Ask question

Ask question

All categories

3 years ago

7

A gas sample of 5 moles, has a volume of 95 L. How many moles of the same gas should I add to obtain a volume of 133 L at the sa

me temperature and pressure.

1 answer:

Dmitry [639]3 years ago

8 0

Answer :
i don’t know cause i’m in 8th grade

You might be interested in

The production of iron and carbon dioxide from iron(iii) oxide and carbon monoxide is an exothermic reaction. which expression c

NARA [144]

Missing question:
A. [3.40 mol Fe2O3 (s) × 26.3 kJ/1 mol Fe2O3 (s)] / 2
<span>B. 3.40 mol Fe2O3 (s) × 26.3 kJ/1 mol Fe2O3 (s) </span>
<span>C. 26.3 kJ/1 mol Fe2O3 (s) / 3.40 mol Fe2O3 (s) </span>
<span>D. 26.3 kJ/1 mol Fe2O3 (s) – 3.40 mol Fe2O3 (s).
</span>Answer is: B.
Chemical reaction: F<span>e</span>₂O₃<span>(s) + 3CO(g) → 2Fe(s) + 3CO</span>₂<span>(g);</span>ΔH = <span>+ 26.3 kJ.
When one mole of iron(III) oxide reacts 26,3 kJ of energy is required and for 3,2 moles of iron(III) oxide 3,2 times more energy is required.</span>

8 0

3 years ago

Read 2 more answers

Name at least three types associated with the microwave

Bond [772]

Hi , if you are asking for the types of energy , here's your answer:

1. Radiative energy , because the microwave is releasing radiation.
2.Light energy , because there is light inside the microwave.
3.electronic energy , because the microwave is using electricity.

8 0

3 years ago

True or false. (Pls help)

grigory [225]

I believe they’re both true.

5 0

2 years ago

Read 2 more answers

How many liters of hydrogen gas will be produced at STP from the reaction of 7.179×10^23 atoms of magnesium with 54.219g of phos

Alexeev081 [22]

Answer: The volume of hydrogen gas produced will be, 12.4 L

Explanation : Given,

Mass of $H_3PO_4$ = 54.219 g

Number of atoms of $Mg$ = $7.179\times 10^{23}$

Molar mass of $H_3PO_4$ = 98 g/mol

First we have to calculate the moles of $H_3PO_4$ and $Mg$ .

$\text{Moles of }H_3PO_4=\frac{\text{Given mass }H_3PO_4}{\text{Molar mass }H_3PO_4}$

$\text{Moles of }H_3PO_4=\frac{54.219g}{98g/mol}=0.553mol$

and,

$\text{Moles of }Mg=\frac{7.179\times 10^{23}}{6.022\times 10^{23}}=1.19mol$

Now we have to calculate the limiting and excess reagent.

The balanced chemical equation is:

$3Mg+2H_3PO_4\rightarrow Mg(PO_4)_2+3H_2$

From the balanced reaction we conclude that

As, 3 mole of $Mg$ react with 2 mole of $H_3PO_4$

So, 0.553 moles of $Mg$ react with $\frac{2}{3}\times 0.553=0.369$ moles of $H_3PO_4$

From this we conclude that, $H_3PO_4$ is an excess reagent because the given moles are greater than the required moles and $Mg$ is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of $H_2$

From the reaction, we conclude that

As, 3 mole of $Mg$ react to give 3 mole of $H_2$

So, 0.553 mole of $Mg$ react to give 0.553 mole of $H_2$

Now we have to calculate the volume of $H_2$ gas at STP.

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

As, 1 mole of hydrogen gas occupies 22.4 L volume of hydrogen gas

So, 0.553 mole of hydrogen gas occupies $0.553\times 22.4=12.4L$ volume of hydrogen gas

Therefore, the volume of hydrogen gas produced will be, 12.4 L

4 0

3 years ago

PLEASE HELP it would be really nice of you to answer this question i would be very grateful <3

Aleksandr-060686 [28]

Answer: D (The abundance percentage of each isotope)

Explanation: hope this helps!

7 0

3 years ago

Read 2 more answers