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Minchanka [31]
3 years ago
7

PbCl2s) + Cl2g - + PbCl40) ΔΗ = ?

Chemistry
1 answer:
ludmilkaskok [199]3 years ago
4 0

Answer:

ichedgepeth

Explanation:

beginner because it's tough than other things

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The half-life of carbon-14 is about ____________ years.<br><br> Fill in the blank.
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The half-life of carbon-14 is about 5730 years
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What is mass per unit volume called?
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Answer:

The correct answer is Density

Explanation:

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3 0
3 years ago
By pipet, 11.00 mL of a 0.823 MM stock solution of potassium permanganate (KMnO4) was transferred to a 50.00-mL volumetric flask
tangare [24]

Answer:

1) 0.18106 M is the molarity of the resulting solution.

2) 0.823 Molar is the molarity of the solution.

Explanation:

1) Volume of stock solution = V_1=11.00 mL

Concentration of stock solution = M_1=0.823 M

Volume of stock solution after dilution = V_2=50.00 mL

Concentration of stock solution after dilution = M_2=?

M_1V_1=M_2V_2 ( dilution )

M_2=\frac{0.823 M\times 11.00 mL}{50 ,00 mL}=0.18106 M

0.18106 M is the molarity of the resulting solution.

2)

Molarity of the solution is the moles of compound in 1 Liter solutions.

Molarity=\frac{\text{Mass of compound}}{\text{Molar mas of compound}\times Volume (L)}

Mass of potassium permanganate = 13.0 g

Molar mass of potassium permangante = 158 g/mol

Volume of the solution = 100.00 mL = 0.100  L ( 1 mL=0.001 L)

Molarity=\frac{13.0 g}{158 g/mol\times 0.100 L}=0.823 mol/L

0.823 Molar is the molarity of the solution.

6 0
2 years ago
What are transferred in an oxidation-reduction reaction (1 point)?
madreJ [45]
electrons are transferred    in a oxidation-reduction  reaction

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3 years ago
Estimate the molar mass of a gas that effuses at 1.6 times the effusion rate of CO2
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To estimate the molar mass of the gas, we use Graham's law of effusion. This relates the rates of effusion of gases with their molar mass. We calculate as follows:

r1/r2 = √(m2/m1)    

where r1 would be the effusion rate of the gas and r2 is for CO2, M1 is the molar mass of the gas and M2 would be the molar mass of CO2 (44.01 g/mol) 

r1 = 1.6r2

1.6 = √(44.01 / m1)
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