Answer:
When the pressure increases to 2.35 atm, the temperature will increase to 378 K
Explanation:
Step 1: Data given
The initial pressure = 1.82 atm
The initial temperature = 293 K
The pressure will be increased to 2.35 atm
Step 2: Calculate the new temperature
P1/T1 = P2/T2
⇒with P1 = the initial pressure = 1.82 atm
⇒with T1 = the initial temperature = 293 K
⇒with P2 = the increased pressure = 2.35 atm
⇒with T2 = the new temperature = TO BE DETERMINED
1.82atm / 293 K = 2.35 atm / T2
T2 = 2.35 atm / (1.82 atm/293 K)
T2 = 2.35 / 0.0062116
T2 = 378 K
When the pressure increases to 2.35 atm, the temperature will increase to 378 K
Answer:
ΔG = 18KJ/mol
Explanation:
Given data:
ΔS = 0.09 Kj/mol.K
ΔH = 27 KJ/mol
Temperature = 100 K
ΔG = ?
Solution:
Formula:
ΔG = ΔH - TΔS
ΔH = enthalpy
ΔS = entropy
by putting values,
ΔG = 27 KJ/mol - 100K(0.09 Kj/mol.K)
ΔG = 27 KJ/mol - 9 KJ/mol
ΔG = 18KJ/mol
Lord Kelvin, were he alive today, would be considered a Thermochemist. Thermochemistry is interested in the role of heat in chemical reactions. This includes the role of heat both as a biproduct of chemical reactions and a facilitator.
Kelvin's description of absolute zero is an important concept in thermochemistry. At absolute zero, there is no movement of molecules, and no energy available facilitate chemical reactions.
Answer:
The volume of water to be added is 0.175 liters of water
Explanation:
The given concentration of the nitric acid = 55% (M/M)
The mass of the nitric acid solution = 100 gm
The concentration solution is to diluted to = 20% (M/M)
The 100 g 55%(M/M) nitric acid solution gives 55g nitric acid in 100 g of solution
Therefore, to have 20% (M/M) nitric acid solution with the 55 g nitric acid, we get
Let "x" represent the volume of the resulting solution, we have;
20% of x = 55 g of nitric acid
∴ 20/100 × x = 55 g
x = 55 g × 100/20 = 275 g
The mass of extra water to be added = The mass of the 20%(M/M) solution solution of nitric acid - The current mass of the 55%(M/M) solution of nitric acid
The mass of extra water to be added = 275 g - 100 g = 175 g
Volume = Mass/Density
The density of water ≈ 1 g/ml
∴ The volume of water to be added that gives 175 g of water = 175 g/(1 g/ml) = 175 ml. = 0.175 l
The volume of water to be added = 0.175 liters of water.
