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Ronch [10]
3 years ago
14

What is the pH of a solution with a concentration of 5.2 × 10–8 M H3O+?

Chemistry
1 answer:
lora16 [44]3 years ago
7 0

Answer : The pH of a solution is, 7.28

Solution : Given,

Concentration of hydronium ion, H_3O^+ = 5.2\times 10^{-8}M

pH : It is defined as the negative logarithm of hydronium ion concentration or hydrogen ion concentration.

pH=-\log [H_3O^+]

Now put the value of hydronium ion concentration in this expression, we get the pH of the solution.

pH=-\log (5.2\times 10^{-8})=7.28

Therefore, the pH of a solution is, 7.28

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Determine the enthalpy change for the decomposition of calcium carbonate. CaCO₃(s) --> CaO(s) + CO₂(g) given the thermochemic
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Answer: c. 179 kJ/mol

Explanation:

According to Hess’s law of constant heat summation, the heat absorbed or evolved in a given chemical equation is the same whether the process occurs in one step or several steps.

According to Hess’s law, the chemical equation can be treated as algebraic expressions and can be added or subtracted to yield the required equation. That means the enthalpy change of the overall reaction is the sum of the enthalpy changes of the intermediate reactions.

Given:

Ca(OH)_2(s)\rightarrow CaO(s)+H_2O (l)   \Delta H_1= 65.2 kJ/mol     (1)

Ca(OH)_2(s)+CO_2(g)\rightarrow CaCO_3(s)+H_2O(l)    \Delta H_2= -113.8 kJ/mol   (2)

C(s)+O_2(g)\rightarrow CO_2(g)    \Delta H_3= -393.5 kJ/mol   (3)

2Ca(s)+O_2(g)\rightarrow 2CaO(s) \Delta H_4=-1270.2 kJ/mol    (4)

On subtracting eq (1) from eq (2) we have:

Ca(OH)_2(s)\rightarrow CaO(s)+H_2O(l) - Ca(OH)_2(s)+CO_2(g)\rightarrow CaCO_3(s)+H_2O(l)

CaCO_3(s)\rightarrow CaO(s)+CO_2(g)

\Delta H=-113.8-(65.2)kJ/mol=-179kJ/mol

Hence the enthalpy change for the raection is 179.0 kJ/mol.

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Ferrous oxide or Iron (II) oxide. Brainliest? if its correct


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A sample of sulfur hexafluoride gas occupies a volume of 5.10 L at 198 ºC. Assuming that the pressure remains constant, what tem
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Answer:

When the volume will be reduced to 2.50 L, the temperature will be reduced to a temperature of 230.9K

Explanation:

Step 1: Data given

A sample of sulfur hexafluoride gas occupies a volume of 5.10 L

Temperature = 198 °C = 471 K

The volume will be reduced to 2.50 L

Step 2 Calculate the new temperature via Charles' law

V1/T2 = V2/T2

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⇒with V2 = the reduced volume of the gas = 2.50 L

⇒with T2 = the new temperature = TO BE DETERMINED

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When the volume will be reduced to 2.50 L, the temperature will be reduced to a temperature of 230.9K

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