To solve this problem, we establish the general energy balance:
ΔE = ΔU + ΔKE + ΔPE
ΔE = Q + W
Q + W = ΔU + ΔKE + ΔPE
In this case, ΔKE and ΔPE are both zero or negligible.
Given:
m = 33.0 grams of CO2
Tsub = 77 K
P = 1 atm
ΔE = Q + W
ΔE = mCpΔT + ΔPV
solve for mCpΔT, find the value of Cp for CO2, then solve for Q. Next, solve for W using the ideal gas law. Add the two values and that will be the value of the delta E.
Answer:fixed and dilted
Explanation:But am pretty sure that right
An inhibitor, which slows down the reaction enough to measure the release of gas.
Even if two samples of matter have the same temperature, it does not necessarily mean they have the same total energy. The more particles a substance has at a given temperature, the more thermal energy it has.
Brainliest Please!
