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3 years ago

How many moles of water are produced if 5.43 mol PbO2 are consumed? ANSWER: 10.9

Chemistry

2 answers:

Arada [10]3 years ago

4 0

Answer: 10.9 mol.

Explanation:

To understand how to solve this problem, we must mention the reaction equation where water produced from PbO₂.

Pb + PbO₂ + 2H₂SO₄ → 2PbSO₄ + 2H₂O

Now, it is a stichiometric oriented problem, that 1 mole of PbO₂ produces 2 moles of H₂O.

Using cross multiplication:

1.0 mole of PbO₂ → 2.0 moles of H₂O

5.43 moles of PbO₂ → ??? moles of water

The moles of water produced = (5.43 x 2.0) = 10.86 moles ≅ 10.9 moles.

Scilla [17]3 years ago

3 0

Answer:

10.9

Explanation:

5.43 x $\frac{2}{1}$ = 10.86 which rounds to 10.9

just did it on edge :)

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True or false

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Answer:

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Explanation:

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How many particles are in 47.7 g of Magnesium? (Round the average

Lynna [10]

Answer:

1.18 × 10²⁴ particles Mg

General Formulas and Concepts:

Chemistry - Atomic Structure

Reading a Periodic Table
Using Dimensional Analysis
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Explanation:

Step 1: Define

47.7 g Mg

Step 2: Identify Conversions

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Molar Mass of Mg - 24.31 g/mol

Step 3: Convert

$47.7 \ g \ Mg(\frac{1 \ mol \ Mg}{24.31 \ g \ Mg} )(\frac{6.022 \cdot 10^{23} \ particles \ Mg}{1 \ mol \ Mg} )$ = 1.18161 × 10²⁴ particles Mg

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2 years ago

The volume of a gas at 2.0 atm is 3.0 L. What is the volume of the gas at 1.5 atm at the same temperature?

Andru [333]

Answer: The final volume of the gas comes out to be 4 L.

Explanation:

To calculate the volume with changing pressure, we use the equation given by Boyle's law.

This law states that pressure is inversely proportional to the volume of the gas at constant temperature and number of moles.

Mathematically,

$P\propto \frac{1}{V}$ (At constant temperature and number of moles)

The equation given by this law is:

$P_1V_1=P_2V_2$

where,

$P_1\text{ and }V_1$ are initial pressure and volume.

$P_2\text{ and }V_2$ are final pressure and volume.

We are given:

$P_1=2.0atm\\V_1=3L\\P_2=1.5atm\\V_2=?L$

Putting values in above equation, we get:

$2atm\times 3L=1.5atm\times V_2\\\\V_2=4L$

Hence, the final volume of the gas will be 4 L.

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3 years ago

Read 2 more answers

Phosphine, an extremely poisonous and highly reactive gas, will react with oxygen to form tetraphosphorus decoxide and water, as

erica [24]

Answer: The amount of $P_4O_{10}$ formed is 469.8 grams.

Explanation:

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ ......(1)

Given mass of phosphine = 225 g

Molar mass of phosphine = 34 g/mol

Putting values in equation 1, we get:

$\text{Moles of phosphine}=\frac{225g}{34g/mol}=6.62mol$

The given chemical reaction follows:

$4PH_3(g)+8O_2(g)\rightarrow P_4O_{10}(s)+6H_2O(g)$

Assuming that oxygen gas is present in excess, it is considered as an excess reagent.

Phosphine is considered as a limiting reagent because it limits the formation of product.

By Stoichiometry of the reaction:

4 moles of phosphine produces 1 mole of $P_4O_{10}$

So, 6.62 moles of phosphine will produce = $\frac{1}{4}\times 6.62=1.655mol$ of $P_4O_{10}$

Now, calculating the mass of $P_4O_{10}$ by using equation 1:

Molar mass of $P_4O_{10}$ = 283.9 g/mol

Moles of $P_4O_{10}$ = 1.655 moles

Putting values in equation 1, we get:

$1.655mol=\frac{\text{Mass of }P_4O_{10}}{283.9g/mol}\\\\\text{Mass of }P_4O_{10}=(1.655mol\times 283.9g/mol)=469.8g$

Hence, the amount of $P_4O_{10}$ formed is 469.8 grams.

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3 years ago