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Gwar [14]
1 year ago
12

If a 1.0x10-5 L sample of agas at 2.0x106 atm isreleased until it is equal to0.275 atm, what would thenew volume of the gas be?

Chemistry
1 answer:
Greeley [361]1 year ago
5 0

Answer:

Explanation:

Here, we want to get the new volume of the gas

What we need to know is the law that connects volume and pressure at constant temperature

The law that supports this is the Boyle's law

It states that the volume of a given mass of gas is inversely proportional to its pressure at constant temperature

Mathematically, we can have this as represented as:

\begin{gathered} P_1V_1=P_2V_2 \\ P_1\text{ = 2.0 }\times10^{6\text{ }}\text{ atm} \\ V_1\text{ = 1.0 }\times10^{-5}\text{ L} \\ P_2\text{ =0.275 atm} \\ V_2\text{ = ?} \end{gathered}

We can proceed mathematically to solve as follows;

undefined

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Molar mass c3h8
maksim [4K]

Answer:

1. 44.11 g

2. 36.03 g

3. 8.08 g

4. 81.7%

5. 18.3%

Explanation:

1. 12.01+12.01+12.01+1.01+1.01+1.01+1.01+1.01+1.01+1.01+1.01=44.11

2. 12.01×3= 36.03

3. 1.01×8= 8.08

4.(36.03/44.11)×100= 81.7%

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2 years ago
How mamy moles of NaCl will be produced from 83.0g of Na, assuming Cl2 is available in excess
Anestetic [448]

Answer:

3.62moles

Explanation:

First let us generate a balanced equation for the reaction

2Na + Cl2 —> 2NaCl

Molar Mass of Na = 23g/mol

Mass of Na from the balanced equation = 2 x 23 = 46g

Molar Mass of NaCl = 23 + 35.5 = 58.5g/mol

Mass of NaCl from the balanced equation = 2 x 58.5 = 117g

From the question,

46g of Na produced 117g of NaCl.

Therefore, 83g of Na will produce = (83 x 117)/46 = 211.11g of NaCl

Converting this mass (211.11g of NaCl) to mole, we obtain:

n = Mass /Molar Mass

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3.62moles

8 0
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Name: _________________ Temperature o fwater_25_degreecent.YOU MUST SHOW ALL CALCULATIONS TO RECEIVE CREDIT FOR THEM! DATA ANALY
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Answer:

1. 0.02 M

2. 0.01 M

3. 4×10⁻⁶

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We know that V₁S₁ = V₂S₂

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Concentration of HCl = 0.05 M

end point comes at = 10 ml

So, concentration of OH⁻(aq) = [OH⁻(aq)] ⇒ (0.05 × 10) ÷ 25 ⇒ 0.02 M

2.

2mol of OH⁻(aq) ≡ 1 mole of Ca²⁺(aq)

[Ca²⁺] = 0.02 ÷ 2 = 0.01 M

3.

K_{sp} = [Ca²⁺(aq)] [OH⁻(aq)]²

Ca(OH)₂ (aq) ⇄ Ca²⁺ (aq) + 2OH⁻ (aq)

K_{sp} = [0.01 × (0.02)²] = 4×10⁻⁶

4.

If reaction is exothermic which means heat energy will get evolved as a result temperature of the reaction media will get increased during the course of the reaction. If temperature is externally increased, the reaction will go backward to accumulate extra heat energy.

5.

K_{sp} value describes the solubility of a particular ionic compound. The higher the K_{sp} value, the higher the Solubility will be.

6.

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