The radius of a helium atom is a 31 PM. What is the radius in nanometers?
1 answer:
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>
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Answer: V=67.2 L
Explanation:
For this problem we will need to use the Ideal Gas Law.
Ideal Gas Law: PV=nRT
P=1.00 atm (STP)
V=?
n=3.00 mol
R=0.08206Latm/Kmol
T=273.15 K (STP)
To find V, we would manipulate the equation to V=nRT/P
With significan figures, our answer is V=67.2 L.
Answer:
0.182 moles of acetic acid are needed, this means 10.93 g.
0.318 moles of sodium acetate are needed, this means 26.08 g.
Explanation:
The Henderson–Hasselbalch (<em>H-H</em>) equation tells us the relationship between the concentration of an acid, its conjugate base, and the pH of a buffer:
pH = pka +
In this case, [A⁻] is the concentration of sodium acetate, and [HA] is the concentration of acetic acid. The pka is a value that can be looked up in literature: 4.76.
From the problem we know that
[A⁻] + [HA] = 250 mM = 0.250 M eq. 1
We use the <em>H-H</em> equation, using the data we know, to describe [A⁻] in terms of [HA]:
5.0 = 4.76 +
eq.2
Now we replace the value of [A⁻] in eq. 1, to calculate [HA]:
1.74 [HA] + [HA] = 0.250 M
[HA] = 0.091 M
Then we calculate [A⁻]:
[A⁻] + 0.091 M = 0.250 M
[A⁻] = 0.159 M
Using the volume, we can calculate the moles of each substance:
- moles of acetic acid = 0.091 M * 2 L = 0.182 moles
- moles of sodium acetate = 0.159 M * 2 L = 0.318 moles
Using the molecular weight, we can calculate the grams of each substance:
- grams of acetic acid = 0.182 mol * 60.05 g/mol = 10.93 g
- grams of sodium acetate = 0.318 mol * 82.03 g/mol = 26.08 g