Answer:1) Volume of 
required is 55.98 mL.
2) 0.62577 grams of 
is produced.
Explanation:
1) Molarity of 
Volume of 
Molarity of 
Volume of 
According to reaction, 1 mole of 
reacts with 3 mole of , then, 0.0041985 moles of will react with:
moles of that is 0.0125955 moles.
Volume of required is 55.98 mL.
2)
Number of moles of 
According to reaction, 3 moles of gives 1 mole of , then 0.004485 moles of will give:
moles of that is 0.001495 moles.
Mass of =
Moles of × Molar Mass of
= 0.001495 moles × 418.58 g/mol = 0.62577 g
0.62577 grams of is produced.
Answer:
The answer to your question is : letter B. 0.25 atm
Explanation:
To solve this problem we need to use the combined gas law:
<u>P₁V₁</u> = <u>P₂V₂</u>
T₁ T₂
Data
P1 = 0.99 atm V1 = 2 l T1 = 273K
P2 = ? V2 = 4 l T2 = 137K
Now, the clear P2 from the equation and we get
P2 = P1V1T2 / T1V2
Substitution P2 = (2 x 0.99 x 137)/(273 x 4)
P2 = 271.26 / 1092
Result P2 = 0.248 atm ≈ 0.25 atm
(the heat bomb calorimeter gained)=(the heat material A gave)
r.h.s=25.57 kJ/⁰C * (26.80-24.33)= 63.16 kJ
2.741g of material A made 63.16kJ
the heat of combustion per gram = 63.16/2.741 kJ/g =24.04kJ/g
Answer:
Compound C is covalent.
Explanation:
Among the properties of a covalent compound is that it has poor conductivity and could possibly be a gas at room temperature.
Answer:
2850 grams I hope this help
