Answer:
1. Yes
2.The solubility of X is 34.55g/L
Explanation:
Solubility of solute refers to how readily a solute will dissolve in a solvent at a particular temperature. Its the amount of moles or grams required to saturate 1dm
or 1 Litre of water.
From the problem, when the liquid was drained off and amount of X which didn't dissolve was measured, it weighed 0.008kg, this means out of 0.027kg, 0.027-0.008 actually dissolved
= 0.019kg*1000 = 19g.
if 19g is required to saturate 550mL at 30°C,
then
will saturate 1L
= 34.545g will saturate 1Litre
The solubility thus is 34.55g/L
Answer:
I think the layers of the atmosphere does temperature increase with increasing height. It is affected by convection because it heats the lower atmosphere. It is affected by conduction because the is the transfer of thermal energy. I guess
Hope this help!
Answer:
4.214 × 10^23 molecules.
Explanation:
Number of molecules in a substance can be calculated by multiplying the number of moles in that substance by Avagadro's number, which is 6.02 × 10^23.
That is, no. of molecule = n × Avagadro constant
In this case, there are 0.7 moles of fructose. Hence;
number of molecules = 0.7 × 6.02 × 10^23
no. of molecule = 4.214 × 10^23 molecules.
Answer:
a. -29.8 kJ/mol-rxn
Explanation:
For a chemical reaction system the forward and reverse rate are equal. The standard molar enthalpy formation of NH3 is -45.9 kJ/mol. For the enthalpy of NH3 (8) the molar enthalpy is -29.8kJ/mol. The molar mass of N2 = 28.02g/mol. Molar enthalpy of formation is standard amount of substance produced in the formation of a reaction. The molar enthalpy is the change in enthalpy due to reaction per mole.
Answer:
W = -10.3 kJ
Explanation:
During combustion, the system performs work and releases heat. Therefore, the change in internal energy is negative, and the change in enthalpy, which is equal to heat at constant pressure, is also negative. Work is then calculated by rearranging the equation for the change in internal energy:
w=ΔE−qp=−5084.3 kJ−(−5074.0 kJ)
The release of heat is much greater than the work performed by the system on its surroundings. The potential energy stored in the bonds of octane explains why considerably large amounts of energy can be lost by the system during combustion.
