According to john dalton, elements consist of tiny particles called atoms. Atoms of each element are similar to each other. Compounds consisted of atoms of different elements combined together.
A proton is the same as an H+ ion, and Arrhenius acids are the ones that release H+ in solution, so the answer is A
Answer:
The equation that gives the overall equilibrium in terms of the equilibrium constants K and Ky is K1 = K^6 * Ky
Explanation:
we have the following balanced reaction:
CaC2 + 2H2O = C2H2 + Ca(OH)2
the value of K for this reaction will be equal to:
K = ([C2H2] * [Ca(OH)2])/([CaC2] * [H2O]^2)
if we multiply the reaction by the value of 6, we have:
6CaC2 + 12H2O = 6C2H2 + 6Ca(OH)2
Again, the value of K for this reaction will be equal to:
K,´ = ([C2H2] ^6 * [Ca(OH)2]^6)/([CaC2]^6 * [H2O]^12) = K^6
For the second reaction:
6C2H2 + 3CO2 + 4H2O = 5CH2CHCO2H
The value of K for this reaction:
K2 = ([CH2CHCO2H]^5)/([C2H2]^6 * [CO2]^3 * [H2O]^4)
we also have:
K1 = ([CH2CHCO2H]^5)/([C2H2]^6 * [CO2]^3 * [H2O]^16)
Thus:
K1 = K^6 * Ky
Answer:
The combustion of 59.7 grams of methane releases 3320.81 kilojoules of energy
Explanation:
Given;
CH₄ + 2O₂ → CO₂ + 2H₂O, ΔH = -890 kJ/mol
From the combustion reaction above, it can be observed that;
1 mole of methane (CH₄) released 890 kilojoules of energy.
Now, we convert 59.7 grams of methane to moles
CH₄ = 12 + (1x4) = 16 g/mol
59.7 g of CH₄ 
1 mole of methane (CH₄) released 890 kilojoules of energy
3.73125 moles of methane (CH₄) will release ?
= 3.73125 moles x -890 kJ/mol
= -3320.81 kJ
Therefore, the combustion of 59.7 grams of methane releases 3320.81 kilojoules of energy
The runner ran a total of 8800 yards.
