Answer:
A or C because they both kinda are the same
Answer: 1.80g
Explanation:
Molar Mass of AlCl3 = 27 + (3x35.5)
= 27 + 106.5 = 133.5g/mol
Number of mole of AlCl3 = 0.0135mol
Mass = 0.0135 x 133.5= 1.80g
Answer:
1. Volume as STP = 755 L
2. Outside temperature = 255 K
3. Percentage yield = 70.5%
Explanation:
1. At STP, pressure = 101.3 kpa, temperature = 0°C or 273.15 K
Using the general gas equation :
P1V1/T1 = P2V2/T2
P1 = 620 kpa
V1 = 140 L
T1 = 37°C or (273.15 + 37) K = 310.15 K
P2 = 101.3 kpa
V2 = ?
T2 = 273.15 K
V2 = P1V1T2/P2T1
V2 = 620 × 140 × 273.15 / 101.3 × 310.15
V2 = 755 L
2. Using Charles' gas law:
V1/T1 = V2/T2
V1 = 2.5 L
T1 = 290 K
V2 = 2.2 L
T2 = ?
T2 = V2T1/VI
T2 = 2.2 × 290 / 2.5
T2 = 255 K
3. Equation of reaction : 2 Al + 3 CuSO4 ---> Al2 (SO4)3 + 3 Cu
From equation of the reaction, 2 moles of Al produces 3 moles of Cu
Molar mass of Al = 27 g; Molar mass of Cu = 63.5 g
2 moles of Al = 2 × 27 g = 54 g; 3 moles of Cu = 3× 63.5 = 190.5 g
54 g of Al produces 190.5 g of Cu
1.87 g of Al will produce 190.5/54 × 1.87 g of Cu = 6.60 g of Cu
Percentage yield = actual yield /theoretical yield × 100%
Percentage yield = 4.65/6.60 × 100%
Percentage yield = 70.5%
Answer:
Hypothesis
Explanation:
The following steps are applicable when we wish to prove a specific fact:
- a hypothesis is made; this is a statement that we provide after some observations and we wish to either prove or deny it;
- multiple experiments are carried out in order to gather significantly substantial amount of data that can be then further analyzed and any tendencies can be noticed;
- based on the data gathered, conclusions are made: we either prove or deny the hypothesis. If hypothesis is proved, it may become a theory over long time.
In the context of this problem, we're at the first step where we make a hypothesis.
Answer:
The combustion of 59.7 grams of methane releases 3320.81 kilojoules of energy
Explanation:
Given;
CH₄ + 2O₂ → CO₂ + 2H₂O, ΔH = -890 kJ/mol
From the combustion reaction above, it can be observed that;
1 mole of methane (CH₄) released 890 kilojoules of energy.
Now, we convert 59.7 grams of methane to moles
CH₄ = 12 + (1x4) = 16 g/mol
59.7 g of CH₄ 
1 mole of methane (CH₄) released 890 kilojoules of energy
3.73125 moles of methane (CH₄) will release ?
= 3.73125 moles x -890 kJ/mol
= -3320.81 kJ
Therefore, the combustion of 59.7 grams of methane releases 3320.81 kilojoules of energy
