I assume what you're asking about is, how does the temperature changes when we increase water's mass, according the formula for heat ?
Well the formula is :
(where Q is heat, m is mass, c is specific heat and
is change in temperature. So according this formula, increasing mass will increase the substance's heat, but won't effect it's temperature since they are not related. Unless, if you want to keep the substance's heat constant, in that case when you increase it's mass you will have to decrease the temperature
Answer : The compound that would be most soluble in water is CH3CH2CH2OH
Explanation :
Water is a polar solvent and can dissolve polar molecules. This is based on the principle "Like dissolves like".
Among the given molecules, CH3CH2CH2CH3 is a hydrocarbon known as butane. All hydrocarbons are non polar. Therefore this compound will not be soluble in water.
The remaining compounds are polar, but Ch3CH2CH2OH shows greater solubility in water owing to presence of hydrogen bonding.
Hydrogen bonding is a type of intermolecular force that gets formed when a compound has hydrogen atom directly attached to highly electro-negative N, F or O atom.
When CH3CH2CH2OH is dissolved in water, it forms hydrogen bonds with water molecules. Due to this hydrogen bonding, the molecule shows greater solubility.
Therefore CH3CH2CH2OH is the most soluble compound in water
Answer:
The answer to your question is: 0.028 kg of NO2
Explanation:
Data
3.7 x 10²⁰ molecules of NO2 in kg
MW of NO2 = 14 + (16 x 2) = 14 + 32 = 46 kg
1 mol of NO2 --------------------- 6.023 x 10 ²³ molecules
x --------------------- 3.7 x 10²⁰ molecules
x = 3.7 x 10²⁰ x 1 / 6.023 x 10 ²³
x = 0.00061 mol
1 mol of NO2 --------------------- 46 kg of NO2
0.00061 mol ------------------ x
x = 0.00061 x 46/1
x = 0.028 kg of NO2
Answer:
W = -120 KJ
Explanation:
Since the piston–cylinder assembly undergoes an isothermal process, then the temperature is constant.
Thus; T1 = T2 = 400K
change in entropy; ΔS = −0.3 kJ/K
Formula for change in entropy is written as;
ΔS = Q/T
Where Q is amount of heat transferred.
Thus;
Q = ΔS × T
Q = -0.3 × 400
Q = -120 KJ
From the first law of thermodynamics, we can find the workdone from;
Q = ΔU + W
Where;
ΔU is Change in the internal energy
W = Work done
Now, since it's an ideal gas model, the change in internal energy is expressed as;
ΔU = m•C_v•ΔT
Where;
m is mass
C_v is heat capacity at constant volume
ΔT is change in temperature
Now, since it's an isothermal process where temperature is constant, then;
ΔT = T2 - T1 = 0
Thus;
ΔU = m•C_v•ΔT = 0
ΔU = 0
From earlier;
Q = ΔU + W
Thus;
-120 = 0+ W
W = -120 KJ
