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3 years ago

A balloon is filled with 3.8 l of helium gas at stp. approximately how many moles of helium are contained in the balloon?

Chemistry

2 answers:

love history [14]3 years ago

7 0

Hope this helps you.

Lelechka [254]3 years ago

3 0

Answer:

0.16964 moles ≈ 0.17 moles

Explanation:

At STP, one mole of a gas has a volume of 22.4 L.

1 mole = 22.4L

x mole = 3.8L

x = ( 3.8 * 1 ) / 22.4

x = 3.8 / 22.4

x = 0.16964 moles ≈ 0.17 moles

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source: http://chemistry.elmhurst.edu/vchembook/207epgeom.html

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3 years ago

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Explanation:

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3 years ago

A marathon covers a distance of 26.2 miles. If 1 mile is equivalent to exactly 1760 yards, what is the distance of the race in y

Contact [7]

Both mile and yard are units of distance thus they are inter convertible.

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Since, 1 mile=1760 yards

Thus, distance can be converted into yards using the following conversion factor:

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8 0

3 years ago

At 850°C, CaCO3 undergoes substantial decomposition to yield CaO and CO2. Assuming that the ΔH o f values of the reactant and pr

Xelga [282]

Answer:

The enthalpy if 68.10 grams of CO2 is produced is -189.04 kJ

Explanation:

Step 1: Data given

temperature = 850 °C

Mass of 68.10 grams of CO2

ΔH°f (CaO) = -635.6 kJ/mol

ΔH°f (CO2) = -693.5 kJ/mol

ΔH°f (CaCO3) =-1206.9 kJ/mol

Step 2: The balanced equation

CaCO3(s) → CaO(s) + CO2(g)

Step 3: Calculate ΔH°reaction

ΔH°reaction = ΣΔH°f (products) - ΣΔH°f (reactants)

ΔH°reaction = (ΔH°f (CaO) + ΔH°f (CO2)) - ΔH°f (CaCO3)

ΔH°reaction = (-635.6 kJ/mol + -693.5 kJ/mol) + 1206.9 kJ/mol

ΔH°reaction = -122.2 kJ /mol

Step 4: Calculate moles of CO2

Moles CO2 = mass CO2 / Molar mass CO2

Moles CO2 = 68.10 grams / 44.01 g/mol

Moles CO2 = 1.547 moles

Step 5: Calculate the enthalpy change for 68.10 grams of CO2

-122.2 kJ/mol * 1.547 moles = -189.04 kJ

The enthalpy if 68.10 grams of CO2 is produced is -189.04 kJ

7 0

3 years ago