Answer: 52.17%
Explanation:
Number of matches in the box = 46
Number of matches that lit on the first strike = 22
Number of matches that did not light on the first strike = 46 - 22 = 24
Therefore, the percentage of the matches in the box did not light on the first strike will be:
= (Number of matches that did not light on the first strike / Number of matches in the box) × 100
= 24/46 × 100
= 52.17%
Therefore, the percentage of the matches in the box that did not light on the first strike is 52.17%.
The reaction between HNO3 and Ba(OH)2 is given by the equation below;
2HNO3 + Ba(OH)2 = Ba(NO3)2 + 2H2O
Moles of Barium hydroxide used;
= 0.200 × 0.039 l
= 0.0078 Moles
The mole ratio of HNO3 and Ba(OH)2 is 2: 1
Therefore; moles of nitric acid used will be;
= 0.0078 ×2 = 0.0156 moles
But; 0.0156 moles are equal to a volume of 0.10
The concentration of Nitric acid will be;
= (0.0156 × 1)/0.1
= 0.156 M
Answer:
ΔH0reaction = [ΔHf0 CO2(g)] - [ΔHf0 CO(g) + ΔHf0 O2(g)]
Explanation:
Chemical equation:
CO + O₂ → CO₂
Balanced chemical equation:
2CO + O₂ → 2CO₂
The standard enthalpy for the formation of CO = -110.5 kj/mol
The standard enthalpy for the formation of O₂ = 0 kj/mol
The standard enthalpy for the formation of CO₂ = -393.5 kj/mol
Now we will put the values in equation:
ΔH0reaction = [ΔHf0 CO2(g)] - [ΔHf0 CO(g) + ΔHf0 O2(g)]
ΔH0reaction = [-393.5 kj/mol] - [-110.5 kj/mol + 0]
ΔH0reaction = [-393.5 kj/mol] - [-110.5 kj/mol]
ΔH0reaction = -283 kj/mol
If Ka for HBrO is 2. 8×10^−9 at 25°C, then the value of Kb for BrO− at 25°C is 3.5× 10^(-6).
<h3>
What is base dissociation constant?
</h3>
The base dissociation constant (Kb) is defined as the measurement of the ions which base can dissociate or dissolve in the aqueous solution. The greater the value of base dissociation constant greater will be its basicity an strength.
The dissociation reaction of hydrogen cyanide can be given as
HCN --- (H+) + (CN-)
Given,
The value of Ka for HCN is 2.8× 10^(-9)
The correlation between base dissociation constant and acid dissociation constant is
Kw = Ka × Kb
Kw = 10^(-14)
Substituting values of Ka and Kw,
Kb = 10^(-14) /{2.8×10^(-9) }
= 3.5× 10^(-6)
Thus, we find that if Ka for HBrO is 2. 8×10^−9 at 25°C, then the value of Kb for BrO− at 25°C is 3.5× 10^(-6).
DISCLAIMER: The above question have mistake. The correct question is given as
Question:
Given that Ka for HBrO is 2. 8×10^−9 at 25°C. What is the value of Kb for BrO− at 25°C?
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Answer:
b.Beta
Explanation:
mass number remains constant while atomic number has been increased by 1 unit . beta is electron like element where its mass number is 0 and atomic number is -1.
