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4 years ago

The chemical equation below shows the combustion of methane (CH4).

Chemistry

2 answers:

Hitman42 [59]4 years ago

8 0

Answer:

Explanation:

edg 2020

STatiana [176]4 years ago

6 0

The correct answer is 6.15g of CO2 will be produced.

The approach to answer this question is to use the molar mass to find the number of moles of oxygen consumed. Then use the chemical equation to determine the number of moles of CO2 that would be produced (half that of oxygen). Then use the molar mass of CO2 to calculate the mass.

Mmass O2 = 32g/mol
Mmass CO2 = 44.01g/mol
mass O2 consumed = 8.94g
mass CO2 produced = ?

moles O2 = mass / molar mass = 8.94 / 32 = 0.279
moles CO2 = half that of O2 = 0.1397

mass CO2 produced = moles CO2 x molar mass = 0.1397 x 44.01 = 6.15 g

:)

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A buffer contains 0.18 mol of propionic acid (C2H5COOH) and 0.26 mol of sodium propionate (C2H5COONa) in 1.20 L. What is the pH

Yuki888 [10]

Answer:

1) pH = 5.05

2) pH = 5.13

3) pH = 4.97

Explanation:

Step 1: Data given

Number of moles of propionic acid = 0.18 moles

Number of moles sodium propionate = 0.26 moles

Volume = 1.20 L

Ka = 1.3 * 10^-5 → pKa = 4.989

Step 2: Calculate concentrations

Concentration = moles / volume

[acid]= 0.18/ 1.2 =0.150 M

[salt]= 0.26/ 1.3 = 0.217 M

pH = 4.89 + log(0.217/0.150)=5.05

What is the pH of the buffer after the addition of 0.02 mol of NaOH?

moles acid = 0.18 - 0.02 = 0.16

[acid]= 0.16/ 1.2=0.133 M

moles salt = 0.26 + 0.02 = 0.28

[salt]= 0.28/ 12=0.233

pH = 4.89 + log 0.233/ 0.133 = 5.13

What is the pH of the buffer after the addition of 0.02 mol of HI?

moles acid = 0.18+ 0.02 = 0.20 moles

[acid]= 0.20/ 1.2 = 0.167 M

[salt]= 0.26 - 0.02= 0.24 moles

[salt]= 0.24/ 1.2 = 0.20 M

pH = 4.89 + log 0.20/ 0.167= 4.97

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