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2 years ago

Anyone can help me please

Chemistry

1 answer:

Vlad1618 [11]2 years ago

5 0

x=3

Use the pythagorean theorem:

a^2 + b^2 = c^2

4^2 + x^2 = 5^2

16 + x^2 = 25

x^2 = 25-16

x^2 = 9

√x^2 = √9

x = 3

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2 years ago

When heated to 150 ºC, CuSO4.5 H2O loses its water of hydration as gaseous H2O. A 2.50 g sample of the compound is placed in a s

Katyanochek1 [597]

Answer:

Water pressure 0.5 atm

Total Pressure= 2.27 atm

Explanation:

To answer this problem, one has to realize that there are two processes that increase the temperature of the sealed vessel.

First, the dry air in the sealed vessel will be heated which will cause its pressure to increase and it can be determined by the equation:

P₁ x T₂ = P₂ x T₁ ∴ P₂ = P₁ x T₂ / T₁

For the second process, we have an amount of n moles of water which will be released when the copper sulfate is heated. In this case, to determine the value of the the water gas we will use the gas law:

PV = nRT ∴ P = nRT/V

n will we calculated from the quantity of sample.

2.50 g CuSo₄ 5H₂O x 1 mol/ 249.69 g = 0.01 mol CuSo₄ 5H₂O

the amount water of hydration is

= 0.01 mol CuSo₄ 5H₂O * 5 mol H₂O / 1 mol CuSo₄ 5H₂O

= 0.05 mo H₂O

pressure of dry air at the final temperature,

P₂ = 1 atm x 500 K/ 300 K = 1.67 atm

Pressure of water :

P (H₂O) 0.05 mol x 0.08206 Latm/kmol x 500 K/ 4 L = 0.5 atm

∴ Total Pressure = 1.67 atm

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2 years ago

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2 years ago

What is the molar mass of water (H2O)?

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Answer:

The molar mass is: 18.02 g/mol.

Explanation:

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1 mole of Hydrogen= 1.01, so if we have 2 moles of it here, that would be 2.02.

1 mole of Oxygen (that's all we have here)= 16.00

Once you add the two together (2.02+16.00), you will get 18.02.

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2 years ago

During the following chemical reaction, 46.3 grams of C3H6O react with 73.2 grams of O2

ra1l [238]

Answer:

a) O2 is the limiting reactant

b) 75.70 grams CO2 (theoretical yield)

c) There remains 12.81 grams of C3H6O

d) The actual yield CO2 is 34.29 grams

Explanation:

Step 1: Data given

Mass of C3H6O = 46.3 grams

Mass of O2 = 73.2 grams

Molar mass of C3H6O = 58.08 g/mol

Molar mass of O2 = 32 g/mol

Step 2: The balanced equation

C3H6O + 4O2 → 3 CO2 + 3H2O

Step 3: Calculate moles C3H6O

Moles C3H6O = mass C3H6O / molar mass C3H6O

Moles C3H6O = 46.3 grams / 58.08 g/mol

Moles C3H6O = 0.793 moles

Step 4: Calculate moles O2

Moles O2 = 73.2 grams / 32 g/mol

Moles O2 = 2.29 moles

Step 5: Calculate limiting reactant

For 1 mol C3H6O we need 4 moles of O2 to produce 3 moles CO2 and 3 moles H2O

O2 is the limiting reactant. It will completely be consumed. (2.29 moles).

C3H6O is in excess. There will react 2.29/4 = 0.5725 moles C3H6O

There will remain 0.793 - 0.5725 = 0.2205 moles C3H6O

This is 0.2205 moles * 58.08 g/mol = 12.81 grams

Step 6: Calculate moles of CO2

For 1 mol C3H6O we need 4 moles of O2 to produce 3 moles CO2 and 3 moles H2O

For 2.29 moles O2 we need 3/4 * 2.29 = 1.72 moles CO2

This is 1,72 moles * 44.01 g/mol = 75.70 grams CO2

Step 7: Calculate actual yield

% yield = 45.3 % = 0.453 = (actual yield / theoretical yield)

actual yield = 0.453 * 75.70 = 34.29 grams

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3 years ago