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Kobotan [32]
3 years ago
12

At what temperature, in Kelvin, will a 250 milliliter cylinder containing 0.100 moles of helium gas have a pressure of 2.50 atmo

sphere? Write your answer as a decimal number to three significant figures. Given: R = 0.08205 liter · atmosphere/mole · Kelvin
Chemistry
2 answers:
lozanna [386]3 years ago
7 0
Use the ideal gas formula-----> PV= nRT

P= 2.50 atm
V= 250 mL= 0.250 L
n= 0.100 moles
R= 0.0821 atmxL/molesxK
T= ?

T= PV/nR

T= (2.50 atm x 0.250 L) / (0.100 moles x 0.0821)= 76.1 K
Gnom [1K]3 years ago
4 0

The correct answer is 72.6

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450g of chromium(iii) sulfate reacts with excess potassium phosphate. How many grams of potassium sulfate will be produced? (ANS
ExtremeBDS [4]

Answer:

1597.959 g  

Explanation:

Given Data:

Amount of Cr₂(SO₄)₃ = 450 g

Amount of potassium phosphate K₃PO₄ = in Excess

grams of potassium sulfate K₂SO₄= ?

Solution

The Reaction will be

                 Cr₂(SO₄)₃ + 2K₃PO₄  ----------> 3K₂SO₄ + 2CrPO₄

Information that we have from reaction

                Cr₂(SO₄)₃ + 2K₃PO₄  ----------> 3K₂SO₄ + 2CrPO₄

                    1 mol          2 mol                   3 mol

we come to know from the above reaction that

1 mole of chromium(iii) sulfate (Cr₂(SO₄)₃) react with 2 mole of potassium phosphate (K₃PO₄) to produce 3 mole of K₂SO₄

We also know that

molar mass of Cr₂(SO₄)₃ = 147 g/mol

molar mass of K₃PO₄ = 212 g/mol

molar mass of K₂SO₄ = 174 g/mol

if we represent mole in grams then

      Cr₂(SO₄)₃             +       2K₃PO₄          ----------> 3K₂SO₄ + 2CrPO₄

       1 mol (147 g/mol)         2 mol  (212 g/mol)      3 mol  (174g/mol)

So, Now we have the following details

          Cr₂(SO₄)₃  +  2K₃PO₄          ----------> 3K₂SO₄ + 2CrPO₄

              147 g         424 g                           522 g

So,

we come to know that 147 g of Cr₂(SO₄)₃ combine with 424 g of 2K₃PO₄ produce  522 g of K₂SO₄

So now we calculate that how many grams of potassium sulfate will be produced

Apply unity formula

              147 g of  Cr₂(SO₄)₃  ≅ 522 g of K₂SO₄

              450 g of  Cr₂(SO₄)₃  ≅ ? g of K₂SO₄

by doing cross multiplication

g of K₂SO₄ =522 g x 450 g / 147 g

g of K₂SO₄ =  1597.959 g

So the write answer is  1597.959 g  

***Note: By calculation it is obvious that the correct answer is  1597.959 g  

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