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PolarNik [594]
3 years ago
14

Help me with this please

Chemistry
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

6 0
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
CaHeK987 [17]
(3) 16, which is Sulfer.
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3 years ago
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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.

6 0
3 years ago
The most concentrated solution is
aev [14]
The most concentrated solution is b
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3 years ago
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