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3 years ago

A gas occupies 900.0 mL at a temperature of 27.0 . What is the volume at 132.0 °C?

1 answer:

7 0

V₁ = 900.0 mL

T₁ = 27.0ºC + 273 = 300 K

V₂ = ?

T₂ = 132.0 + 273 = 405 K

V₁ / T₁ = V₂ / T₂

900.0 / 300 = V₂ / 405

300 x V₂ = 405 x 900.0

300 x V₂ = 364500

V₂ = 364500 / 300

V₂ = 1215 mL

hope this helps!

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Concentrated hydrochloric acid is made by pumping hydrogen chloride gas into distilled water. If concentrated HCl contains 439 g

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Answer:

12.10 mol / 1 L

Explanation:

Molarity of a substance , is the number of moles present in a liter of solution .

M = n / V

M = molarity ( unit = mol / L or M )

V = volume of solution in liter ( unit = L ),

n = moles of solute ( unit = mol ),

Moles is denoted by given mass divided by the molecular mass ,

Hence ,

n = w / m

n = moles ,

w = given mass ,

m = molecular mass .

From the question ,

The data given is as follows -

w = 439 g

As , we known for HCl ,

m = 36.46 g/mol

V = 1 L

From the above data ,

Moles are given as -

n = w / m

n = 439 / 36.26 = 12.10 mol ,

Now , the molarity is given as ,

M = n / V

M = 12.10 mol / 1 L

M = 12.10 mol /L

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A compound whose water solution conducts electricity and turnsphenolphthalein pink is(1) HCl (3) NaOH(2) HC2H3O2 (4) CH3OH

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A strong base turns phenolphthalein pink, so the answer is (3) NaOH

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A student borrows $10,000 from a bank but ends up paying $13,000 on the

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The student paid an interest of $3,000 over the next five years.

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The interest on a borrowed amount of money can be calculated by subtracting the principal from the amount paid over time. That is;

Interest = principal - amount

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Therefore, the student paid an interest of $3,000 over the next five years.

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Liquid octane (CH3(CH2)CH3) reacts with gaseous oxygen gas(O2) to produce gaseous carbon dioxide(CO2) and gaseous water (H2O). W

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Answer:

The theoretical yield of carbon dioxide formed is 68.4 grams CO2

Explanation:

Step 1: Data given

Liquid octane = C8H18

gaseous oxygen gas = O2

gaseous carbon dioxide = CO2

gaseous water = H2O

Mass of octane = 27.4 grams

Molar mass of octane = 114.23 g/mol

Mass of oxygen = 77.8 grams

Molar mass of oxygen = 32.0 g/mol

Step 2: The reaction

2C8H18(l) + 25O2(g) → 16CO2(g) + 18H2O(g)

Step 3: Calculate number of moles

Moles = mass / molar mass

Moles of octane = 27.4 grams / 114.23 g/mol

Moles of octane = 0.240 moles

Moles of oxygen = 77.8 grams / 32.0 g/mol

Moles of oxygen = 2.43 moles

Step 4: Calculate the moles of the products

For 2 moles of octane we need 25 moles of O2 to produce 16 moles of CO2 and 18 moles H2O

O2 is the limiting reactant. It will completely react. (2.43 moles).

There will react 2.43/12.5 = 0.194 moles

There will remain 0.240 - 0.194 = 0.046 moles of octane.

There will be produced:

16/25 * 2.43 = 1.555 moles of CO2

18/25 * 2.43 = 1.750 moles of H2O

This is:

1.56 moles * 44.01 g/mol = 68.4 grams CO2

1.750 moles * 18.02 g/mol = 31.5 grams H2O

The theoretical yield of carbon dioxide formed is 68.4 grams CO2

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