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2 years ago

Why do you use 6.02*10^23 here?? #21???

Chemistry

2 answers:

almond37 [142]2 years ago

6 0

6.02*10^23 is Avagadro's number representing the number of molecules per mole of substance.

Sindrei [870]2 years ago

3 0

In order to convert from moles to molecules, you have to use Avogadro’s number, 6.022*10^23mol^-1.

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Answer: with? i can help:))

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2 years ago

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How much energy is needed to raise 70 g of paper 40 degrees celsius?

Natali5045456 [20]

Explanation: Heat capacity is the amount of heat required to change the temperature of a ... Therefore, specific heat is measured in Joules per g times degree Celsius

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2 years ago

If a 95.27 mL sample of acetic acid (HC2H3O2) is titrated to the equivalence point with 79.06 mL of 0.113 M KOH, what is the pH

KATRIN_1 [288]

Answer:

8.73

Explanation:

The concentration of acetic acid can be determined as follows:

$M_1V_1 = M_2V_2\\(KOH) = (CH_3COOH)$

$M_{KOH}=0.113 M\\V_{KOH}=79.06 mL$

$V_{CH_3COOH}=95.27 \\\\M_{CH_3COOH)=?????$

$M_{CH3COOH} = \frac{M_{KOH}*V_{KOH}}{V_{CH_3COOH}}$

$M_{CH3COOH} = \frac{0.113*79.06}{95.27}$

$M_{CH3COOH} = 0.094 M$

Moles of $CH_3COOH$ = $95.27* 10^{-3}* 0.094$

=0.0090 moles

Moles of $KOH = 79.06*10^{-3}*0.113$

= 0.0090 moles

The equation for the reaction can be expressed as :

$CH_3COOH$ $+$ $KOH$ -----> $CH_3COO^{-}K^+$ $+$ $H_2O$

Concentration of $CH_3COO^{-$ ion = $\frac{0.0090}{Total volume (L)}$

= $\frac{0.0090}{(95.27+79.06)} *1000$

= 0.052 M

Hydrolysis of $CH_3COO^{-$ ion:

$CH_3COO^{-$ $+$ $H_2O$ -----> $CH_3COOH$ $+$ $OH^-$

$K = \frac{K_w}{K_a} = \frac{x*x}{0.052-x}$

⇒ $\frac{10^{-14}}{1.82*10^{-5}}= \frac{x*x}{0.052-x}$

= $0.5494*10^{-9}= \frac{x*x}{0.052-x}$

As K is so less, then x appears to be a very infinitesimal small number

0.052-x ≅ x

$0.5494*10^{-9}= \frac{x^2}{0.052}$

$x^2 = 0.5494*10^{-9}*0.052$

$x^2 = 0.286*10^{-10$

$x = \sqrt{0.286*10^{-10$

$x =0.535*10^{-5}M$

$[OH] = x =0.535*10^{-5}$

$pOH = -log[OH^-]$

$pOH = -log[0.535*10^{-5}]$

$pOH = 5.27$

pH = 14 - pOH

pH = 14 - 5.27

pH = 8.73

Hence, the pH of the titration mixture = 8.73

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2 years ago

Which parameters can you change to affect the rate of the reaction? in alphabetical order, list all that apply to this system.?

postnew [5]

Catalyst
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Temperature

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3 years ago

The pKa of formic acid is 3.75. At pH of 5.00, _______ .

AlladinOne [14]

The pKa of formic acid is 3.75. At pH of 5.00, b) [formate] > [formic acid].

Formic acid is a weak acid. Thus, together with its conjugate base (formate) they form a buffer system. We can calculate the pH of a buffer system using Henderson-Hasselbach's equation.

$pH = pKa + log \frac{[formate]}{[formic\ acid]} \\\\5.00 = 3.75 + log \frac{[formate]}{[formic\ acid]}\\\\\frac{[formate]}{[formic\ acid]} = 17.8$

As expected, at a pH above the pKa, the concentration of formate is higher than that of the formic acid.

The pKa of formic acid is 3.75. At pH of 5.00, b) [formate] > [formic acid].

Learn more: brainly.com/question/22821585

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2 years ago