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2 years ago

What is the mass of 0.28 moles of iron

Chemistry

2 answers:

Xelga [282]2 years ago

4 0

1 mole has mass = 56 g

0.28 mol has mass = 56 x 0.28 = <u>15.68 g</u>

Hope this helps!

Greeley [361]2 years ago

4 0

<h2>Answer:</h2>

Given data:

Number of moles of iron = 0.28 M

Mass of iron = ?

Solution:

Formula : Mass of iron = Number of moles of iron * Molecular mass of iron

Putting values in above formula:

Mass of iron = 0.28 * 55.84 = 15.63

Mass of iron = 15.63

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Calculate the pressure of the CO₂ (g) in the container at 425 K.

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The pressure of the CO₂ = 0.995 atm

<h3>Further explanation</h3>

The complete question

<em>A student is doing experiments with CO2(g). Originally, a sample of gas is in a rigid container at 299K and 0.70 atm. The student increases the temperature of the CO2(g) in the container to 425K.</em>

<em>Calculate the pressure of the CO₂ (g) in the container at 425 K.</em>

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Gay Lussac's Law

When the volume is not changed, the gas pressure is proportional to its absolute temperature

$\tt \dfrac{P_1}{T_1}=\dfrac{P_2}{T_2}$

P₁=0.7 atm

T₁=299 K

T₂=425 K

$\tt P_2=\dfrac{P_1\times T_2}{T_1}\\\\P_2=\dfrac{0.7\times 425}{299}=0.995 `atm$

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"To determine the amount of heroin in the mixture, you dissolve 1.00 g of the white powdery mixture in water in a 100.0-mL volum

UkoKoshka [18]

Explanation:

Formula to calculate osmotic pressure is as follows.

Osmotic pressure = concentration × gas constant × temperature( in K)

Temperature = $25^{o} C$

= (25 + 273) K

= 298.15 K

Osmotic pressure = 531 mm Hg or 0.698 atm (as 1 mm Hg = 0.00131)

Putting the given values into the above formula as follows.

0.698 = $C \times 0.082 \times 298.15 K
$

C = 0.0285

This also means that,

$\frac{\text{moles}}{\text{volume (in L)}}$ = 0.0285

So, moles = 0.0285 × volume (in L)

= 0.0285 × 0.100

= $2.85 \times 10^{-3
}$

Now, let us assume that mass of $C_{12}H_{23}O_{5}N$ = x grams

And, mass of $C_{12}H{22}O_{11}$ = (1.00 - x)

So, moles of $C_{12}H_{23}O_{5}N = \frac{mass}{\text{molar mass}}$

= $\frac{x}{369}$

Now, moles of $C_{12}H_{22}O_{11} = \frac{(1.00 - x)}{342}$

= $\frac{x}{369} + \frac{(1.00 - x)}{342}$

= $2.85 \times 10^{-3}$

= x = 0.346

Therefore, we can conclude that amount of $C_{12}H_{23}O_{5}N$ present is 0.346 g and amount of $C_{12}H_{22}O_{11}$ present is (1 - 0.346) g = 0.654 g.