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3 years ago

If you burn 29.4 g of hydrogen and produce 263 g of water, how much oxygen reacted?

Chemistry

1 answer:

vovikov84 [41]3 years ago

7 0

The reaction between hydrogen and oxygen to form water is given as:

$H_2 + O_2 \rightarrow H_2O$

The balanced reaction is:

$2H_2 + O_2 \rightarrow 2H_2O$

According to the balanced reaction,

4 g of hydrogen ( $4\times 1$ ) reacts with 32 g of oxygen ( $2\times 16$ ).

So, oxygen reacted with 29.4 g of hydrogen is:

$\frac{29.4\times 32}{4} = 235.2 g$

Hence, the mass of oxygen that is reacted with 29.4 g of hydrogen is 235.2 g.

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A gas mixture with a total pressure of 745 mmHg contains each of the following gases at the indicated partial pressures: CO2, 24

Setler79 [48]

Answer:

For Part A: The partial pressure of Helium is 218 mmHg.

For Part B: The mass of helium gas is 0.504 g.

Explanation:

For Part A:

We are given:

$p_{CO_2}=245mmHg\\p_Ar}=119mmHg\\p_{O_2}=163mmHg\\P=745mmHg$

To calculate the partial pressure of helium, we use the formula:

$P=p_{CO_2}+p_{Ar}+p_{O_2}+p_{He}$

Putting values in above equation, we get:

$745=245+119+163+p_{He}\\p_{He}=218mmHg$

Hence, the partial pressure of Helium is 218 mmHg.

For Part B:

To calculate the mass of helium gas, we use the equation given by ideal gas:

PV = nRT

or,

$PV=\frac{m}{M}RT$

where,

P = Pressure of helium gas = 218 mmHg

V = Volume of the helium gas = 10.2 L

m = Mass of helium gas = ? g

M = Molar mass of helium gas = 4 g/mol

R = Gas constant = $62.3637\text{ L.mmHg }mol^{-1}K^{-1}$

T = Temperature of helium gas = 283 K

Putting values in above equation, we get:

$218mmHg\times 10.2L=\frac{m}{4g/mol}\times 62.3637\text{ L.mmHg }mol^{-1}K^{-1}\times 283K\\\\m=0.504g$

Hence, the mass of helium gas is 0.504 g.

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3 years ago

Which best describes the transfer of heat energy through conduction? A. Heat energy is transferred from molecule to molecule by

TiliK225 [7]

It should be option (A) because conduction is the transfer of heat energy from one place to another by vibrations/motion of the molecules. I hoped I helped.

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? Answer the question below. Type your response in the space provided. What volume of a 2.5 M stock solution of acetic acid (HC2

Novay_Z [31]

Data:
$M_{concentrated} = 2.5\:mol$
$V_{concentrated} = ?$
$M_{dilute} = 0.50\:mol$
$V_{dilute} = 100\:mL\to0.100\:L$

Formula: Dilution Calculations

 $M_{concentrated} * V_{concentrated} = M_{dilute} * V_{dilute}$

Solving:


$M_{concentrated} * V_{concentrated} = M_{dilute} * V_{dilute}$
$2.5 * V_{concentrated} = 0.50 * 0.100$
$2.5V_{concentrated} = 0.05$
$V_{concentrated} = \frac{0.05}{2.5}$
$\boxed{\boxed{V_{concentrated} = 0.02\:L\:or\:20\:mL}} \end{array}}\qquad\quad\checkmark$

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