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erastova [34]
4 years ago
5

How many protons are in one atom of 37Ar

Chemistry
1 answer:
Archy [21]4 years ago
7 0

There are 18 protons in one atom of ³⁷Ar.

The number 37 is its <em>mass number</em> — the number of <em>protons + neutrons</em>.

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9. Using the balanced equation from Question #8, how many grams of lead will be produced if 2.54 grams of PbS is burned with 1.8
MissTica

Answer: 2.24 grams of Pb

Explanation:

<u>Step 1</u>

Balanced chemical reaction;

2PbS + 3O2 → 2Pb + 2SO3

<u>Step 2</u>

Moles of both PbS and O2

Moles = mass / molar mass

Moles of PbS = 2.54 g / 239.3 g/mol = 0.0108 moles

Moles of O2 = 1.88 / 32 g/mol = 0.0588 moles

<u>Step 3</u>

Finding the limiting reactant.

Limiting reactant, is that reactant which is completely used in the reaction;

If we assume that PbS is the limiting reactant;

We have 0.0588 moles of O2. This needs ( 0.0588 * 2) / 3 = 0.0392 moles of PbS to fully react. But we have only 0.0108 moles of PbS available. That means that the PbS will be completely consumed hence the limiting reactant

If we assume O2 is the limiting reactant;

We have 0.0108 moles of PbS. That needs ( 0.0108 * 3) / 2 = 0.0162 moles of O2. But we have 0.0588 moles of O2 which is in excess further confirming that PbS is the limiting reactant since it will be depleted in the reaction.

<u>Step 4</u>

Moles of lead

For this step we apply the mole ratios with the limiting reactant;

Mole ratio of PbS : Pb = 2 : 2 = 1 : 1

Therefore;

Moles of Pb = (0.0108 moles  * 1 ) 1

Moles of Pb =0.0108 moles

<u>Step 5</u>

Mass of Pb

Mass = moles * molar mass

Mass of Pb =0.0108 moles * 207.2 g/mol

Mass of Pb = 2.24 grams

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Commercial manufacturers produce nitric acid (HNO3) by the Ostwald process. The process requires three steps:
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Nitric acid and nitrogen monoxide react to form nitrogen dioxide and water, like this: At a certain temperature, a chemist finds
poizon [28]

Answer:

K = 3.3

Explanation:

Nitric acid, HNO3, reacts with nitrogen monoxide, NO, to produce nitrogen dioxide, NO2 and water H2O as follows:

2HNO3(g) + NO(g) → 3NO2(g) + H2O(g)

Where equilibrium constant, K, is:

K = [NO2]³[H2O] / [HNO3]²[NO]

<em>[] is the molar concentration of each species at equilibrium.</em>

To solve this question we need to find molarity of each gas and replace these in the equation as follows:

<em>[NO2] -Molar mass NO2-46.0g/mol-</em>

18.6g * (1mol/46.0g) = 0.404mol / 7.7L = 0.0525M

<em>[H2O] -Molar mass:18.01g/mol- </em>

236.7g * (1mol/18.01g) = 13.14 moles / 7.7L = 1.707M

<em>[HNO3] -Molar mass:53.01g/mol-</em>

16.2g * (1mol/53.01g) = 0.3056 moles / 7.7L = 0.0397M

<em>[NO] -Molar mass: 30.0g/mol-</em>

11.0g * (1mol/30.0g) = 0.367 moles / 7.7L = 0.0476M

Replacing:

K = [NO2]³[H2O] / [HNO3]²[NO]

K = [0.0525M]³[1.707M] / [0.0397M]²[0.0476M]

<h3>K = 3.3</h3>

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4 0
3 years ago
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