How many milligrams of AgNO3 is required to completely react with 81.5 mg LiOH?
1 answer:
Answer:
Explanation:
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In this case, according to the given chemical reaction, it is first necessary to compute the moles of reacting LiOH given its molar mass:
Thus, since there is a 1:1 mole ratio between lithium hydroxide and silver nitrate (169.87 g/mol) the resulting milligrams turn out to be:
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The molar concentration of the original HF solution : 0.342 M
Further explanation
Given
31.2 ml of 0.200 M NaOH
18.2 ml of HF
Required
The molar concentration of HF
Solution
Titration formula
M₁V₁n₁=M₂V₂n₂
n=acid/base valence (amount of H⁺/OH⁻, for NaOH and HF n =1)
Titrant = NaOH(1)
Titrate = HF(2)
Input the value :
Half-life is the time taken for the concentration of the substance to reduce by 50%. The original sample of strontium had a mass of 1.6 gms. Thus, option d is correct.
<h3>What is half-life?</h3>The half-life of any radioactive substance is the time period at which the concentration will get reduced to half the initial amount. The initial mass of Sr-90 is calculated as,
Given,
Quantity of the remaining substance N (t) = 0.40 gm
Initial radioactive substance quantity =?
Time duration (t) = 56 years
Half-life = 28 years
Substituting values above:
= 1.6 gm
Therefore, option d. the initial mass of Sr is 1.6 gm.
Learn more about half-life here:
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