Answer:
The ratio of the temperature of helium to that of hydrogen gas is 2:1.
Explanation:
Atomic mass of hydrogen = M
Temperature of hydrogen gas =T
Pressure of the hydrogen gas = P
Mass of the hydrogen gas = m
Moles of the hydrogen gas = [tex]n=\frac{m}{2M}[/tex]
Volume of the hydrogen gas = V
Using an ideal gas equation:
[tex]PV=nRT=PV=\frac{mRT}{M}[/tex]...(1)
Temperature of helium gas =T'
Pressure of the helium gas = P'= P
Mass of the helium gas = m' =m
Moles of the helium gas = [tex]n'=\frac{m}{M'}=\frac{m}{4M}[/tex]
Volume of the helium gas = V' = V
Using an ideal gas equation:
[tex]P'V'=n'RT'=\frac{mRT'}{4M}[/tex]...(2)
Divide (2) by (1)
[tex]\frac{P'V'}{PV}=\frac{\frac{mRT'}{4M}}{\frac{mRT}{2M}}[/tex]
[tex]\frac{T'}{T}=\frac{2}{1}[/tex]
The ratio of the temperature of helium to that of hydrogen gas is 2:1.
You heat 51 grams of magnesium over a Bunsen burner for several minutes until it reacts with oxygen in the air. Then you weigh the resulting product and see that it is now 53 grams. How does this happen without breaking the Law of Conservation of Mass?
Final answer:
The increase in mass of magnesium after burning is due to its combination with oxygen to form magnesium oxide, which is in accordance with the Law of Conservation of Mass. No mass is lost or created; the oxygen simply adds to the original mass of the magnesium metal.
Explanation:
The observation that a piece of magnesium gains mass when burnt can be explained without violating the Law of Conservation of Mass. This law states that mass is neither created nor destroyed in a chemical reaction. When magnesium burns in the presence of oxygen, a chemical change occurs, resulting in the formation of magnesium oxide, a white crumbly powder. The reaction can be depicted by the word equation magnesium + oxygen → magnesium oxide. The increase in mass from 51 grams to 53 grams is due to the addition of oxygen from the air. When magnesium (Mg) reacts with oxygen (O₂), the oxygen atoms combine with magnesium to form magnesium oxide (MgO), thus accounting for the increase in mass.
For example, if you heat 10.0 grams of calcium carbonate (CaCO₃) and produce 4.4 g of carbon dioxide (CO₂) and 5.6 g of calcium oxide (CaO), the total mass of the products equals the original mass of the reactants, which is in agreement with the Conservation of Mass. Similarly, in the magnesium burning experiment, the mass of the magnesium and the oxygen that combined during burning would equal the mass of the magnesium oxide produced, if all the reactants and products could be contained and measured.
Ideal gas (n 2.388 moles) is heated at constant volume from T1 299.5 K to final temperature T2 369.5 K. Calculate the work and heat during the process and the change of entropy of the gas. The isobaric heat capacity of the gas is Cp.m 28.253 J-K1-mol1.
Answer : The work, heat during the process and the change of entropy of the gas are, 0 J, 3333.003 J and -10 J respectively.
Explanation :
(a) At constant volume condition the entropy change of the gas is:
[tex]\Delta S=-n\times C_v\ln \frac{T_2}{T_1}[/tex]
We know that,
The relation between the [tex]C_p\text{ and }C_v[/tex] for an ideal gas are :
[tex]C_p-C_v=R[/tex]
As we are given :
[tex]C_p=28.253J/K.mole[/tex]
[tex]28.253J/K.mole-C_v=8.314J/K.mole[/tex]
[tex]C_v=19.939J/K.mole[/tex]
Now we have to calculate the entropy change of the gas.
[tex]\Delta S=-n\times C_v\ln \frac{T_2}{T_1}[/tex]
[tex]\Delta S=-2.388\times 19.939J/K.mole\ln \frac{369.5K}{299.5K}=-10J[/tex]
(b) As we know that, the work done for isochoric (constant volume) is equal to zero. [tex](w=-pdV)[/tex]
(C) Heat during the process will be,
[tex]q=n\times C_v\times (T_2-T_1)=2.388mole\times 19.939J/K.mole\times (369.5-299.5)K= 3333.003J[/tex]
Therefore, the work, heat during the process and the change of entropy of the gas are, 0 J, 3333.003 J and -10 J respectively.