Mass of PH3= 6.086 g
<h3>Further explanation</h3>
Given
6.0 L of H2
Required
mass of PH3
Solution
Reaction
P4 + 6H2 → 4PH3
Assumed at STP ( 1 mol gas=22.4 L)
Mol of H2 for 6 L :
= 6 : 22.4 L
= 0.268
From the equation, mol PH3 :
= 4/6 x moles H2
= 4/6 x 0.268
= 0.179
Mass PH3 :
= 0.179 x 33,99758 g/mol
= 6.086 g
Answer:
1.35 g
Explanation:
water is h2o, so the molar mass is 1.01x2+16.00=18.02. divide 4.5 x 10^22 by 6.022 x 10^23 to get 7.5 x 10^-2 (2 sig figs). 18.02 x 7.5 x 10^-2 is 1.35 g
Answer:
3.07 Cal/g
Explanation:
Step 1: Calculate the heat absorbed by the calorimeter
We will use the following expression.
Q = C × ΔT
where,
- C: heat capacity of the calorimeter (37.60 kJ/K = 37.60 kJ/°C)
- ΔT: temperature change (2.29 °C)
Q = 37.60 kJ/°C × 2.29 °C = 86.1 kJ
According to the law of conservation of energy, the heat released by the candy has the same magnitude as the heat absorbed by the calorimeter.
Step 2: Convert 86.1 kJ to Cal
We will use the conversion factor 1 Cal = 4.186 kJ.
86.1 kJ × 1 Cal/4.186 kJ = 20.6 Cal
Step 3: Calculate the number of Cal per gram of candy
20.6 Cal/6.70 g = 3.07 Cal/g