Water freezes at 0°C, 32°F, and 273°K. The only temperature warmer than the freezing point is 1°C.
Answer:
1.0 ° C
Explanation:
The molar mass for Sodium Nitrate NaNO₃ = (23+14+(16×3)) = 85
Number of moles of NaNO₃ = mass of NaNO₃ /molar mass of NaNO₃
⇒ 17/85 = 1.38 moles
Since 1 mole of NaNO₃ dissolved in 1 cubic decimeter of water, 40 kJ of heat energy is absorbed.
when 1.38 mole of NaNO₃ dissolved in 1 cubic decimeter of water, x kJ of heat energy is absorbed..
Then; x kJ of 1.38 mole of NaNo₃ = 1.38 × 40 kJ =55.2 kJ of heat absorbed.
Using the relation : Q = mcΔT to determine the temperature drop ; we get:
55.2 = 17 × 4 (ΔT)
55.2 = 68 ΔT
ΔT= 0.8 ° C
ΔT ≅ 1.0 ° C
Therefore, the drop in temperature when 17.0g of sodium nitrate is dissolved in 1 cubic decimeter of water is 1.0 ° C
Answer:
All other units are gotten from them
