Answer:
[K₂CrO₄] → 8.1×10⁻⁵ M
Explanation:
First of all, you may know that if you dilute, molarity must decrease.
In the first solution we need to calculate the mmoles:
M = mmol/mL
mL . M = mmol
0.0027 mmol/mL . 3mL = 0.0081 mmoles
These mmoles of potassium chromate are in 3 mL but, it stays in 100 mL too.
New molarity is:
0.0081 mmoles / 100mL = 8.1×10⁻⁵ M
Answer:
A. 0.0440 moles/day
Explanation:
First, let's figure out how many moles 33.23 grams of silver is. We do this by dividing the number of grams by the molar mass of silver, which is 107.87 g/mol:
33.23 g Ag ÷ 107.87 g/mol = 0.3081 mol Ag
Now, let's divide this by 7 to get the rate per day:
0.3081 mol Ag ÷ 7 days = 0.0440 mol/day
Thus, the answer is A.
Answer:
a) pH = 4.213
b) % dis = 2 %
Explanation:
Ch3COONa → CH3COO- + Na+
CH3COOH ↔ CH3COO- + H3O+
∴ Ka = 1.8 E-5 = ([ CH3COO- ] * [ H3O+ ]) / [ CH3COOH ]
mass balance:
⇒ <em>C</em> CH3COOH + <em>C</em> CH3COONa = [ CH3COOH ] + [ CH3COO- ]
<em>∴ C </em>CH3COOH = 3.40 mM = 3.4 mmol/mL * ( mol/1000mmol)*(1000mL/L)
∴ <em>C</em> CH3COONa = 1.00 M = 1.00 mol/L = 1.00 mmol/mL
⇒ [ CH3COOH ] = 4.4 - [ CH3COO- ]
charge balance:
⇒ [ H3O+ ] + [ Na+ ] = [ CH3COO- ] + [ OH- ]....is negligible [ OH-], comes from water
⇒ [ CH3COO- ] = [ H3O+ ] + 1.00
⇒ Ka = (( [ H3O+ ] + 1 )* [ H3O+ ]) / ( 3.4 - [ H3O+])) = 1.8 E-5
⇒ [ H3O+ ]² + [ H3O+ ] = 6.12 E-5 - 1.8 E-5 [ H3O+ ]
⇒ [ H3O+ ]² + [ H3O+ ] - 6.12 E-5 = 0
⇒ [ H3O+ ] = 6.12 E-5 M
⇒ pH = - Log [ H3O+ ] = 4.213
b) (% dis)* mol acid = <em>C</em> CH3COOH = 3.4
∴ mol CH3COOH = 500*3.4 = 1700 mmol = 1.7 mol
⇒ % dis = 3.4 / 1.7 = 2 %
The charge would be -2 because there are 2 more electrons than protons, and electrons have a negative charge.
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