Which of the following is and example of a chemical change

How do i calculate the mass of CO2 emitted per Kj of heat produced in a combustion reaction?

Alexxandr [17]

methanol:

1 mole CH3 OH --> produces --> 1 mole CO2

1 mole CO2 has a molar mass of 44.01 gh/mole

your set up is:
(44.01 g CO2) / -726.5kJ = 0.06058g

your answer 0.06058 grams of CO2 produced per kJ released.

6 0

4 years ago

When 28 g of nitrogen and 6 g of hydrogen react, 34 g of ammonia are produced. If 100 g of nitrogen react with 6 g of hydrogen,

love history [14]

Answer:

34 g

Explanation:

Let's consider the following balanced equation.

N₂ + 3 H₂ → 2 NH₃

The theoretical mass ratio of N₂ to H₂ is 28g N₂ : 6g H₂ = 4.6g N₂ : 1g H₂.

The experimental mass ratio of N₂ to H₂ is 100g N₂ : 6g H₂ = 16.6g N₂ : 1g H₂.

As we can see, hydrogen is the limiting reactant.

According to the task, we 6 g of H₂ react completely, 34 g of ammonia are produced.

3 0

3 years ago

A balloon filled with helium has a volume of 6.9 L. What is the mass, in grams, of helium in the balloon?

balandron [24]

Answer : The mass of helium gas in the balloon is, 1.23 grams.

Explanation : Given,

Volume of helium gas = 6.9 L

First we have to calculate moles of helium gas at STP.

As we know that, 1 mole of substance occupy 22.4 L volume of gas.

As, 22.4 L volume of helium gas present in 1 mole of helium

So, 6.9 L volume of helium gas present in $\frac{6.9L}{22.4L}\times 1mole=0.308mole$ of helium

Now we have to calculate the mass of helium gas.

$\text{Mass of He gas}=\text{Moles of He gas}\times \text{Molar mass of He gas}$

Molar mass of He gas = 4 g/mol

$\text{Mass of He gas}=0.308mol\times 4g/mol=1.23g$

Thus, the mass of helium gas in the balloon is, 1.23 grams.

6 0

3 years ago

Given: There are 39.95 grams of Argon (39.95 g/1 mole) and one mole has a volume of 22.4 Liters (1 mole/22.4 L). What is the vol

marta [7]

Answer:

$V_2=19.23L$

Explanation:

Hello,

In this case, by using the Avogadro's law which allows us to understand the volume-moles behavior as a directly proportional relationship:

$\frac{V_2}{n_2} =\frac{V_1}{n_1}$

We can compute the volume of 34.3 g of argon by representing it in mole as shown below:

$n_1=1 mol\\\\n_2=34.3g*\frac{1mol}{39.95g} =0.859mol$

Thus, we find:

$V_2=\frac{V_1*n_2}{n_1}=\frac{22.4L*0.859mol}{1mol} \\\\V_2=19.23L$

Best regards.

6 0

3 years ago

Radioactive isotopes have been used commercially in:

Law Incorporation [45]

<span>Radioactive isotopes have been used commercially in all these applications.
The last option (D) is your answer

</span>

3 0

3 years ago