Answer:
-2.79 × 10³ cal
Explanation:
Step 1: Given data
- Mass of water (m): 35.0 g
- Latent heat of fusion of water (L): -79.7 cal/g
Step 2: Calculate the heat required to freeze 35.0 g of water
We have 35.0 g of liquid water and we want to freeze it, that is, to convert it in 35.0 g of ice (solid water), at 0 °C (melting point). We can calculate the heat (Q) that must be released using the following expression.
Q = L × m
Q = -79.7 cal/g × 35.0 g
Q = -2.79 × 10³ cal
Answer:
there are no mutiple choices
The Answer would be e: 3.2 g/L
<span>At STP : p = 1 atm and T = 273.15 K
molar mass Cl2 = 70.906 g/mol
d = molar mass x p/RT = 70.906 x 1 atm / 0.08206 x 273.15=3.16 g/L</span>
Answer: but thx for the points
Explanation:
Here's link to the answer:
bit.
ly/3a8Nt8n
