The equation that shows the formation of chromium (ii) ion from neutral chromium atom is as follow
Cr ---> cr^2+ + 2e-
Cr^2+ is the chromium ion with oxidation state of two which is one of the common ion of chromium. Other common ion of chromium include chromium of oxidation state 6 and 3
Answer:
Part A = The mass of sulfur is 6.228 grams
Part B = The mass of 1 silver atom is 1.79 * 10^-22 grams
Explanation:
Part A
Step 1: Data given
A mixture of carbon and sulfur has a mass of 9.0 g
Mass of the product = 27.1 grams
X = mass carbon
Y = mass sulfur
x + y = 9.0 grams
x = 9.0 - y
x(molar mass CO2/atomic mass C) + y(molar mass SO2/atomic mass S) = 22.6
(9 - y)*(44.01/12.01) + y(64.07/32.07)
(9-y)(3.664) + y(1.998)
32.976 - 3.664y + 1.998y = 22.6
-1.666y = -10.376
y = 6.228 = mass sulfur
x = 9.0 - 6.228 = 2.772 grams = mass C
The mass of sulfur is 6.228 grams
Part B
Calculate the mass, in grams, of a single silver atom (mAg = 107.87 amu ).
Calculate moles of 1 silver atom
Moles = 1/ 6.022*10^23
Moles = 1.66*10^-24 moles
Mass = moles * molar mass
Mass = 1.66*10 ^-24 moles *107.87
Mass = 1.79 * 10^-22 grams
The mass of 1 silver atom is 1.79 * 10^-22 grams
Answer: After 4710 seconds, 1/8 of the compound will be left
Explanation:
Using the formulae
Nt/No = (1/2)^t/t1/2
Where
N= amount of the compound present at time t
No= amount of compound present at time t=0
t= time taken for N molecules of the compound to remain = 4710 seconds
t1/2 = half-life of compound = 1570 seconds
Plugging in the values, we have
Nt/No = (1/2)^(4710s/1570s)
Nt/No = (1/2)^3
Nt/No= 1/8
Therefore after 4710 seconds, 1/8 molecules of the compound will be left
