The answer is true because if you look at the question what does it say
Answer : The mass of sodium bromide added should be, 18.3 grams.
Explanation :
Molality : It is defined as the number of moles of solute present in kilograms of solvent.
Formula used :

Solute is, NaBr and solvent is, water.
Given:
Molality of NaBr = 0.565 mol/kg
Molar mass of NaBr = 103 g/mole
Mass of water = 315 g
Now put all the given values in the above formula, we get:


Thus, the mass of sodium bromide added should be, 18.3 grams.
.5 mol of A will be left over since 1.5 mol of A will be used for every 3 mol of B due to the 2:1 ratio established by the formula.
Answer:
6,8 g
Explanation:
c = 4.18 J/(g * °C) = 4180 J / (kg * °C)
= 25 °C
= 36,4 °C
Q = 325 J
The formula is: Q = c * m * (
)
m =
Calculating:
m = 325 / 4180 * (36,4 - 25) ≈ 0,0068 kg = 6,8 g