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The molar ratio of HPO42- to H2PO4- in a solution is 1.4. Calculate the pH of the solution. Phosphoric acid (H3PO4) is a triprot

Bezzdna [24]

Given is the ratio of conjugate base and conjugate acid of phosphoric acid. pH of a substance is the concentration of the hydrogen ions in its solution and higher this concentration lower is the value of pH.

pKa value is a measure of the strength of acid, it is the negative log of acid dissociation constant Ka.

7 0

3 years ago

g Ammonia has been studied as an alternative "clean" fuel for internal combustion engines, since its reaction with oxygen produc

Gala2k [10]

$\text{Ammonia has been studied as an alternative "clean" fuel for internal combustion}$

$\text{engines, since its reaction with oxygen produces only nitrogen and water vapor,}$

$\text{and in the liquid form it is easily transported. An industrial chemist studying this}$

$\text{reaction fills a} \ \mathbf{100 \ L }\ \text{tank with} \ \mathbf{8.6 \ mol} \ \text{of ammonia gas and} \ \mathbf{28 \ mol} \ \ \text{of oxygen gas, }$

$\text{to be} \ \mathbf{2.6\ mol} \ .\ \text{Calculate the concentration equilibrium constant for the combustion of}$

$\text{ammonia at the final temperature of the mixture. Round your answer to 2 significant digits.}$

Answer:

Explanation:

From the correct question above:

The reaction can be represented as:

$\mathbf{4 NH_3_{(g)}+ 3O_{2(g)} \iff 2N_{2(g)}+ 6H_2O_{(g)} }$

From the above reaction; the ICE table can be represented as:

$\mathbf{4 NH_3_{(g)}+ 3O_{2(g)} \iff 2N_{2(g)}+ 6H_2O_{(g)} }$

I (mol/L) 0.086 0.28 0 0

C -4x -3x +2x +6x

E 0.086 - 4x 0.28 - 3x +2x +6x

At equilibrium;

The water vapor = $\dfrac{2.6 \ mol}{100 \ L} = 6x$

$x = \dfrac{2.6}{100} \times \dfrac{1}{6}$

$x = 0.00433$

$\text{equilibrium constant} ({k_c}) = \dfrac{ [N_2]^2 [H_2O]^6 }{ [[NH_3]^4] [O_2]^3 }$

$\implies \dfrac{(2x)^2 (6x)^6}{(0.086-4x)^4\times (0.28-3x)^3} \\ \\$

Replacing the value of x, we have:

$K_c = \dfrac{4 \times 46,656 \times x^8}{(0.086-4x)^4\times (0.28 -3x)^3} \\ \\ K_c = \dfrac{4 \times 46656 \times (0.00433)^8}{(0.06868)^4(0.26701)^3} \\ \\ K_c = \mathbf{5.4446 \times 10^{-8}}$

$K_c = \mathbf{5.5 \times 10^{-8} \ to \ 2 \ significant \ figures}$

5 0

3 years ago

Which regions on the periodic table can adopt positive and negative oxidation numbers?

Dennis_Churaev [7]

<span>The region(s) of the periodic table which are made up of elements that can adopt both positive and negative oxidation numbers are the “non-metal” region. As we can see on the periodic table, the elements situated at the right side of the table have two oxidation states, one positive and the other a negative. </span>

7 0

3 years ago

Weight in grams of NaCl

Allisa [31]

Answer: 58.44g

Explanation: The molar mass of NaCl is 58.44g.

7 0

3 years ago

Read 2 more answers

What is the mass of 0.46 mol of MgCl2

Ronch [10]

Answer:

43.7

Explanation:

mole is equal mass concentration/ moler mass of MgCl²

8 0

3 years ago

Help help help :/ Please