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3 years ago

14

Help me with this please

1 answer:

Doss [256]3 years ago

6 0

= 0.2+560.66JK-2eF

use this app : Photomath

It will help you :)

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What is the charge on an atom after it gains two electrons during the formation of a bond?

scoundrel [369]

Answer:

negative or neutral

Explanation:

electrons have a negative charge

the atom can be balanced by the negative charge if it was at two protons before

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4 years ago

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At STP, an element that is a brittle solid and a poor conductor of heat and electricity could have an atomic number of

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(3) 16, which is Sulfer.

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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

8 0

3 years ago

Calculate the molarity of a 10.0% (by mass) aqueous solution of hydrochloric acid.

olga2289 [7]

The question is incomplete,the complete question :

Calculate the molality of a 10.0% (by mass) aqueous solution of hydrochloric acid:

a) 0.274 m

b) 2.74 m

c) 3.05 m

d) 4.33 m

e) the density of the solution is needed to solve the problem

Answer:

The molality of a 10.0% (by mass) aqueous solution of hydrochloric acid is 3.05 mol/kg.

Explanation:

10.0% (by mass) aqueous solution of hydrochloric acid.

10 grams of HCl is present in 100 g of solution.

Mass of HCl = 10 g

Mass of solution = 100 g

Mass of solution = Mass of solute + Mass of water

Mass of water = 100 g - 10 g = 90 g

Moles of HCl = $\frac{10 g}{36.5 g/mol}=0.2740 mol$

Mass of water in kilograms = 0.090 kg

Molality = $\frac{0.2740 mol}{0.090 kg}=3.05 mol/kg$

The molality of a 10.0% (by mass) aqueous solution of hydrochloric acid is 3.05 mol/kg.

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3 years ago

The most concentrated solution is

aev [14]

The most concentrated solution is b

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