Why was the morning session stopped unsuccessfully?

How many hydrogen atoms are needed to form five water molecules

BARSIC [14]

2 Hydrogen Atoms Are Needed To Form Five Water Molecules

8 0

3 years ago

Oxygen gas generated in the thermal decomposition of potassium chlorate is collected over water. The volume of gas collected is

Leya [2.2K]

The mass of oxygen collected from the thermal decomposition of potassium chlorate at a temperature of 297 K and 762 mmHg is 0.16 g

<h3>How to determine the mole of oxygen produced </h3>

We'll begin by obtaining the number of mole of oxygen gas produced from the reaction. This can be obtained by using the ideal gas equation as illustrated below:

Volume (V) = 0.128 L
Temperature (T) = 297 K
Pressure (P) = 762 – 22.4 = 739.6 mmHg
Gas constant (R) = 62.363 mmHg.L/Kmol
Number of mole (n) =?

PV = nRT

739.6 × 0.128 = n × 62.363 × 297

Divide both sides by 62.363 × 297

n = (739.6 × 0.128) / (62.363 × 297)

n = 0.0051 mole

Thus, the number of mole of oxygen gas produced is 0.0051 mole

<h3>How to determine the mass of oxygen collected</h3>

Haven obtain the number of mole of oxygen gas produced, we can determine the mass of the oxygen produced as follow:'

Mole = 0.0051 mole
Molar mass of oxygen gas = 32 g/mole
Mass of oxygen =?

Mole = mass / molar mass

0.0051 = mass of oxygen / 32

Cross multiply

Mass of oxygen = 0.0051 × 32

Mass of oxygen = 0.16 g

Thus, we can conclude that the mass of oxygen gas collected is 0.16 g

Learn more about ideal gas equation:

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5 0

1 year ago

How many molecules are in 97.21 grams of Sodium Chloride (NaCl)?

professor190 [17]

1.66 is the answer because it’s is

6 0

3 years ago

Saftey goggles are worn in the lab-

Harrizon [31]

Answer:

Yes

Explanation:

Is this a question or what?

4 0

3 years ago

The theoretical yield of a reaction is the amount of product obtained if the limiting reactant is completely converted to produc

Elis [28]

Answer: 9.9 grams

Explanation:

To calculate the moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text {Molar mass}}$

a) moles of $H_2$

$\text{Number of moles}=\frac{8.150g}{2g/mol}=4.08moles$

b) moles of $C_2H_4$

$\text{Number of moles}=\frac{9.330g}{28g/mol}=0.33moles$

$H_2(g)+C_2H_4(g)\rightarrow C_2H_6(g)$

According to stoichiometry :

1 mole of $C_2H_4$ combine with 1 mole of $H_2$

Thus 0.33 mole of $C_2H_4$ will combine with = $\frac{1}{1}\times 0.33=0.33$ mole of $H_2$

Thus $C_2H_4$ is the limiting reagent as it limits the formation of product.

As 1 mole of $C_2H_4$ give = 1 mole of $C_2H_6$

Thus 0.33 moles of $C_2H_4$ give = $\frac{1}{1}\times 0.33=0.33moles$ of $C_2H_6$

Mass of $C_2H_6=moles\times {\text {Molar mass}}=0.33moles\times 30g/mol=9.9g$

Thus theoretical yield (g) of $C_2H_6$ produced by the reaction is 9.9 grams

7 0

3 years ago