Answer:
2. 
3. 
Explanation:
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2. In this case, we can evidence the problem by which volume and temperature are involved, so the Charles' law is applied to:

Thus, considering the temperatures in kelvins and solving for the final volume, V2, we obtain:

Therefore, we plug in the given data to obtain:

3. In this case, it is possible to realize that the 3.7 moles of neon gas are at 273 K and 1 atm according to the STP conditions; in such a way, considering the ideal gas law (PV=nRT), we can solve for the volume as shown below:

Therefore, we plug in the data to obtain:

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8,700 B.C.
No clue who discovered it.
Answer:
B
Explanation:
conduction is the transfer of heat between objects that touch
The empirical formula for the caproic acid, given the combustion analysis data is C₃H₆O
We'll begin bey obtaining the mass of carbon, hydrogen and oxygen in the compound. This is illustrated below:
How to determine the mass of C
- Mass of CO₂ = 9.78 g
- Molar mass of CO₂ = 44 g/mol
- Molar of C = 12 g/mol
- Mass of C =?
Mass of C = (12 / 44) × 9.78
Mass of C = 2.67 g
How to determine the mass of H
- Mass of H₂O = 20.99 g
- Molar mass of H₂O = 18 g/mol
- Molar of H = 2 × 1 = 2 g/mol
- Mass of H =?
Mass of H = (2 / 18) × 4
Mass of H = 0.44 g
How to determine the mass of O
- Mass of compound = 4.30 g
- Mass of C = 2.67 g
- Mass of H = 0.44 g
- Mass of O =?
Mass of O = (mass of compound) – (mass of C + mass of H)
Mass of O = 4.30 – (2.67 + 0.44)
Mass of O = 1.19 g
<h3>How to determine the empirical formula </h3>
The empirical formula of the compound can be obtained as follow:
- C = 2.67 g
- H = 0.44 g
- O = 1.19 g
- Empirical formula =?
Divide by their molar mass
C = 2.67 / 12 = 0.2225
H = 0.44 / 1 = 0.44
O = 1.19 / 16 = 0.074
Divide by the smallest
C = 0.2225 / 0.074 = 3
H = 0.44 / 0.0744 = 6
O = 0.074 / 0.074 = 1
Thus, the empirical formula of the compound is C₃H₆O
Learn more about empirical formula:
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The reaction is:
Cl2 + 2 KBr --> 2 KCl + Br2
Moles of KCl is
n = m /M = 12 /74 = 0.16 mol
As, twice the moles of KCl is producing from 1 mol of chlorine
mole of Cl2 = 0.16 /2 = 0.08 mol
Mass of Cl2
m /70 = 0.08 = 5.6 g
Hence, 5.6 g mol Cl2 consumed to produce KCl