The given question is incorrect. The correct question is as follows.
If 20.0 g of 
and 4.4 g of 
are placed in a 5.00 L container at 
, what is the pressure of this mixture of gases?
Explanation:
As we know that number of moles equal to the mass of substance divided by its molar mass.
Mathematically, No. of moles = 
Hence, we will calculate the moles of oxygen as follows.
No. of moles =
Moles of = 
= 0.625 moles
Now, moles of 
= 0.1 moles
Therefore, total number of moles present are as follows.
Total moles = moles of + moles of
= 0.625 + 0.1
= 0.725 moles
And, total temperature will be:
T = (21 + 273) K = 294 K
According to ideal gas equation,
PV = nRT
Now, putting the given values into the above formula as follows.
P = 
= 
= 
atm
= 3.498 atm
or, = 3.50 atm (approx)
Therefore, we can conclude that the pressure of this mixture of gases is 3.50 atm.
The climate would become colder...
Answer:
It would move either left or right
Explanation: Taking assumption that,
Fructose + ATP fructose - 6 - phosphate + ADP (The standard free energy of hydrolysis for fructose-6-phosphate is - 15.9 kJ/mol.) 3 - phosphoglycerate + ATP 1,3 - bisphosphoglycerate + ADP (The standard free energy of hydrolysis for 1,3-bisphosphoglycerate is - 4 9.3 kJ/mol.) pyruvate + ATP phosphoenolpyruvate + ADP (The standard free energy of hydrolysis for phosphoenolpyruvate -is -61.9 kJ/mol.)
Answer:
Explanation:
Given that:
Half life = 30 min
Where, k is rate constant
So,
The rate constant, k = 0.0231 min⁻¹
Using integrated rate law for first order kinetics as:
![[A_t]=[A_0]e^{-kt}](https://tex.z-dn.net/?f=%5BA_t%5D%3D%5BA_0%5De%5E%7B-kt%7D)
Where,
![[A_t]](https://tex.z-dn.net/?f=%5BA_t%5D)
is the concentration at time t
![[A_0]](https://tex.z-dn.net/?f=%5BA_0%5D)
is the initial concentration
Given that:
The rate constant, k = 0.0231 min⁻¹
Initial concentration = 7.50 mg
Final concentration = 0.25 mg
Time = ?
Applying in the above equation, we get that:-
