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iris [78.8K]
3 years ago
14

How would the addition of protons affect the concentration of CH3COOH? How would the addition of OH– affect the amount of CH3COO

H present? How would the addition of CH3COO– affect the concentration of protons? What would happen to [H+] if [CH3COOH] were increased?
Chemistry
2 answers:
Deffense [45]3 years ago
8 0

Answer:

1) increase concentration

2) decrease the amount

3) decrease the concentration

4) it would increase

Explanation: edge 2021

fiasKO [112]3 years ago
7 0

Answer:

1) increase concentration

2) decrease the amount

3) decrease the concentration

4) it would increase

Explanation:

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13.50 grams of Pb(NO3)4 are dissolved in enough water to make 250 mL of solution. What is the molar its of the resulting solutio
likoan [24]
Data:
M (molarity) = ? (M or Mol/L)
m (mass) = 13.50 g
V (volume) = 250 mL → 0.25 L
MM (Molar Mass) of Lead(IV) Nitrate Pb(NO_3)_4
Pb = 1*207 = 207 amu
N = (1*14)*4 = 14*4 = 56 amu
O = (3*16)*4 = 48*4 = 192 amu
------------------------------------
MM of Pb(NO_3)_4 = 207+56+192 = 455 g/mol

Formula:
M =  \frac{m}{MM*V}

Solving:
M = \frac{m}{MM*V}
M =  \frac{13.50}{455*0.25}
M =  \frac{13.50}{113.75}
M = 0.118681318...\:\:\to\:\:\boxed{\boxed{M \approx 0.119\:Mol/L}}\end{array}}\qquad\quad\checkmark

Answer:
<span>B. 0.119 M</span>
5 0
3 years ago
True or false:
monitta
I think this maybe false
3 0
3 years ago
Label the parts of the atom.<br><br> Which atomic particle determines the identity of the atom?
zavuch27 [327]
A: Electron
B: Neutron
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3 0
3 years ago
Concentrated hydrochloric acid, HCl, comes with an approximate molar concentration of 12.1 M. If you are instructed to prepare 3
Semmy [17]

Answer:

28.20 mL of the stock solution.

Explanation:

Data obtained from the question include the following:

Molarity of stock solution (M1) = 12.1 M

Volume of diluted solution (V2) = 350.0 mL

Molarity of diluted solution (M2) = 0.975 M

Volume of stock solution needed (V1) =..?

The volume of stock solution needed can be obtained by using the dilution formula as shown below:

M1V1 = M2V2

12.1 x V1 = 0.975 x 350

Divide both side by 12.1

V1 = (0.975 x 350)/12.1

V1 = 28.20 mL.

Therefore, 28.20 mL of the stock solution will be needed to prepare 350.0 mL of 0.975 M HCl solution.

6 0
3 years ago
What is the percent by mass of carbon in acetone, c 3 h 6 o?
Tatiana [17]
In the problem, we are tasked to solved for the amount of carbon (C) in the acetone having a molecular formula of C 3 H 6 O. We need to find first the molecular weight if Carbon (C), Hydrogen (H), Oxygen (O).

Molecular Weight:
C=12 g/mol
H=1 g/mol
O=16 g/mol

To calculate for the percent by mass of acetone, we assume 1 mol of acetone.
%C=( \frac{3(12 g/mol)}{3(12 g/mol)+6(1 g/mol)+16 g/mol} ) x 100%
%C=62.07%
Therefore, the percent by mass of carbon in acetone is 62.07%
5 0
2 years ago
Read 2 more answers
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