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Svetach [21]
3 years ago
12

A gas occupies a volume of 67.54 liters at 158°C and 4.87 atm pressure. Calculate the number of moles of this gas.

Chemistry
1 answer:
Katarina [22]3 years ago
6 0

Answer:

The number of moles of the gas is 9.295 moles or 9.30 moles

Explanation:

We use PV = nRT

Where P = 4.87 atm;

V = 67.54 L

R= 0.0821Latm/molK

T = 158 C = 158 +273 K = 431 K

the number of moles can be obtained by substituting the values in the respective columns and solve for n

n = PV / RT

n = 4.87 * 67.54 / 0.0821 * 431

n = 328.9198 / 35.3851

n = 9.295moles

The number of moles is approximately 9.30moles.

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I believe the answer is background radiatin
6 0
3 years ago
A 3.452 g sample containing an unknown amount of a Ce(IV) salt is dissolved in 250.0-mL of 1 M H2SO4. A 25.00 mL aliquot is anal
SOVA2 [1]

Answer:

1,812 wt%

Explanation:

The reactions for this titration are:

2Ce⁴⁺ + 3I⁻ → 2Ce³⁺ + I₃⁻

I₃⁻ + 2S₂O₃⁻ → 3I⁻ + S₄O₆²⁻

The moles in the end point of S₂O₃⁻ are:

0,01302L×0,03428M Na₂S₂O₃ = 4,463x10⁻⁴ moles of S₂O₃⁻. As 2 moles of S₂O₃⁻ react with 1 mole of I₃⁻, the moles of I₃⁻ are:

4,463x10⁻⁴ moles of S₂O₃⁻×\frac{1molI_{3}^-}{2molS_{2}O_{3}^-} = 2,2315x10⁻⁴ moles of I₃⁻

As 2 moles of Ce⁴⁺ produce 1 mole of I₃⁻, the moles of Ce⁴⁺ are:

2,2315x10⁻⁴ moles of I₃⁻×\frac{2molCe^{4+}}{1molI_{3}^-} = 4,463x10⁻⁴ moles of Ce(IV). These moles are:

4,463x10⁻⁴ moles of Ce(IV)×\frac{140,116g}{1mol} = <em>0,0625 g of Ce(IV)</em>

As the sample has a 3,452g, the weight percent is:

0,0625g of Ce(IV) / 3,452g × 100 = <em>1,812 wt%</em>

I hope it helps!

5 0
3 years ago
How much heat is absorbed when 63.7 g H2O(l) at 100 degrees Celsius and 101.3kPa is converted to steam at 100
Alona [7]

Answer:

Q = 143,921 J = 143.9 kJ.

Explanation:

Hello there!

In this case, according to the given information, it turns out possible for us to calculate the absorbed heat by considering this is a process involving sensible heat associated to the vaporization of water, which is isothermic and isobaric; and thus, the heat of vaporization of water, with a value of about 2259.36 J/g, is used as shown below:

Q=m*\Delta _{vap}H

Thus, we plug in the mass and the aforementioned heat of vaporization of water to obtain the following:

Q=63.7g*2259.36J/g\\\\Q=143,921J=143.9kJ

Regards!

3 0
3 years ago
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Which element in period 3 has three times the electronegativity value compared to that of li in period 2
victus00 [196]

Answer:

Nitrogen

Explanation:

Elements in period two includes lithium, beryllium, boron, carbon, nitrogen, oxygen, fluorine and neon.

According to periodic trends, the electro negativity values are expected to increase across the period up to fluorine. Hence, as we go right wards, we encounter elements with higher electronegative values.

While lithium has an electronegative value of 1 , the electronegative value of element nitrogen is thrrr times this which is equal to three

7 0
3 years ago
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