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GenaCL600 [577]
3 years ago
12

Calculate the number of moles of an ideal gas if it occupies 1750 dm3 under 125,000 Pa at a temperature of 127 C.

Chemistry
2 answers:
andrey2020 [161]3 years ago
6 0

Answer:

65.81

Explanation:

Educere/ Founder's Education Answer

Phoenix [80]3 years ago
5 0

Hey there!


* Converts 1750 dm³ in liters :


1 dm³ = 1 L so 1750 dm³ = 1750 liters



* Convertes 125,000 Pa in atm :


1 Pa = 9.86*10⁻⁶ atm so 9.86*10⁻⁶ / 125,000 => 1.233 atm


* Convertes 127ºC in K :


127 + 273.15 => 400.15 K


R = 0.082 atm.L/mol.K


Finally, it uses an equation of clapeyron :


p * V = n * R * T


1.233 * 1750 = n * 0.082 * 400.15


2157.75 = n * 32.8123


n = 2157.75 / 32.8123


n = 65.76 moles



hope this helps!



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1) What is the mass of 6.2 mol of K2CO3?
lorasvet [3.4K]

Answer:

\boxed {\boxed {\sf 856.8648 \ grams \ of \ K_2CO_3}}

Explanation:

To convert from moles to grams, we must find the molar mass.

1. Molar Mass

First, identify the elements in the compound. K₂CO₃ It has potassium, carbon, and oxygen. Find these elements and their masses on the Periodic Table.

  • K: 39.098 g/mol
  • C: 12.011 g/mol
  • O: 15.999 g/mol

Note the subscript of 2 after K and 3 after O. We must multiply oxygen's molar mass by 2, then oxygen's by 3, and add carbon.

  • 2(39.098 g/mol) + 3(15.999 g/mol) + 12.011 g/mol= 138.204 g/mol

2. Convert Moles to Grams

Use the molar mass as a fraction.

\frac {138.204 \ g \ K_2CO_3}{1 \ mol \ K_23CO_3}

Multiply by the given number of moles: 6.2

6.2 \ mol \ K_2CO_3 *\frac {138.204 \ g \ K_2CO_3}{1 \ mol \ K_23CO_3}

6.2  *\frac {138.204 \ g \ K_2CO_3}{1 }

6.2  * {138.204 \ g \ K_2CO_3}

856.8648 \ g \ K_2CO_3

There are <u>856.8648 grams</u> of potassium carbonate in 6.2 moles.

4 0
3 years ago
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