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babymother [125]
2 years ago
15

After looking at the schematics of the bubble, he finds that it holds 0.750 moles of gas at Standard Temperature and Pressure. U

nfortunately for Rooster, helium gas is not sold by the mole, only by the liter. He needs to know how many liters of gas 0.750 moles of helium is.
Chemistry
1 answer:
Lina20 [59]2 years ago
6 0

Answer:

16.8 L

Explanation:

From the question given above, the following data were obtained:

Number of mole of He = 0.750 mole

Volume of He =?

Recall:

1 mole of any gas occupy 22.4 L at STP. This also implies that 1 mole of He occupies 22.4 L at STP.

Finally, we shall determine the volume occupied by 0.750 mole of He. This can be obtained as follow:

1 mole of He occupied 22.4 L at STP.

Therefore, 0.750 mole of He will occupy = 0.750 × 22.4 = 16.8 L at STP.

Thus, 0.750 mole of He is equivalent to 16.8 L

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Coal and petroleum are useful because the both have ___ stored in them.
gizmo_the_mogwai [7]
I think the answer is 4 carbon dioxide
5 0
3 years ago
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The iodine "clock reaction" involves the following sequence of reactions occurring in a reaction mixture in a single beaker. 1.
Mars2501 [29]

C: 0.012 mol.

<h3>Explanation</h3>

Start with 0.0020 moles of iodate ions {\text{IO}_{3}}^{-}.

How many moles of iodine \text{I}_2 will be produced?

{\text{IO}_{3}}^{-} converts to \text{I}_2 in the first reaction. The coefficient in front of \text{I}_2 is three times the coefficient in front of {\text{IO}_{3}}^{-}. In other words, each mole of {\text{IO}_{3}}^{-} will produce three moles of \text{I}_2. 0.0020 moles of {\text{IO}_{3}}^{-} will convert to 0.0060 moles of \text{I}_2.

How many moles of thiosulfate ions {\text{S}_2\text{O}_3}^{2-} are required?

\text{I}_2 reacts with {\text{S}_2\text{O}_3}^{2-} in the second reaction. The coefficient in front of \text{I}_2 is twice the coefficient in front of {\text{S}_2\text{O}_3}^{2-}. How many moles of {\text{S}_2\text{O}_3}^{2-} does each mole of \text{I}_2 consume? Two. 0.0060 moles of \text{I}_2 will be produced. As a result, 2 \times 0.0060 = 0.0120 moles of {\text{S}_2\text{O}_3}^{2-} will be needed.

6 0
3 years ago
What is that the theoretical yield of aluminum oxide I if 3.20 mol of aluminum metal is exposed to 2.70 mole of oxygen
photoshop1234 [79]

Answer:

163.2g

Explanation:

First let us generate a balanced equation for the reaction. This is shown below:

4Al + 3O2 —> 2Al2O3

From the question given, were were told that 3.2moles of aluminium was exposed to 2.7moles of oxygen. Judging by this, oxygen is excess.

From the equation,

4moles of Al produced 2moles of Al2O3.

Therefore, 3.2moles of Al will produce = (3.2x2)/4 = 1.6mol of Al2O3.

Now, let us covert 1.6mol of Al2O3 to obtain the theoretical yield. This is illustrated below:

Mole of Al2O3 = 1.6mole

Molar Mass of Al2O3 = (27x2) + (16x3) = 54 + 48 =102g/mol

Mass of Al2O3 =?

Number of mole = Mass /Molar Mass

Mass = number of mole x molar Mass

Mass of Al2O3 = 1.6 x 102 = 163.2g

Therefore the theoretical of Al2O3 is 163.2g

8 0
3 years ago
Do states need to be written in a chemical equation
Reptile [31]
Yes. because equation balances the number of particles.
4 0
3 years ago
Hydrogen and iodine react to form hydrogen iodide, like this: H_2 (g) + I_2 (g) rightarrow 2 HI(g) Also, a chemist finds that at
telo118 [61]

This is an incomplete question, here is a complete question.

Hydrogen and iodine react to form hydrogen iodide, like this:

H_2(g)+I_2(g)\rightarrow 2HI(g)

Also, a chemist finds that at a certain temperature the equilibrium mixture of hydrogen, iodine, and hydrogen iodide has the following composition:

Compound            Pressure at equilibrium

H_2                                   61.8 atm

I_2                                    46.5 atm

HI                                  52.3 atm

Calculate the value of the equilibrium constant K_p for this reaction. Round your answer to 2 significant digits.

Answer : The value of equilibrium constant K_p for this reaction is, 0.952

Explanation :

The given chemical reaction :

H_2(g)+I_2(g)\rightarrow 2HI(g)

The expression of K_p for above reaction follows:

K_p=\frac{(P_{HI})^2}{P_{H_2}\times P_{I_2}}

We are given:

P_{H_2}=61.8atm

P_{I_2}=46.5atm

P_{HI}=52.3atm

Putting values in above equation, we get:

K_p=\frac{(52.3)^2}{61.8\times 46.5}\\\\K_p=0.952

Therefore, the value of equilibrium constant K_p for this reaction is, 0.952

7 0
3 years ago
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