For a comparison of the nucleus 5626fe, the density of the nucleus 112 48cd is mathematically given as the same.
n(Cd) / n(Fe)=1
<h3>What is the density of the nucleus 112 48cd?</h3>
Generally, the equation for the density is mathematically given as
d=\frac{A}{4/3}\piR^3
Therefore
n(Cd) / n(Fe) = [A (Cd) / (A Fe) ] * [ R (Fe) / R (Cd)]^3
n(Cd) / n(Fe)= (112 / 56 ) * (1/1.26)3
n(Cd) / n(Fe)=1
In conclusion, The ratio of n(Cd) = n(Fe) is 1, hence same
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brainly.com/question/14010194
Answer:
d/t=s
(divide 16)16/t= 8(divide 16)
t= 0.5 km/h²
Explanation:
Distance: 16 kilometres
Speed: 8 km/h
Time: ?
Answer:
Below
Explanation:
Balanced form;
1.Benzene + Dioxygen = Carbon Dioxide + Water
2.Tricalcium phosphate +Carbon = Calcium phosphide + carbon monoxide
3.Nitrous acid react with oxygen to produce nitric acid.
4.This means that the carbon dioxide and limewater react to produce calcium carbonate and water.
5.Potassium react with bromine to produce potassium bromide
6. An aqueous solution of ferrous sulphate reacts with aqueous solution of sodium hydroxide to form a precipitate of ferrous hydroxide and sodium sulphate remains in the solution.
When we can get Pka for K2HPO4 =6.86 so we can determine the Ka :
when Pka = - ㏒ Ka
6.86 = -㏒ Ka
∴Ka = 1.38 x 10^-7
by using ICE table:
H2PO4- → H+ + HPO4
initial 0.4 m 0 0
change -X +X +X
Equ (0.4-X) X X
when Ka = [H+][HPO4] / [H2PO4-]
by substitution:
1.38 X 10^-7 = X^2 / (0.4-X) by solving for X
∴X = 2.3x 10^-4
∴[H+] = X = 2.3 x 10^-4
∴PH = -㏒[H+]
= -㏒ (2.3 x 10^-4)
∴PH = 3.6
