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3 years ago

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A sulfuric acid solution containing 571.3 g of h2so4 per liter of aqueous solution has a density of 1.329 g/cm3. Part a calculat

e the mass percentage of h2so4 in this solution. Express your answer using four significant figures. G

1 answer:

loris [4]3 years ago

3 0

Mass percentage of a solution is the amount of solute present in 100 g of the solution.

Given data:

Mass of solute H2SO4 = 571.3 g

Volume of the solution = 1 lit = 1000 ml

Density of solution = 1.329 g/cm3 = 1.329 g/ml

Calculations:

Mass of the given volume of solution = 1.329 g * 1000 ml/1 ml = 1329 g

Therefore we have:

571.3 g of H2SO4 in 1329 g of the solution

Hence, the amount of H2SO4 in 100 g of solution= 571.3 *100/1329 = 42.987

Mass percentage of H2SO4 (%w/w) is 42.99 %

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The equilibrium constant for the reaction AgBr(s) Picture Ag+(aq) + Br− (aq) is the solubility product constant, Ksp = 7.7 × 10−

barxatty [35]

Answer:

The reaction will be non spontaneous at these concentrations.

Explanation:

$AgBr(s)\rightarrow Ag^+(aq) + Br^- (aq)$

Expression for an equilibrium constant $K_c$ :

$K_c=\frac{[Ag^+][Br^-]}{[AgCl]}=\frac{[Ag^+][Br^-]}{1}=[Ag^+][Br^-]$

Solubility product of the reaction:

$K_{sp}=[Ag^+][Br^-]=K_c=7.7\times 10^{-13}$

Reaction between Gibb's free energy and equilibrium constant if given as:

$\Delta G^o=-2.303\times R\times T\times \log K_c$

$\Delta G^o=-2.303\times R\times T\times \log K_{sp}$

$\Delta G^o=-2.303\times 8.314 J/K mol\times 298 K\times \log[7.7\times 10^{-13}]$

$\Delta G^o=69,117.84 J/mol=69.117 kJ/mol$

Gibb's free energy when concentration $[Ag^+] = 1.0\times 10^{-2} M$ and $[Br^-] = 1.0\times 10^{-3} M$

Reaction quotient of an equilibrium = Q

$Q=[Ag^+][Br^-]=1.0\times 10^{-2} M\times 1.0\times 10^{-3} M=1.0\times 10^{-5}$

$\Delta G=\Delta G^o+(2.303\times R\times T\times \log Q)$

$\Delta G=69.117 kJ/mol+(2.303\times 8.314 Joule/mol K\times 298 K\times \log[1.0\times 10^{-5}])$

$\Delta G=40.588 kJ/mol$

For reaction to spontaneous reaction: $\Delta G$ .
For reaction to non spontaneous reaction: $\Delta G>0$ .

Since ,the value of Gibbs free energy is greater than zero which means reaction will be non spontaneous at these concentrations

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Within the nuclear us of individual Adams of the same element, the proton number

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the same

Explanation:

Within the nucleus of individual atoms of the same element, the proton number is the same.

All atoms of the same kind from the same element have the same number of protons.

The number of protons does not change, it remains fixed after a chemical reaction.

Different elements have different number of protons that typifies and makes them unique.

But atoms of the same element have the same number of protons in them.

learn more;

Atomic number brainly.com/question/5425825

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The diagram below shows three types of earthquake waves, labeled A, B, and C at different time intervals.

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The diagram below shows three types of earthquake waves, labeled A, B, and C at different time intervals. The statement that is correct about the three waves is <span>Wave A and Wave B are absorbed by liquid medium. The answer is letter B.</span>

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The acid-dissociation constant of hydrocyanic acid (hcn) at 25.0 °c is 4.9 ⋅ 10−10. what is the ph of an aqueous solution of 0.0

vodomira [7]

According to the reaction equation:

and by using ICE table:

CN- + H2O ↔ HCN + OH-

initial 0.08 0 0

change -X +X +X

Equ (0.08-X) X X

so from the equilibrium equation, we can get Ka expression

when Ka = [HCN] [OH-]/[CN-]

when Ka = Kw/Kb

= (1 x 10^-14) / (4.9 x 10^-10)

= 2 x 10^-5

So, by substitution:

2 x 10^-5 = X^2 / (0.08 - X)

X= 0.0013

∴ [OH] = X = 0.0013

∴ POH = -㏒[OH]

= -㏒0.0013

= 2.886

∴ PH = 14 - POH

= 14 - 2.886 = 11.11

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A tank of gas has partial pressures of nitrogen and oxygen equal to 1.61 Ã— 104 kPa and 4.34 Ã— 103 kPa , respectively.What is t

Alchen [17]

Answer:

$20.44\times 10^3 kPa$ is the total pressure of the tank.

Explanation:

Partial pressures of nitrogen = $p_{N_2}=1.61\times 10^4 kPa$

Partial pressure of oxygen = $p_{O_2}=4.34\times 10^3 kPa$

Total pressure of gases in the tank = P

Applying Dalton's law of partial pressures :

$P=p_{N_2}+p_{O_2}=1.61\times 10^4 kPa+4.34\times 10^3 kPa$

$P=20.44\times 10^3 kPa$

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