Answer:
400°C
Explanation:
22,000 cal / (0.11 cal/g°C x 500 g) = 400°C
Answer: Enthalpy of combustion (per mole) of 
is -2657.5 kJ
Explanation:
The chemical equation for the combustion of butane follows:
The equation for the enthalpy change of the above reaction is:
![\Delta H^o_{rxn}=[(8\times \Delta H^o_f_{CO_2(g)})+(10\times \Delta H^o_f_{H_2O(g)})]-[(1\times \Delta H^o_f_{C_4H_{10}(g)})+(4\times \Delta H^o_f_{O_2(g)})]](https://tex.z-dn.net/?f=%5CDelta%20H%5Eo_%7Brxn%7D%3D%5B%288%5Ctimes%20%5CDelta%20H%5Eo_f_%7BCO_2%28g%29%7D%29%2B%2810%5Ctimes%20%5CDelta%20H%5Eo_f_%7BH_2O%28g%29%7D%29%5D-%5B%281%5Ctimes%20%5CDelta%20H%5Eo_f_%7BC_4H_%7B10%7D%28g%29%7D%29%2B%284%5Ctimes%20%5CDelta%20H%5Eo_f_%7BO_2%28g%29%7D%29%5D)
We are given:
Putting values in above equation, we get:
![\Delta H^o_{rxn}=[(8\times -393.5)+(10\times -241.82)]-[(2\times -125.6)+(4\times 0)]\\\\\Delta H^o_{rxn}=-5315kJ](https://tex.z-dn.net/?f=%5CDelta%20H%5Eo_%7Brxn%7D%3D%5B%288%5Ctimes%20-393.5%29%2B%2810%5Ctimes%20-241.82%29%5D-%5B%282%5Ctimes%20-125.6%29%2B%284%5Ctimes%200%29%5D%5C%5C%5C%5C%5CDelta%20H%5Eo_%7Brxn%7D%3D-5315kJ)
Enthalpy of combustion (per mole) of is -2657.5 kJ
Since there is one mole of Ca^2+ in calcium acetate, its concentration is 0.80 mol/L.
<h3>What is concentration?</h3>
The term concentration has to do with the amount of substance in solution. The concentration can be measured in several units. Generally, concentration is expressed in molarity, molality, mass concentration units or percentage.
Now we are asked to find the amount concentration of calcium ions and acetate ions in a 0.80 mol/L solution of calcium acetate. The formula of calcium acetate is Ca(CH3COO)2.
Thus;
Ca(CH3COO)2(aq) ----> Ca^2+(aq) + 2CH3COO^-(aq)
It then follows that since there is one mole of Ca^2+ in calcium acetate, its concentration is 0.80 mol/L.
Learn more about concentration:brainly.com/question/10725862
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Momentum = (mass) x (velocity) = (1,100) x (30) =
33,000
kg-m/sec due east
Answer:
convection
Explanation:
as air is heated, it rises up and away from the heat source, cools then falls closer to the heat source over and over again.
