The amount of energy that was used to heat Cu is 27.6 cal
<u><em>calculation</em></u>
Heat (Q) = M ( mass) x c(specific heat capacity) x ΔT( change in temperature)
where;
Q=?
M = 5.0 g
C=0.092 cal/g°C
ΔT = 80°c-20°c=60°C
Q is therefore = 5.0 g x 0.092 cal / g°c x 60°c =27.6 cal
Your answer is B. Rust does not rust.
Hi there!
Zinc: Is qualitative
Chlorine: is quantitative
Gallium: is neither
Nitrogen: is quantitative
Aluminum: is quantitative
If you need an explanation, please let me know !
Hope this helps and have a good day :) !
~Angel
Molarity = number of moles / volume in liters solution
Answer (4)
hope this helps!
Actually, you do NOT need to calculate the number of moles, so you do NOT need to convert the volume and pressure. (You DO have to convert the temperature.) That's because for this problem, pV/T is constant, so the units cancel (but the temperature scale matters).
<span>200kPa * 15L / 373K = 101kPa * V / 273K </span>
<span>V = 21.7 L </span>
