Answer:
Thus, the radius of the helium atom in nanometers is - 0.031 nm
Explanation:
Given that:-
The radius of the helium atom = 31 pm
Considering the conversion of length in pm to the length in nm as:-
1 pm = 0.001 nm
So,
Applying the above conversion factor in the radius of helium atom as:-
Radius =
nm = 0.031 nm
<u>Thus, the radius of the helium atom in nanometers is - 0.031 nm</u>
Answer:
17.04 g/mol
Explanation:
Molar Mass of NH₃
we know that
Nitrogen has 14.01 gram/mol
And Hydrogen has 1.01 gram/mol
but we have 3 Hydrogens So we multiply
1.01 by 3 i.e., 3.03
Now, add
14.01
+<u> </u><u>3</u><u>.</u><u>0</u><u>3</u>
17.04
So, The molar mass of ammonia, NH₃ is
17.04 g/mol
<u>-TheUnknown</u><u>Scientist</u>
Answer:
Initial concentration of HI is 5 mol/L.
The concentration of HI after
is 0.00345 mol/L.
Explanation:

Rate Law: ![k[HI]^2 ](https://tex.z-dn.net/?f=k%5BHI%5D%5E2%0A)
Rate constant of the reaction = k = 
Order of the reaction = 2
Initial rate of reaction = 
Initial concentration of HI =![[A_o]](https://tex.z-dn.net/?f=%5BA_o%5D)
![1.6\times 10^{-7} mol/L s=(6.4\times 10^{-9} L/mol s)[HI]^2](https://tex.z-dn.net/?f=1.6%5Ctimes%2010%5E%7B-7%7D%20mol%2FL%20s%3D%286.4%5Ctimes%2010%5E%7B-9%7D%20L%2Fmol%20s%29%5BHI%5D%5E2)
![[A_o]=5 mol/L](https://tex.z-dn.net/?f=%5BA_o%5D%3D5%20mol%2FL)
Final concentration of HI after t = [A]
t = 
Integrated rate law for second order kinetics is given by:
![\frac{1}{[A]}=kt+\frac{1}{[A_o]}](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B%5BA%5D%7D%3Dkt%2B%5Cfrac%7B1%7D%7B%5BA_o%5D%7D)
![\frac{1}{[A]}=6.4\times 10^{-9} L/mol s\times 4.53\times 10^{10} s+\frac{1}{[5 mol/L]}](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B%5BA%5D%7D%3D6.4%5Ctimes%2010%5E%7B-9%7D%20L%2Fmol%20s%5Ctimes%204.53%5Ctimes%2010%5E%7B10%7D%20s%2B%5Cfrac%7B1%7D%7B%5B5%20mol%2FL%5D%7D)
![[A]=0.00345 mol/L](https://tex.z-dn.net/?f=%5BA%5D%3D0.00345%20mol%2FL)
The concentration of HI after
is 0.00345 mol/L.
The answer for the following problem is mentioned below.
- <u><em>Therefore 298.44 grams of mercuric oxide is needed to produce 0.692 moles of oxygen molecule </em></u>
Explanation:
Given:
no of moles of the oxygen gas = 0.692
Also given:
2 HgO → 2 Hg + 
where,
HgO represents mercuric oxide
Hg represents mercury
represents oxygen
To calculate:
Molar mass of HgO:
Molar mass of HgO = 216 grams
molar mass of mercury (Hg) = 200 grams
molar mass of oxygen (O) =16 grams
HgO = 200 +16 = 216 grams
We know;
2×216 grams of HgO → 1 mole of oxygen molecule
? → 0.692 moles of oxygen molecule
= 
= 298.944 grams of HgO
<u><em>Therefore 298.44 grams of mercuric oxide is needed to produce 0.692 moles of oxygen molecule </em></u>
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