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aalyn [17]
4 years ago
7

CHM 17 Practice Final Exam

Chemistry
1 answer:
serious [3.7K]4 years ago
7 0

Answer:

moneyyyyyyyyfght

Explanation:

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Chemistry is the study of matter true or false
mario62 [17]
The answer is "True"
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3 years ago
Read 2 more answers
What was the biggest challenge you faced in getting to where you are today and how did you overcome it?
Alchen [17]
It’s all based on you but yours could be anything hard you went through and how you overcame it. Also say how it made you the person you are today
4 0
3 years ago
Gaseous ethane will react with gaseous oxygen to produce gaseous carbon dioxide and gaseous water . Suppose 7.82 g of ethane is
NeTakaya

Answer:

4.79 g of water

Explanation:

From the reaction equation;

C2H6(g) + 7/2O2(g) ----> 2CO2(g) + 3H2O(g)

Next we convert the given masses of reactants to moles of reactants.

For ethane; number of moles = mass/molar mass= 7.82g/ 30gmol-1= 0.261 moles

For oxygen; number of moles= 9.9 g/32gmol-1= 0.31 moles

Next we determine the limiting reactant, the limiting reactant yields the least amount of product.

For ethane;

From the reaction equation,

1 mole of ethane yields 3 moles of water

0.261 moles of Ethan yields 0.261 ×3 = 0.783 moles of water

For oxygen;

3.5 moles of oxygen yields 3 moles of water

0.31 moles of oxygen yields 0.31 × 3/3.5 = 0.266 moles of water

Hence oxygen is the limiting reactant.

Mass of water produced = 0.266 moles of water × 18gmol-1 = 4.79 g of water

4 0
3 years ago
N2O4(g) + 4H2(g) → N2(g) + 4H2O(g), solve using standard enthalpies of formation
aev [14]

The standard enthalpy of formation (ΔH_f) is a measure of the energy released or consumed when one mole of a substance is created under standard conditions from its pure elements.

Standard enthalpies (ΔH_f) of formation for given reaction is 978.3 kJ

<h3>What is Standard enthalpies of formation?</h3>

The standard enthalpy of formation is defined as the enthalpy change when one mole of a substance in the standard state (1 atm of pressure and 298.15 K) is formed from its pure elements under the same conditions.

Given reaction ;

N_2O_4(g) + 4H_2 (g) - > N_2(g) + 4H_2O(g)

To Find : ΔH_f

ΔH_f = ∑np ΔH_f (products) – ∑np ΔH_f (reactants)

ΔH_f = [1(ΔH_f N_2) + 4(ΔH_f H_2O)] – [1(ΔH_f N_2O_4) + 4(ΔH_f H_2)]

ΔH_f = [1(0) + 4(-241.8)] – [1(+9.16) + 4(0)]

ΔH_f  = [4(-241.8)] – [1(+9.16)] = 978.3 kJ

Learn more about Enthalpy here ;
brainly.com/question/16720480

#SJF1

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2 years ago
What is 0.00 degrees Kelvin known as
vodka [1.7K]
-273.15 degrees celsius
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