Answer:
ΔU = 25.8 J
Explanation:
The gas absorbs 33.3 J of heat, that is, Q = 33.3 J.
The work (W) of expansion can be calculated using the following expression:
W = -P. ΔV
where,
P is the external pressure
ΔV is the change in volume
W = -1.45 × 10⁴ N . m⁻² × (8.40 × 10⁻⁴ m³ - 3.24 × 10⁻⁴ m³) = -7.48 J
The change in the internal energy (ΔU) is:
ΔU = Q + W
ΔU = 33.3 J + (-7.48 J) = 25.8 J
I’m pretty sure it’s abode and cathode
Answer:
250 mL (total solution) = 104 mL (stock solution) + 146 mL (water)
Explanation:
Data Given
M1 = 6.00 M
M2 = 2.5 M
V1 = 250 mL
V2 = ?
Solution:
As the chemist needs to prepare 250 mL of solution from 6.00 M ammonium hydroxide solution to prepare a 2.50 M aqueous solution of ammonium hydroxide.
Now
first he have to determine the amount of ammonium hydroxide solution that will be taken from6.00 M ammonium hydroxide solution
For this Purpose we use the following formula
M1V1=M2V2
Put values from given data in the formula
6 x V1 = 2.5 x 250
Rearrange the equation
V1 = 2.5 x 250 /6
V1 = 104 mL
So 104 mL is the volume of the solution which we have to take from the 6.00 M ammonium hydroxide solution to prepare 2.5 M aqueous solution of ammonium hydroxide
But we have to prepare 250 mL of the solution.
so the chemist will take 104 mL from 6.00 M ammonium hydroxide solution and have to add 146 mL water to make 250 mL of new solution.
in this question you have to tell about the amount of water that is 146 mL
250 mL (total solution) = 104 mL (stock solution) + 146 mL (water)
The answer is Rutherford.
Answer:
see explanation
Explanation:
To determine limiting reactant divide mole quantities of reactants by the respective coefficient in the balanced equation. The smaller value is the limiting reactant.
P₄ + 5O₂ => 2P₂O₅
12/1 = 12 15/5 = 3
O₂ is the limiting reactant. P₄ will be in excess when rxn stops.
