The percentage of glucose given is m/v. This means that the given percentage of volume consists of mass.
In this solution, percentage of glucose is 5.5% m/v.
This means that 5.5% of the volume is the mass of glucose.
Given volume is 285 mL.
Therefore mass of glucose is 5.50% of 285 mL = (5.5*285)/100
mass of glucose = 15.67 g
Answer:
T½ = 16hours
Explanation:
Final mass (N) = 10g
Initial mass (No) = 20g
Time (t) = 16hours
T½ = ?
T½ = In2 / λ
But λ = ?
In(N/No) = -λt
In(10/20) = -(λ * 16)
In(0.5) = -16λ
-0.693 = -16λ
λ = 0.693 / 16
λ = 0.0433
Note : λ is known as the disintegration constant
T½ = In2 / λ
T½ = 0.693 / 0.0433
T½ = 16hours
The half-life of the sample is 16hours
Given the percentage composition of HC as C → 81.82 % and H → 18.18 %
So the ratio of number if atoms of C and H in its molecule can will be:
C : H = 81.82 12 : 18.18 1 C : H = 6.82 : 18.18 = 6.82 6.82 : 18.18 6.82 = 1 : 2.66 ≈ 3 : 8
So the Empirical Formula of hydrocarbon is:
C 3 H 8
As the mass of one litre of hydrocarbon is same as that of C O 2 The molar mass of the HC will be same as that of C O 2 i.e 44 g mol
Now let Molecular formula of the HC be ( C 3 H 8 ) n
Using molar mass of C and H the molar mass of the HC from its molecular formula is:
( 3 × 12 + 8 × 1 ) n = 44 n So 44 n = 44 ⇒ n = 1
Hence the molecular formula of HC is C 3 H 8
Does that help?
Answer: Using more fossil fuels
Explanation: Burning fossil fuels releases Green House gasses into the atmosphere, such as carbon dioxide. The carbon dioxide in the atmosphere traps in heat, which causes global temperatures to increase.
The nucleus!! This is made up of protons and neutrons that each weigh about 1 amu.
Electrons are not found in the nucleus and weigh almost nothing so chemistry in school doesn’t bother with them :)
