Answer:
= 61.25 g
= 88.75 g
Explanation:
= 
= 50 g
⇒ 
= 
= 1.25 (moles)
2NaOH + H2SO4 ⇒ Na2SO4 + 2H2O
2 : 1 : 1 : 2
1.25 (moles)
⇒ 
= 1.25 × 1 ÷ 2 = 0.625 (moles) ⇒ 
= 0.625 × 98 = 61.25 g
= 1.25 × 1 ÷ 2 = 0.625 (moles) ⇒ = 0.625 × 142 = 88.75 g
The solid form of a substance is usually more dense than its
liquid and gaseous forms. Similarly the liquid form is usually more dense than
the gaseous form. Ice floating in water is an exception that breaks the general
density rule. So option “A” is the correct option in regards to the given
question. In case of ice formation, actually the density of water decreases by
about 9%. This is the main reason behind ice floating in water. Pure water has
the maximum density at 4 degree centigrade.
Answer:
1. Hydrogen will diffuse faster.
2. The ratio of diffusion of hydrogen gas to that of the unknown gas is 4 : 1
Explanation:
Let the rate of diffusion of hydrogen gas, H2 be R1
Let the molar mass of H2 be M1
Let the rate of diffusion of the unknown gas be R2.
Let the molar mass of the unknown gas be M2.
Molar mass of H2 (M1) = 2x1 =2g/mol
Molar mass of unknown gas (M2) = 16 times that of H2
= 16 x 2 = 32g/mol
1. Determination of the gas that will diffuse faster. This is illustrated below:
R1/R2 = √(M2/M1)
R1/R2 = √(32/2)
R1/R2 = √16
R1/R2 = 4
Cross multiply
R1 = 4R2
From the above calculations, we can see that the rate of diffusion H2 (R1) is four times the rate of diffusion of the unknown gas (R2).
Therefore, hydrogen will diffuse faster.
2. Again, from the calculations made above, the ratio of diffusion of hydrogen (R1) to that of the unknown gas (R2) is given by;
R1/R2 = 4
Therefore, the ratio of diffusion of hydrogen (R1) to that of the unknown gas (R2) is:
4 : 1
I think it's 6?...........
Answer:
1. Options A and B
2. Options B and C
3.. B. Net ∆G = -16.7 KJ/mol; C. Net ∆G = -14.2 KJ/mol
Explanation:
1. The spontaneity of a chemical reaction depends on its standard free energy change, ∆G. If ∆G is negative, the reaction is favourable, but when it is positive, the reaction is unfavorable.
Therefore, since reaction A and B have ∆G to be positive, they are unfavorable
2. Coupling an unfavorable reaction to a favourable reaction can help the reaction to proceed in the forward direction as long as the net free energy change is negative.
Coupling reaction A and C, as well as reaction B and C will make the reactions to become favourable as net ∆G is negative in both instances.
3. A and C: net ∆G = 13.8 - 30.5 = -16.7 KJ/mol
B and C: net ∆G = 16.3- 30.5 = -14.2 KJ/mol
