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quester [9]
3 years ago
10

Need someone to answer this question as soon as possible

Chemistry
1 answer:
denpristay [2]3 years ago
8 0
Student C: The total mass before and after the reaction is the same.
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What is the pH when 10.0 mL of 0.20 M potassium hydroxide is added to 30.0 mL of 0.10 M cinnamic acid, HC9H7O2 (Ka = 3.6 × 10–5)
astra-53 [7]

Answer:-

Solution:- As is clear from the given Ka value, Cinnamic acid is a weak acid. let's calculate the moles of acid and KOH added to it from their given molarities and mL.

For KOH,  10.0mL(\frac{1L}{1000mL})(\frac{0.20mol}{1L})

= 0.002 mol

For Cinnamic acid,  30.0mL(\frac{1L}{1000mL})(\frac{0.10mol}{1L})

= 0.003 mol

Acid and base react as:

HC_9H_7O_2(aq)+KOH(aq)\rightleftharpoons KC_9H_7O_2(aq)+H_2O(l)

The reaction takes place in 1:1 mol ratio. Since the moles of acid are in excess, the acid is still remaining when all the kOH is used.

0.002 moles of KOH react with 0.002 moles of Cinnamic acid to form 0.002 moles of potassium cinnamate. Excess moles of Cinnamic acid = 0.003 - 0.002 = 0.001

As the solution have weak acid and it's salt(or we could say conjugate base), it is a buffer solution and the pH of the buffer solution could easily be calculated using Handerson equation:

pH=pKa+log(\frac{base}{acid})

pKa could be caluted from given Ka value using the formula:

pKa = - log Ka

pKa=-log3.6*10^-^5

pKa = 4.44

let's plug in the values in Handerson equation and calculate the pH:

pH=4.44+log(\frac{0.002}{0.001})

pH = 4.44+0.30

pH = 4.74

So, the first choice is correct, pH is 4.74.

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Mercury(II) oxide (HgO) decomposes to form mercury (Hg) and oxygen (O2). The balanced chemical equation is shown below.2HgO Righ
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Answer: 15.62 moles of HgO are needed to produce 250.0 g of O_2

Explanation:

To calculate the moles :

\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}    

\text{Moles of} O_2=\frac{250.0g}{32g/mol}=7.812moles

2HgO\rightarrow 2Hg+O_2

According to stoichiometry :

1 mole of O_2 is produced from decomposition of =  2 moles of HgO

Thus 7.812 moles of O_2 will be produced from decomposition of  =\frac{2}{1}\times 7.812=15.62moles of HgO

Thus 15.62 moles of HgO are needed to produce 250.0 g of O_2

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