<h3>
Answer:</h3>
5.6 L
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Explanation:</h3>
We are given;
- Initial volume, V1 = 3.5 L
- Initial pressure, P1 = 0.8 atm
- Final pressure, P2 = 0.5 atm
We are required to calculate the final volume;
- According to Boyle's law, the volume of a fixed mass of a gas and the pressure are inversely proportional at a constant temperature.
- That is; P α 1/V
- Mathematically, P=k/V
- At two different pressure and volume;
P1V1 = P2V2
In this case;
Rearranging the formula;
V2 = P1V1 ÷ P2
= (0.8 atm × 3.5 L) ÷ 0.5 atm
= 5.6 L
Therefore, the resulting volume is 5.6 L
sodium cloride is salt created from sodium Na and chlorine Ci
Na-sodium Ca- calcium
Ci-chlorine FL- flerovium
Ca- calcium Br-bromine
H- hydrogen He-helium
Answer:
- 13.56 g of sodium chloride are theoretically yielded.
- Limiting reactant is copper (II) chloride and excess reactant is sodium nitrate.
- 0.50 g of sodium nitrate remain when the reaction stops.
- 92.9 % is the percent yield.
Explanation:
Hello!
In this case, according to the question, it is possible to set up the following chemical reaction:

Thus, we can first identify the limiting reactant by computing the yielded mass of sodium chloride, NaCl, by each reactant via stoichiometry:

Thus, we infer that copper (II) chloride is the limiting reactant as it yields the fewest grams of sodium chloride product. Moreover the formed grams of this product are 13.56 g. Then, we take 13.56 g of sodium chloride to compute the consumed mass sodium nitrate as it is in excess:

Therefore, the leftover of sodium nitrate is:

Finally, the percent yield is computed via:

Best regards!
Answer:
1.63 kilo Joules of energy will be produced when 14.8 g of silver nitrate react.
Explanation:

Mass of silver nitrate = 14.8 g
Moles of silver nitrate = 
According to reaction, when 1 mole of silver nitrate reacts with 1 mole of calcium it gives 18.7 kJ of heat.
Then amount of heat released when 0.08706 moles of silver nitrate reacts :

1.63 kilo Joules of energy will be produced when 14.8 g of silver nitrate react.