Answer:
Explanation:
Increasing the temperature increases reaction rates because of the disproportionately large increase in the number of high energy collisions.
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Answer: -
The concentration is 0.91 mol/ L
Explanation: -
The osmotic pressure = 23 atm
Absolute temperature = 309 K
Universal gas constant = 0.08206 L atm mol⁻¹ K⁻¹
We know the formula
Osmotic pressure = molar concentration x universal gas constant x absolute temperature
Plugging in the values
23 atm = molar concentration x 0.08206 L atm mol⁻¹ K⁻¹ x 309 K
Molar concentration =
mol/ L
= 0.91 mol / L
Answer : The work, heat during the process and the change of entropy of the gas are, 0 J, 3333.003 J and -10 J respectively.
Explanation :
(a) At constant volume condition the entropy change of the gas is:

We know that,
The relation between the
for an ideal gas are :

As we are given :



Now we have to calculate the entropy change of the gas.


(b) As we know that, the work done for isochoric (constant volume) is equal to zero. 
(C) Heat during the process will be,

Therefore, the work, heat during the process and the change of entropy of the gas are, 0 J, 3333.003 J and -10 J respectively.
Answer:
Explanation:
Not Many
1 mol of CO has a mass of
C = 12
O = 16
1 mol = 28 grams.
1 mol of molecules = 6.02 * 10^23
x mol of molecules = 3.14 * 10^15 Cross multiply
6.02*10^23 x = 1 * 3.14 * 10^15 Divide by 6.02*10^23
x = 3.14*10^15 / 6.02*10^23
x = 0.000000005 mols
x = 5*10^-9
1 mol of CO has a mass of 28
5*10^-9 mol of CO has a mass of x Cross Multiply
x = 5 * 10^-9 * 28
x = 1.46 * 10^-7 grams
Answer: there are 1.46 * 10-7 grams of CO if only 3.14 * 10^15 molecules are in the sample