Answer:
16 g/L
Explanation:
Given that:-
Mass of sucrose added = 1.2 g
Volume of water = 75 mL = 0.075 L ( 1mL = 0.001 L )
%w/v is defined as the mass of the solution in 1 L of the solution.
So,
It will also be halved... but the relationship is with Kelvin not c. So 273 +273 = 546 K. half of that is 273 K, then subtract 273... you get 0 degrees c
I’m pretty sure the immediate answer is 2. Using the equation 60N=(30kg)x2(whatever unit of measure) also it will change because if you change the mass or the Net Force, your going to have to redo the equation based on the new information.
Answer:
305 litres of NO gas will be produced from 916 L of NO₂
Explanation:
Given the balanced equation of the chemical reaction as follows:
3 NO₂ (g) + H₂O( l) —— 2 HNO₃ (l) + NO (g)
Under standard conditions, 3 moles of No₂ will react with 1 mole of water to produce 1 mole of NO gas.
Molar volume of a gas at STP is 22.4 L
Number of moles of NO₂ gas present in 916 L = 916/22.4 = 40.893 moles of NO₂ gas
From the mole ratio of NO₂ to NO in the equation of reaction,
Number of moles of NO that will be produced = 1/3 × 40.893 moles = 13.631 moles of NO gas
Volume of 13.631 moles of NO gas = 13.631 × 22.4
Volume of NO gas produced = 305.334L
Therefore, Volume of NO gas produced from the reaction of 916 L of NO₂ with water = 305 L
Depression in freezing point (Δ
) =
×m×i,
where,
= cryoscopic constant =
,
m= molality of solution = 0.0085 m
i = van't Hoff factor = 2 (For
)
Thus, (Δ
) = 1.86 X 0.0085 X 2 =
Now, (Δ
) =
- T
Here, T = freezing point of solution
= freezing point of solvent =
Thus, T =
- (Δ
) = -
