Explanation:
From the knowledge of law of multiple proportions,
mass ratio of S to O in SO:
mass of S : mass of O
= 32 : 16
= 32/16
= 2/1
mass ratio of S to O in SO2:
= mass of S : 2 × mass of O
= 32 : 2 × 16
= 32/32
= 1/1
ratio of mass ratio of S to O in SO to mass ratio of S to O in SO2:
= 2/1 ÷ 1/1
= 2
Thus, the S to O mass ratio in SO is twice the S to O mass ratio in SO2.
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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984 grams of strontium will be recovered from 9.84x10^8 cubic meter of seawater.
Explanation:
From the question data given is :
volume of strontium in sea water= 9.84x10^8 cubic meter
(1 cubic metre = 1000000 ml)
so 9 .84x10^8 cubic meter
= 984 ml.
density of sea water = 1 gram/ml
from the formula mass of strontium can be calculated.
density = 
mass = density x volume
mass = 1 x 984
= 984 grams of strontium will be recovered.
98400 centigram of strontium will be recovered.
Strontium is an alkaline earth metal and is highly reactive.
