Answer:
Step-by-step explanation:
Here we have the mass of CO₂ added = 340 g
From
![Number \, of \, moles = (Mass)/(Molar \, mass)](https://img.qammunity.org/2021/formulas/chemistry/high-school/qnpxqvhocxz6cwsou8cpxgfjjnov61cssq.png)
We have, where the molar mass of CO₂ is 44.01 g/mol
Therefore,
![Number \, of \, moles = (340)/(44.01) = 7.73 \, \, \, moles](https://img.qammunity.org/2021/formulas/chemistry/high-school/ca0cef4dzn6m35x8wufaozfvgs9atute3p.png)
71. Included drawing attached
72. Here we have the pressure of the gas given by Charles law which can be resented as follows;
![(P_1)/(P_2) =(T_1)/(T_2)](https://img.qammunity.org/2021/formulas/chemistry/high-school/w753rna1c9e2dsdt0yx38v3ue6b2ygm69j.png)
Where:
P₁ = Initial pressure = 6.1 atmospheres
P₂ = Final pressure
T₁ = Initial Temperature = 293 K
T₂ = Initial Temperature = 313 K
Therefore,
![P_2= T_2 * (P_1)/(T_1) = 313 * (6.1)/(293) = 3.312 \, \, \, atmospheres](https://img.qammunity.org/2021/formulas/chemistry/high-school/cfzad3q6ezp4e5wu8zo0l63vj4ow16zlsl.png)