Answer: 64.6 mmHg
Explanation:
Given that:
Volume of gas V = 3.47L
(since 1 liter = 1dm3
3.47L = 3.47dm3)
Temperature T = 85.0°C
Convert Celsius to Kelvin
(85.0°C + 273 = 358K)
Pressure P = ?
Number of moles of gas N = 0.100 mole
Note that Molar gas constant R is a constant with a value of 0.0082 ATM dm3 K-1 mol-1
Then, apply ideal gas equation
pV = nRT
p x 3.47dm3 = 0.10 x (0.0082 atm dm3 K-1 mol-1 x 358K)
p x 3.47dm3 = 0.29 atm dm3
p = (0.29 atm dm3 / 3.47 dm3)
p = 0.085 atm
Recall that pressure of the gas is required in mm hg, so convert 0.085 atm to mm Hg
If 1 atm = 760 mm Hg
0.085atm = 0.085 x 760
= 64.6 mm Hg
Thus, the pressure of the gas is 64.6 mm hg
The top number of a portion is called its numerator whereas the bottom number is its denominator. So a Fraction is the quantity of shaded parts separated by the quantity of a balance of as demonstrated as follows: number of shaded parts is the numerator over the whole part which is the denominator.
Answer:- C. H
Explanations:- Reduction is gain of electron. In other words we could say that decrease in oxidation number is reduction.
As per the rules, oxidation number of hydrogen in its compounds is +1(except metal hydrides) and the oxidation number of oxygen in its compounds is -2.
The oxidation number in elemental form is zero.
In 
, the oxidation number of H is +1 and oxidation number of O is -2. Oxidation number of Cl in 
is -1. On product side, the oxidation number of hydrogen in 
is zero and in 
the oxidation number of H is +1 and that of O is -2. Oxidation number of Cl in 
is 0.
From above data, Oxidation number of O is -2 on both sides so it is not reduced.
Oxidation number of Cl is changing from -1 to 0 which is oxidation.
Oxidation number of H is changing from +1 to 0 which is reduction.
So, the right choice is C.H
Answer:
8.68 moles of BaI₂
Explanation:
Given data:
Number of moles of BaI₂ = ?
Number of formula units = 5.23× 10²⁴
Solution:
By using Avogadro number,
1 mole of any substance contain 6.022× 10²³ formula units.
5.23× 10²⁴ formula units of BaI₂ × 1 mol / 6.022× 10²³ formula units
0.868 × 10¹ moles of BaI₂
8.68 moles of BaI₂
Thus, 5.23× 10²⁴ formula units of BaI₂ contain 8.68 moles of BaI₂
In total dominance, the phenotype sees only one allele in the genotype. Both alleles in the genotype are seen in the phenotype during codominance. In incomplete dominance, in the phenotype, a mixture of the alleles in the genotype is seen.
