Loss of an electron is oxidation!
Answer:
Check explanation
Explanation:.
NOTE: kindly check for attached file/picture for the graph.
From the graph of absorbance against concentration from the question. We can see that the 0.3 mark absorbance is equivalent to 0.15 M. So, the concentration of CuSO4 is 0.15 M.
The concentration can also be calculated using the Beer-lambert equation for absorbance. The equation is given below;
A= ɛ×C×l --------------------------------------------------------------------------------------(1).
Where A= absorbance, ɛ= molar absorptivity, C= concentration and l= length.
Therefore, the concentration,C will now be; C= A/ ɛ×l. -------------------------------------------------------------------------(2).
Assuming the length,l is 1cm.
Hence, C= 0.300/ ɛ×1.
C= (0.300/ ɛ) M.
B. The third shell would be empty so that the eight electrons in the second level would be outermost after the atom loses one electron.
The questions you can answer are
1) what is the mass of one mole of raindrops?, and
2) How many moles of raindrops are in the pacific ocean
Solutions:
1) what is the mass of one mole of raindrops?
mass = number of rain drops * mass of on rain drop
The number of rain drops in one mole of rain drops is 6.02 * 10^23
So, the mass of one mole of rain drops is
6.02 * 10 ^ 23 rain drops * 50. mg * (1 kg / 1,000,000 mg) = 3.01 * 10^ 19 kg
The correct number of significant digits is 2, because 50. mg has two signficant digits, so the answer must be shown as 3.0 * 10^ 19 kg.
2) How many moles of raindrops are in the pacific ocean
Use the proportion 1 mol / 3.0 * 10^19 kg = x mol / 7.08 * 10^20 kg
And you solve for x:
x = (7.08 * 10^20 kg) * 1 mol / (3.0 * 10^19 kg) = 2.36 * 10 = 23.6 moles
Which rounded to two significant digits is 24 moles of rain drops.
Answer:
0.3 mole
Explanation:
number of moles grams
one 14+(4×1)+14+(3×16)
1 80
? 24
Therefore, 24×1÷80 = 0.3 moles of ammonium nitrate
