Answer:
47.9 g of ethanol
Explanation:
Combustion is a chemical reaction in which a substance reacts with oxygen to produce heat and light. Combustion reactions have been very useful as a source of energy. Ethanol is now burnt for energy purposes as a fuel. Ethanol has even been proposed as a possible alternative to fossil fuels.
Since 1 mole of ethanol when combusted releases 1367 kJ/mol of energy
x moles of ethanol releases 1418 kJ/mol.
x= 1 × 1418 kJ/mol/ 1367 kJ/mol
x= 1.04 moles of ethanol.
Mass of ethanol = number of moles × molar mass
Molar mass of ethanol = 46.07 g/mol
Mass of ethanol = 1.04 moles × 46.07 g/mol
Mass of ethanol= 47.9 g of ethanol
Answer:
[SO2Cl2] = = 0.015 M
[SO2] = = 0.0027 M
[Cl2] = = 0.0027 M
Q = = = 4.8 × 10−4
No. Q < Kc, so reaction will shift to the right.
Explanation:
Answer:
Las bebidas gaseosas como las gaseosas están hechas de un soluto de dióxido de carbono gaseoso en un líquido. La solubilidad del dióxido de carbono en el líquido depende de la presión y la temperatura de la lata de refresco, y también de agitar la lata de refresco que introduce burbujas que permanecen ocultas hasta que se abre la lata antes de que burbujee.
Por lo tanto, dado que la presión en la lata de refresco permanece constante, elevar la temperatura, agitar la lata de refresco o congelar el refresco, lo que aumenta la cantidad de dióxido de carbono en la porción líquida, hará que el refresco forme espuma y se derrame.
Explanation:
Answer:
See Explanation
Explanation:
Let us consider the first two reactions, the initial concentration of CO was held constant and the concentration of Hbn was doubled.
2.68 * 10^-3/1.34 * 10^-3 = 6.24 * 10^-4/3.12 * 10^-4
2^1 = 2^1
The rate of reaction is first order with respect to Hbn
Let us consider the third and fourth reactions. The concentration of Hbn is held constant and that of CO was tripled.
1.5 * 10^-3/5 * 10^-4 = 1.872 * 10^-3/6.24 * 10^-4
3^1 = 3^1
The reaction is also first order with respect to CO
b) The overall order of reaction is 1 + 1=2
c) The rate equation is;
Rate = k [CO] [Hbn]
d) 3.12 * 10^-4 = k [5 * 10^-4] [1.34 * 10^-3]
k = 3.12 * 10^-4 /[5 * 10^-4] [1.34 * 10^-3]
k = 3.12 * 10^-4/6.7 * 10^-7
k = 4.7 * 10^2 mmol-1 L s-1
e) The reaction occurs in one step because;
1) The rate law agrees with the experimental data.
2) The sum of the order of reaction of each specie in the rate law gives the overall order of reaction.
