Answer:
See explanation
Explanation:
Molar mass is obtained as the sum of relative atomic masses as follow;
For CaBr = 40.08 + 79.90 = 119.98 g/mol
For BeBr = 9.012 + 79.90 = 88.912 g/mol
For CdBr2 = 112.41 + 2(79.90) = 272.21 g/mol
For CuBr2 = 63.55 + 2(79.90) = 223.35 g/mol
The answer would be 425.599 because 1 ATM is 760 mmHg.
Answer:
A. The rate of heat transfer through the material would increase.
Explanation:
To calculate the heat transfer in a heat exchanger you decide that there is not heat leakage to the surroundings, that means that magnitude of the two transfer rates will be equal. Any heat lost by the hot fluid, is gained by the cold fluid. The equation that describes this is Q = m×Cp×dT
Where:
heat = mass flow ×specific heat capacity × temperature difference
So if we increase the rate of flow of cooling water and the other variables that ypu can control remain the same, the result is that the rate of heat transfer through the material would increase, as it is stated in option a.
calculate moles of both reagents given and the moles of FeS that each of them would form if they were in excess
moles = mass / molar mass
moles Fe = 7.62 g / 55.85 g/mol
= 0.1364 moles
1 mole Fe produces 1 mole FeS
Therefore 7.62 g Fe can form 0.1364 moles FeS
moles S = 8.67 g / 32.07 g/mol
= 0.2703 moles S
1 mole S can from 1 moles FeS
So 8.67 g S can produce 0.2703 moles FeS
The limiting reagent is the one that produces the least product. So Fe is limiting.
The maximum amount of FeS possible is from complete reaction of all the limiting reagent.
We have already determined that the Fe can form up to 0.1364 moles of FeS, so this is max amount of FeS you can get.
Convert to mass
hope this helps :)
Answer:
The correct answer is 10.939 mol ≅ 10.94 mol
Explanation:
According to Avogadro's gases law, the number of moles of an ideal gas (n) at constant pressure and temperature, is directly proportional to the volume (V).
For the initial gas (1), we have:
n₁= 1.59 mol
V₁= 641 mL= 0.641 L
For the final gas (2), we have:
V₂: 4.41 L
The relation between 1 and 2 is given by:
n₁/V₁ = n₂/V₂
We calculate n₂ as follows:
n₂= (n₁/V₁) x V₂ = (1.59 mol/0.641 L) x 4.41 L = 10.939 mol ≅ 10.94 mol
