The question is incomplete,the complete question :
Calculate the molality of a 10.0% (by mass) aqueous solution of hydrochloric acid:
a) 0.274 m
b) 2.74 m
c) 3.05 m
d) 4.33 m
e) the density of the solution is needed to solve the problem
Answer:
The molality of a 10.0% (by mass) aqueous solution of hydrochloric acid is 3.05 mol/kg.
Explanation:
10.0% (by mass) aqueous solution of hydrochloric acid.
10 grams of HCl is present in 100 g of solution.
Mass of HCl = 10 g
Mass of solution = 100 g
Mass of solution = Mass of solute + Mass of water
Mass of water = 100 g - 10 g = 90 g
Moles of HCl = 
Mass of water in kilograms = 0.090 kg
Molality = 
The molality of a 10.0% (by mass) aqueous solution of hydrochloric acid is 3.05 mol/kg.
Its C. It has to be equivalent to the number divided or multiplied by 5.
The answer is color as the least reliable
Answer:
Water - H2O
Ammonia - NH3
Sulfur dioxide - SO2
Hydrogen sulfide - H2S
Ethanol - C2H6O
Explanation:
Those are some atoms with polar covalent bonds. Hope this helps!!
Answer:
(B) 0.038 M
Explanation:
Kc = [H2][I2]/[HI]^2
Let the equilibrium concentration of H2 be y M
From the equation of reaction, mole ratio of H2 to I2 formed is 1:1, therefore equilibrium concentration of I2 is also y M
Also, from the equation of reaction, mole ratio of HI consumed to H2 formed is 2:1, therefore equilibrium concentration of HI is (1 - 2y) M
1.6×10^-3 = y×y/(1 - 2y)^2
y^2/1-4y+4y^2 = 0.0016
y^2 = 0.0016(1-4y+4y^2)
y^2 = 0.0016 - 0.0064y + 0.0064y^2
y^2-0.0064y^2+0.0064y-0.0016 = 0
0.9936y^2 + 0.0064y - 0.0016 = 0
The value of y must be positive and is obtained by using the quadratic formula
y = [-0.0064 + sqrt(0.0064^2 - 4×0.9936×-0.0016)] ÷ 2(0.9936) = 0.0736 ÷ 1.9872 = 0.038 M
