What Le Chatelier’s factors cause an increase in the concentration of Hb(CO)4(aq) in the Hb(CO)4(aq) + 4O2(g)Hb(O2)4(aq) + 4CO(g
1 answer:
You might be interested in
Answer : The specific heat of this substance is, 1.13 J/g.K
Explanation :
Molar heat capacity : It is defined as the amount of heat absorbed by one mole of a substance to raise its temperature by one degree Celsius.
As we are given that the chemical formula of compound is,
First we have to calculate the molar mass of given compound.
Molar mass of =
Molar mass of = 78 g/mol
Now we have to calculate the specific heat of this substance.
Specific heat of this substance =
Specific heat of this substance =
Specific heat of this substance = 1.13 J/g.K
Thus, the specific heat of this substance is, 1.13 J/g.K
<span>Molar mass (MM) of benzene C6H6
C = 6 * 12 = 72u
H = 6 * 1 = 6u
MM C6H6 = 72 + 6 = 78 g / mol
Benzene - Molar Mass = 78 g --------- 1 mol
Of A Mix has 468 g -------------- x
78x = 468
X = 468/78
X = 6 moles
Molar mass (MM) of Hydrochloric Acid HCl H = 1 * 1 = 1u
CI = 1 * 35 = 35u
MM HCl = 1 + 35 = 36 g / mol
Hydrochloric Acid - Molar Mass = 36 g ---------- 1 mol
Of A Mix has 72 g ------------ y
36y = 72
Y = 72/36
Y = 2 moles
Thus, a mixture has a total of 8 moles (6 mol + 2 mol).
Dividing One Mole Amount of Each Substance by the Number of Total Mole Amounts, Then we will obtain a Molar Fraction of Each:
Molar fraction make benzene = (6/8) simplify 2 = 3/4
Molar Fraction to make Hydrochloric Acid = (2/8) = simplify 2 = 1/4
Note:. The sum of the molar fractions of the always give goes 1, we have: 3/4 + 1/4 = 1
ANSWER:
</span>
C = 6 * 12 = 72u
H = 6 * 1 = 6u
MM C6H6 = 72 + 6 = 78 g / mol
Benzene - Molar Mass = 78 g --------- 1 mol
Of A Mix has 468 g -------------- x
78x = 468
X = 468/78
X = 6 moles
Molar mass (MM) of Hydrochloric Acid HCl H = 1 * 1 = 1u
CI = 1 * 35 = 35u
MM HCl = 1 + 35 = 36 g / mol
Hydrochloric Acid - Molar Mass = 36 g ---------- 1 mol
Of A Mix has 72 g ------------ y
36y = 72
Y = 72/36
Y = 2 moles
Thus, a mixture has a total of 8 moles (6 mol + 2 mol).
Dividing One Mole Amount of Each Substance by the Number of Total Mole Amounts, Then we will obtain a Molar Fraction of Each:
Molar fraction make benzene = (6/8) simplify 2 = 3/4
Molar Fraction to make Hydrochloric Acid = (2/8) = simplify 2 = 1/4
Note:. The sum of the molar fractions of the always give goes 1, we have: 3/4 + 1/4 = 1
ANSWER:
</span>