Answer:
Average atomic mass = 3.9 amu
Explanation:
Given data:
Percent abundance of He-2 = 0.420%
Percent abundance of He-3 = 2.75%
Percent abundance of He-4 = 96.83%
Average atomic mass = ?
Solution:
Average atomic mass = (abundance of 1st isotope × its atomic mass) +(abundance of 2nd isotope × its atomic mass) + (abundance of 3rd isotope × its atomic mass) / 100
Average atomic mass = (0.420×2)+(2.75×3) +(96.83×4)/100
Average atomic mass = 0.84 + 8.25 +387.32 / 100
Average atomic mass = 396.41 / 100
Average atomic mass = 3.9 amu.
Answer:
NaOH(aq)
Explanation:
NaOH(aq) is known to precipitate Mn^2+ ions according to the following reaction; Mn^2+(aq)+2OH^−(aq)↽−−⇀Mn(OH)2(s)
Hence, manganese(II) oxide reacts more readily with NaOH(aq) under ordinary conditions precipitating the metal hydroxide solid. This is one of the characteristic reactions of Mn^2+.
4 NH₃ + 3O₂ --> 2N₂ + 6H₂O
First, make sure that this is a balanced equation.
There are 4 moles of nitrogen on the left side, and 4 moles of nitrogen on the right side.
There are 12 moles of hydrogen on the left side, and 12 moles of hydrogen on the right side.
There are 6 moles of oxygen on the left side, and 6 moles of oxygen on the right side.
The equation is therefore balanced, and we may proceed.
a) the mole ratio for NH₃ to N₂ is 4 to 2, which can be simplified to 2:1 or 2/1.
b) the mole ratio for H₂O to O₂ is 6 to 3, which can be simplified to 2:1 or 2/1.
Answer:
7.2
Explanation:
you first have to find the number of moles of nitrogen dioxide by using the number of moles for calcium nitrate and the mole to mole ratios
number of moles of calcium nitrate=mass/mm
=16.4/102
=0.16g/mol
then you use the mole to mole ratios
2 : 4
0.16: x
2x/2=0.64/2
x=0.32g/moles of nitrogen dioxide
then you use the formula for the volume
v=22.4n
=22.4×0.32
=7.2
I hope this helps
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