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2 years ago

Hich formula can be used to calculate the molar mass of hydrogen peroxide (H2O2)? (5 points)

Chemistry

2 answers:

Alexxandr [17]2 years ago

6 0

Answer:

Answers are 34.6 grams and objective d.

mina [271]2 years ago

4 0

Answer:

which formula can be used to calculate the molar mass of hydrogen peroxide (H2O2)?

» d.2 x molar mass of H + 2 x molar mass of O

Explanation:

Since we've two hydrogen atoms and two oxygen atoms, we multiply each mass by 2

$= (2 \times hydrogen) + (2 \times oxygen) \\ = (2 \times 1) + (2 \times 16) \\ = 2 + 32 \\ = 34 \: g$

Which of the following statements best defines the actual yield of a reaction?

» d.The ratio of measured yield over stoichiometric yield

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A rigid tank contains 0.66 mol of oxygen (O2). Find the mass of oxygen that must be withdrawn from the tank to lower the pressur

dsp73

Answer:

12.8 g of $O_{2}$ must be withdrawn from tank

Explanation:

Let's assume $O_{2}$ gas inside tank behaves ideally.

According to ideal gas equation- $PV=nRT$

where P is pressure of $O_{2}$ , V is volume of $O_{2}$ , n is number of moles of $O_{2}$ , R is gas constant and T is temperature in kelvin scale.

We can also write, $\frac{V}{RT}=\frac{n}{P}$

Here V, T and R are constants.

So, $\frac{n}{P}$ ratio will also be constant before and after removal of $O_{2}$ from tank

Hence, $\frac{n_{before}}{P_{before}}=\frac{n_{after}}{P_{after}}$

Here, $\frac{n_{before}}{P_{before}}=\frac{0.66mol}{43atm}$ and $P_{after}=17atm$

So, $n_{after}=\frac{n_{before}}{P_{before}}\times P_{after}=\frac{0.66mol}{43atm}\times 17atm=0.26mol$

So, moles of $O_{2}$ must be withdrawn = (0.66 - 0.26) mol = 0.40 mol

Molar mass of $O_{2}$ = 32 g/mol

So, mass of $O_{2}$ must be withdrawn = $(32\times 0.40)g=12.8g$

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melamori03 [73]

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Explanation:

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Answer:

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Explanation:

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