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yawa3891 [41]
3 years ago
13

What is the mass of oxygen in 10.0 g of water?

Chemistry
2 answers:
iris [78.8K]3 years ago
7 0
The answer is 16.00 amu.
natita [175]3 years ago
6 0

Answer:

The answer is 16.00 amu.

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Halons contain halogens, which are highly reactive with oxygen. true or false
Natalija [7]

Answer:

true

Explanation:

6 0
3 years ago
Read 2 more answers
The conversion of cyclopropane to propene occurs with a first-order rate constant of . How long will it take for the concentrati
Gennadij [26K]

This is an incomplete question, here is a complete question.

The conversion of cyclopropane to propene occurs with a first-order rate constant of 2.42 × 10⁻² hr⁻¹. How long will it take for the concentration of cyclopropane to decrease from an initial concentration 0.080 mol/L to 0.053 mol/L?

Answer : The time taken will be, 17.0 hr

Explanation :

Expression for rate law for first order kinetics is given by:

t=\frac{2.303}{k}\log\frac{a}{a-x}

where,

k = rate constant  = 2.42\times 10^{-2}\text{ hr}^{-1}

t = time passed by the sample  = ?

a = initial concentration of the reactant  = 0.080 M

a - x = concentration left = 0.053 M

Now put all the given values in above equation, we get

t=\frac{2.303}{2.42\times 10^{-2}}\log\frac{0.080}{0.053}

t=17.0\text{ hr}

Therefore, the time taken will be, 17.0 hr

5 0
3 years ago
A reaction where Oxygen is a reactant.
Goryan [66]

Answer:

chemical

Explanation:

6 0
3 years ago
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The combustion of 0.1240 kg of propane in the presence of excess oxygen produces 0.3110 kg of carbon dioxide. What is the limiti
nadezda [96]

Answer:

The limiting reactant is the propane gas, C₃H₈ while the percentage yield is 83.77%

Explanation:

Here we have

Propane gas with molecular formula C₃H₈, molar mass  = 44.1 g/mol combining with O₂ as follows

C₃H₈ + 5O₂ → 3CO₂ + 4H₂O

Therefore, 1 mole of C₃H₈  combines with 5 moles of O₂ to produce 3 moles CO₂ and 4 moles of H₂O

Mass of propane = 0.1240 kg = 124.0 g

Number of moles of propane = mass of propane/(molar mass of propane)

The number of moles of propane = 124/44.1 = 2.812 moles

The molar mass of CO₂ = 44.01 g/mol

Mass of CO₂ = 0.3110 kg = 311.0 g

Therefore, number of moles of CO₂ = mass of CO₂/(molar mass of CO₂)

The number of moles of CO₂ = 311.0 kg/ 44.01 g/mol = 7.067 moles

Therefore, since 1 mole of propane produces 3 moles of CO₂, 2.812 moles of propane will produce 3 × 2.812 moles or 8.44 moles of CO₂

Therefore;

The limiting reactant is the propane gas, C₃H₈, since the oxygen is in excess

Hence

The \ percentage \ yield = \frac{Actual \, yield}{Theoretical \, yield} \times 100 = \frac{7.067}{8.44} \times 100 = 83.77 \%

The percentage yield = 83.77%.

7 0
3 years ago
A 6.0 L container of gas is 2.5atm. What would be the pressure if the volume is 12.0L
Scorpion4ik [409]
This works because it demonstrates that as volume increases, pressure decreases (inverse relationship)

4 0
3 years ago
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