Answer:
a) 38.2 % mass
b) 61.8 g solute/100 g solvent
c) 1.65 g/mL
Explanation:
Given the data:
mass of solute = 17.5 g
mass of solvent= 28.3 g
total solution volume= 27.8 mL
a)- mass percent= mass of solute/mass of solution x 100
mass of solution = mass solute + mass solvent = 17.5 g + 28.3 g = 45.8 g
mass % = 17.5 g/45.8 g x 100 = 38.2 % mass
b)- solubility = grams of solute/ 100 g solvent
= 17.5 g x (100 g /28.3 g solvent) = 61.8 g solute/100 g solvent
c)- density = massof solution/total volumesolution = 45.8 g/27.8 mL = 1.65 g/mL
1.04gK*1molK/39.01g K= 0.0267 mol K
0.70gCr*1mol/52.0g Cr = <span>0.0135 mol Cr
0.86 gO* 1 mol/16.0 g O = 0.0538 mol O
</span>0.0267 mol K/0.0135 = 2 mol K
0.0135 mol Cr /0.0135= 1 mol Cr
0.0538 mol O/0.035= 4 mol Cr
K2CrO4
Answer:
The final pressure is approximately 0.78 atm
Explanation:
The original temperature of the gas, T₁ = 263.0 K
The final temperature of the gas, T₂ = 298.0 K
The original volume of the gas, V₁ = 24.0 liters
The final volume of the gas, V₂ = 35.0 liters
The original pressure of the gas, P₁ = 1.00 atm
Let P₂ represent the final pressure, we get;



∴ The final pressure P₂ ≈ 0.78 atm.
Answer:
Equilibrium constant for
is 0.5
Equilibrium constant for decomposition of
is 
Explanation:
dissociates as follows:

initial 0.72 mol 0 0
at eq. 0.72 - 0.40 0.40 0.40
Expression for the equilibrium constant is as follows:
![k=\frac{[PCl_3][Cl_2]}{[PCl_5]}](https://tex.z-dn.net/?f=k%3D%5Cfrac%7B%5BPCl_3%5D%5BCl_2%5D%7D%7B%5BPCl_5%5D%7D)
Substitute the values in the above formula to calculate equilibrium constant as follows:
![k=\frac{[0.40/1][0.40/1]}{0.32/1} \\=\frac{0.40 \times 0.40}{0.32} \\=0.5](https://tex.z-dn.net/?f=k%3D%5Cfrac%7B%5B0.40%2F1%5D%5B0.40%2F1%5D%7D%7B0.32%2F1%7D%20%5C%5C%3D%5Cfrac%7B0.40%20%5Ctimes%200.40%7D%7B0.32%7D%20%5C%5C%3D0.5)
Therefore, equilibrium constant for
is 0.5
Now calculate the equilibrium constant for decomposition of 
It is given that
is decomposed.
decomposes as follows:

initial 1.0 M 0 0
at eq. concentration of
is:
![[NO_2]_{eq}=1-(0.000066) = 0.999934\ M](https://tex.z-dn.net/?f=%5BNO_2%5D_%7Beq%7D%3D1-%280.000066%29%20%3D%200.999934%5C%20M)
![[NO]_{eq}=6.6 \times 10^{-5}\ M](https://tex.z-dn.net/?f=%5BNO%5D_%7Beq%7D%3D6.6%20%5Ctimes%2010%5E%7B-5%7D%5C%20M)
Expression for equilibrium constant is as follows:
![K=\frac{[NO]^2[O_2]}{[NO_2]^2}](https://tex.z-dn.net/?f=K%3D%5Cfrac%7B%5BNO%5D%5E2%5BO_2%5D%7D%7B%5BNO_2%5D%5E2%7D)
Substitute the values in the above expression
![K=\frac{[6.6\times 10^{-5}]^2[3.3 \times 10^{-5}]}{[0.999934]^2} \\=1.79\times 10^{-14}](https://tex.z-dn.net/?f=K%3D%5Cfrac%7B%5B6.6%5Ctimes%2010%5E%7B-5%7D%5D%5E2%5B3.3%20%5Ctimes%2010%5E%7B-5%7D%5D%7D%7B%5B0.999934%5D%5E2%7D%20%5C%5C%3D1.79%5Ctimes%2010%5E%7B-14%7D)
Equilibrium constant for decomposition of
is 
The wording makes this question difficult, but HCl is found in stomach acid. Hope this helps!