Because that is the outer ring.
We can use two equations for this problem.<span>
t1/2 = ln
2 / λ = 0.693 / λ
Where t1/2 is the half-life of the element and λ is
decay constant.
20 days = 0.693 / λ
λ = 0.693 / 20 days
(1)
Nt = Nο eΛ(-λt) (2)
Where Nt is atoms at t time, No is the initial amount of substance, λ is decay constant and t is the time
taken.
t = 40 days</span>
<span>No = 200 g
From (1) and (2),
Nt = 200 g eΛ(-(0.693 / 20 days) 40 days)
<span>Nt = 50.01 g</span></span><span>
</span>Hence, 50.01 grams of isotope will remain after 40 days.
<span>
</span>
Answer: The mass of given amount of copper (II) cyanide is 462.4 g
Explanation:
To calculate the number of moles, we use the equation:
We are given:
Moles of copper (II) cyanide = 4 moles
Molar mass of copper (II) cyanide = 115.6 g/mol
Putting values in above equation, we get:
Hence, the mass of given amount of copper (II) cyanide is 462.4 g
Answer:
Explanation:
Hello!
In this case, according to the chemical reaction:
We can evidence the 2:1 mole ratio between hydrogen and tin, thus, we perform the following stoichiometric setup to obtain the mass of produced tin:
Best regards!
The percentage by mass of water in Na2CO3.10H2O from the calculation is 62.9%
<h3>What is percentage by mass?</h3>
The term percentage by mass refers to the amount of a particular moiety by mass in a molecule.
In this case, we know that the molar mass of Na2CO3.10H2O is 286 g/mol the molar mass of the 10 moles of water is 180 g/mol hence the percentage by mass of water in the compound is; 180 g/mol /286 g/mol * 100/1 = 62.9%
Learn more about percentage by mass: brainly.com/question/16885872
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