Answer:
NH3
Explanation:
2NH3(aq)+CO2(aq)→CH4N2O(aq)+H2O(l)
So for two moles of NH3 we need one mole of CO2. So let's count moles for each reagent.
n(NH3)=m(NH3)/M(NH3)=135700/17,03=7968.29 mol
n(CO2)=m(CO2)/M(CO2)=211400/44.01=4803.45 mol
From equation we have to divide n(NH3) by 2 because we need two equivalent per one CO2. That will be 3984.145. So the limiting agent is NH3 because it's not enough of it to react with all CO2
Answer:
We can use heat = mcΔT to determine the amount of heat, but first we need to determine ΔT. Because the final temperature of the water is 55°C and the initial temperature is 20.0°C, ΔT is as follows:
ΔT = Tfinal − Tinitial = 55.0°C − 20.0°C = 35.0°C
given the specific heat of water as 1 cal/g·°C. Substitute the known values into heat = mcΔT and solve for amount of heat:
= heat=(75.0 g)(1 cal/ g· °C )(35.0°C) =
= 75x1x35=2625 cal
Thick liquid lava
It's right
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Answer:
Kp = 8.76×10⁻³
Explanation:
We determine the carbamate decomposition in equilibrium:
NH₄CO₂NH₂ (s) ⇄ 2NH₃(g) + CO₂(g)
Let's build the expression for Kp
Kp = (Partial pressure NH₃)² . Partial pressure CO₂
We do not consider, the carbamate because it is solid and we only need the partial pressure from gases
Kp = (0.370atm)² . 0.0640 atm
Kp = 8.76×10⁻³
Remember Kp does not carry units
