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Alekssandra [29.7K]
3 years ago
8

-

Chemistry
1 answer:
Dovator [93]3 years ago
3 0

Answer:

5 × 10^-4 L

Explanation:

The equation of the reaction is;

2KClO3 = 2KCl + 3O2

Number of moles of KClO3 = 13.5g/122.5 g / mol = 0.11 moles

From the stoichiometry of the reaction;

2 moles of KClO3 yields 3 moles of O2

0.11 moles of KClO3 yields 0.11 × 3/2 = 0.165 moles of oxygen gas

From the ideal gas equation;

PV= nRT

P= 85.4 × 10^4 KPa

V=?

n= 0.165

R= 8.314 J K-1 mol-1

T= 40+273 = 313K

V= 0.165 ×8.134 × 313/85.4 × 10^4

V=429.4/85.4 × 10^4

V= 5 × 10^-4 L

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The molar mass of the protein is 45095 g/mol.

The mass of a sample of a chemical compound divided by the quantity, or number of moles in the sample, measured in moles, is known as the molar mass of that compound.

The expression of molar mass of protein is

M₂ = (W₂/P) (RT/V)

Given;

W₂ = 1.31g

P = 4.32 torr = 5.75 X 10⁻³ bar

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Putting all the values in the above formula

M₂= (1.31 g/5.75 X 10⁻³ bar) X (0.083 Lbar/mol/K X 2)/0.125 L)

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Learn more about the Molar mass with the help of the given link:

brainly.com/question/22997914

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What is the maximum amount of water (in grams) that can be removed from 15ml of toluene by the addition?
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Complete Question

Magnesium sulfate forms a hydrate with the formula MgSO_4. 7H_20. What is the maximum amount of water (in grams) that can be removed from 15 ml of toluene by the addition of 200 mg of anhydrous magnesium sulfate? The molar mass of MgSO_4 is 120.4 g/mol; H20 = 18 g/mol.

Answer:

The value  is  z =  0.2093 \  g of  H_2O

Explanation:

From the question we are told that

   The volume of toluene is  V = 15 mL

    The mass of  anhydrous magnesium sulfate is  m =  200m g  = 200 *10^{-3} \  g

   The formula of the hydrate is   MgSO_4. 7H_20

    The molar mass of   MgSO_4  is  z =120.4 \ g/mol

From the formula given we see that

  1 mole of  Mg SO_4 wil remove  7 moles of H_2O to for the given formula

Hence

  120.4 g (1 mole) will remove  7 moles (7 * 18 g = 126 g  ) of  H_2O to for the given formula

Therefore 1 g of  Mg SO_4  x g  of  H_2O  

So

     x  =  \frac{x]126 *  1}{ 120.4 }

=>     x  =  1.0465 \  g

From our calculation we obtained that

  1 g of Mg SO_4 will remove  x  =  1.0465 \  g  of  H_2O  

Then  

   200 *10^{-3} \  g of Mg SO_4 will remove z g of  x  =  1.0465 \  g  of  H_2O  

So

   z =  200 *10^{-3} *  1.0465

=>z =  200 *10^{-3} *  1.0465

=>z =  0.2093 \  g of  H_2O  

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