Answer:
0.78 atm
Explanation:
Step 1:
Data obtained from the question. This includes:
Mass of CO2 = 5.6g
Volume (V) = 4L
Temperature (T) =300K
Pressure (P) =?
Step 2:
Determination of the number of mole of CO2.
This is illustrated below:
Mass of CO2 = 5.6g
Molar Mass of CO2 = 12 + (2x16) = 12 + 32 = 44g/mol
Number of mole CO2 =?
Number of mole = Mass/Molar Mass
Number of mole of CO2 = 5.6/44
Number of mole of CO2 = 0.127 mole
Step 3:
Determination of the pressure in the container.
The pressure in the container can be obtained by applying the ideal gas equation as follow:
PV = nRT
The gas constant (R) = 0.082atm.L/Kmol
The number of mole (n) = 0.127 mole
P x 4 = 0.127 x 0.082 x 300
Divide both side by 4
P = (0.127 x 0.082 x 300) /4
P = 0.78 atm
Therefore, the pressure in the container is
this statement is true
Answer:
<em>Hope </em><em>it </em><em>helps </em><em>u</em>
Answer:
Molar heat enthalpy of KBr = -19.89 kJ/mol
Explanation:
Change in temperature (Δt) = 0.370 K
Heat capacity = 2.71 kJ ⋅ K^-1
Heat absorbed by calorimeter = heat capacity × change in temperature
= 2.71× 0.370
= 1.0027 kJ
Molar mass of KBr = 119 g/mol
No. of moles of KBr = 6.00/119
= 0.0504 mol
Heat absorbed by the calorimeter is given by KBr.
Now calculate the heat released by per mol of KBr as follows:
Heat released by per mol of KBr = 1.0027 kJ / 0.0504 mol
=19.89 kJ/mol
Heat is released therefore, sign will be negative.
Molar heat enthalpy of KBr = -19.89 kJ/mol
Answer:
0.221M
Explanation:
From the question ,
The Molarity of AgNO₂ = 0.310 M
Hence , the concentration of Ag⁺ = 0.301 M
The volume of AgNO₂ = 250 mL
and,
The Molarity of Sodium chromate = 0.160 M
The volume of Sodium chromate = 100 mL.
As the solution is mixed the final volume becomes ,
250mL +100mL = 350mL
Now, using the formula , to find the final molarity of the mixture ,
M₁V₁ ( initial ) = M₂V₂ ( final )
substituting the values , in the above equation ,
0.310M * 250ml = M₂ * 350ml
M₂ = 0.221M
Hence , the concentration of the silver in the final solution = 0.221M
In all chemical equations, regardless of the type of reaction that is occurring, the reagents that are on the left side of the arrow are the reactants and on the right your products.
B. Is the solution.
