A first-order reaction is 81omplete in 264s.The half-life for this reaction (i) t 1/2 = =3.465×10 −3 s.to reach 95% Completion = 285 s.
To measure reaction rates, chemists initiate the reaction, measure the concentration of the reactant or product at different times as the reaction progresses,
For a 0-order response, the mathematical expression that may be employed to determine the half of life is: t1/2 = [R]0/2k. For a first-order reaction, the half of-existence is given by: t1/2 = zero.693/ok. For a 2d-order response, the method for the half-life of the response is: 1/okay[R]0
The 1/2-life of a response (t1/2), is the quantity of time needed for a reactant concentration to lower via half of compared to its initial awareness. Its software is used in chemistry and medicine to are expecting the awareness of a substance over time
Half of the lifestyles is the time required for exactly 1/2 of the entities to decay 50%.
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Answer:
7650
Explanation:
formula- multiply the volume value by 1e+6
Answer:
Concentration AgBr at saturation = 7.07 x 10⁻⁷M
Explanation:
Given AgBr(s) => Ag⁺(aq) + Br⁻(aq) ; Ksp = 5 x 10⁻¹³ = [Ag⁺][Br⁻]
I --- 0 0
C --- +x +x
E --- x x
[Ag⁺][Br⁻] = (x)(x) = x² = 5 x 10⁻¹³ => x = SqrRt(5 x 10⁻¹³) = 7.07 x 10⁻⁷M
Answer:
Theoretical yield of C6H10 = 3.2 g.
Explanation:
Defining Theoretical yield as the quantity of product obtained from the complete conversion of the limiting reactant in a chemical reaction. It can be expressed as grams or moles.
Equation of the reaction
C6H11OH --> C6H10 + H2O
Moles of C6H11OH:
Molar mass of C6H110H = (12*6) + (1*12) + 16
= 100 g/mol
Mass of C6H10 = 3.8 g
number of moles = mass/molar mass
=3.8/100
= 0.038 mol.
Using stoichoimetry, 1 moles of C6H110H was dehydrated to form 1 mole of C6H10 and 1 mole of water.
Therefore, 0.038 moles of C6H10 was produced.
Mass of C6H10 = molar mass * number of moles
Molar mass of C6H10 = (12*6) + (1*10)
= 82 g/mol.
Mass = 82 * 0.038
= 3.116 g of C6H10.
Theoretical yield of C6H10 = 3.2 g
