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Korolek [52]
3 years ago
9

Gustave-Gaspard Coriolanus described this type of energy

Chemistry
1 answer:
Paha777 [63]3 years ago
8 0
I believe it’s Chemical energy but please correct me if i’m wrong
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How many moles of helium are needed to fill a balloon to a volume of 6.3 L at 28 °C and 320
Ray Of Light [21]

Answer:

.11 mol

Explanation:

Convert mmHg to atms by dividing by 760. Then multiply 6.3 by the atms and divide by .08206*(273+28) to get mol

5 0
2 years ago
Read 2 more answers
Given the equation:
My name is Ann [436]

Answer:

this isnt even a question...

Explanation: what the heck

6 0
3 years ago
The theoretical yield of a reaction is the amount of product obtained if the limiting reactant is completely converted to produc
Elis [28]

Answer: 9.9 grams

Explanation:

To calculate the moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text {Molar mass}}

a) moles of H_2

\text{Number of moles}=\frac{8.150g}{2g/mol}=4.08moles

b) moles of C_2H_4

\text{Number of moles}=\frac{9.330g}{28g/mol}=0.33moles

H_2(g)+C_2H_4(g)\rightarrow C_2H_6(g)

According to stoichiometry :

1 mole of C_2H_4 combine with 1 mole of H_2

Thus 0.33 mole of C_2H_4 will combine with =\frac{1}{1}\times 0.33=0.33 mole of H_2

Thus C_2H_4 is the limiting reagent as it limits the formation of product.

As 1 mole of C_2H_4 give =  1 mole of C_2H_6

Thus 0.33 moles of C_2H_4 give =\frac{1}{1}\times 0.33=0.33moles  of C_2H_6

Mass of C_2H_6=moles\times {\text {Molar mass}}=0.33moles\times 30g/mol=9.9g

Thus theoretical yield (g) of C_2H_6 produced by the reaction is 9.9 grams

7 0
3 years ago
How many moles of KNO3 are in 500 mL of 2.0 M KNO3?<br> ___ mol KNO3
weeeeeb [17]

Answer: 1mole

Explanation:

Mole = concentration× Volume (dm3)

Mole = 2× 500/1000

4 0
3 years ago
Read 2 more answers
A 32.8 g sample of gas occupies 22.414 L at STP. What is the molecular weight of this gas?
olasank [31]

Answer:

32.8g/mole

Explanation:

Given parameters:

Mass of sample of gas = 32.8g

Volume  = 22.4L

Unknown:

Molecular weight  = ?

Solution:

To solve this problem we must understand that at rtp;

             1 mole of gas occupies a volume of 22.4L

 

Number of mole of the gas  = 1 mole

 Now;

   Mass  = number of moles x molecular weight

    molecular weight  = \frac{mass}{molecular weight}    = \frac{32.8g}{1mole }      = 32.8g/mole

8 0
3 years ago
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