Answer:
5
Explanation:
Given that the formula is;
1/λ= R(1/nf^2 - 1/ni^2)
λ = 93.7 nm or 93.7 * 10^-9 m
R= 1.097 * 10^7 m-1
nf = ?
ni = 1
From;
ΔE = hc/λ
ΔE = 6.63 * 10^-34 * 3* 10^8/93.7 * 10^-9
ΔE = 21 * 10^-19 J
ΔE = -2.18 * 10^-18 J (1/nf^2 - 1/ni^2)
21 * 10^-19 J = -2.18 * 10^-18 J (1/nf^2 - 1/ni^2)
21 * 10^-19/-2.18 * 10^-18 = (1/nf^2 - 1/1^2)
-0.963 = (1/nf^2 - 1)
-0.963 + 1 = 1/nf^2
0.037 = 1/nf^2
nf^2 = (0.037)^-1
nf^2 = 27
nf = 5
Answer:
<em>3.27·10²³ atoms of O</em>
Explanation:
To figure out the amount of oxygen atoms in this sample, we must first evaluate the sample.
The chemical formula for sodium sulfate is <em>Na₂SO₄, </em>and its molar mass is approximately 142.05
.
We will use stoichiometry to convert from our mass of <em>Na₂SO₄ </em>to moles of <em>Na₂SO₄</em>, and then from moles of <em>Na₂SO₄ </em>to moles of <em>O </em>using the mole ratio; then finally, we will convert from moles of <em>O </em>to atoms of <em>O </em>using Avogadro's constant.
19.3g <em>Na₂SO₄</em> ·
·
·
After doing the math for this dimensional analysis, you should get a quantity of approximately <em>3.27·10²³ atoms of O</em>.
Answer:- 0.138 M
Solution:- The buffer pH is calculated using Handerson equation:

acts as a weak acid and
as a base which is pretty conjugate base of the weak acid we have.
The acid hase two protons(hydrogen) where as the base has only one proton. So, we could write the equation as:

Phosphoric acid gives protons in three steps. So, the above equation is the second step as the acid has only two protons and the base has one proton.
So, we will use the second pKa value. The acid concentration is given as 0.10 M and we are asked to calculate the concentration of the base to make a buffer of exactly pH 7.00.
Let's plug in the values in the equation:



Taking antilog:


On cross multiply:
[base] = 1.38(0.10)
[base] = 0.138
So, the concentration of the base that is
required to make the buffer is 0.138M.
Answer:
The plane with aluminium can lift more mass of passangers than the plane of steel.
Explanation:
The total mass the airplane canc lift is:

For aluminium:


and
![V_{fuselage}=\frac{\pi *L}{4}*[D^2-(D-e)^2]](https://tex.z-dn.net/?f=V_%7Bfuselage%7D%3D%5Cfrac%7B%5Cpi%20%2AL%7D%7B4%7D%2A%5BD%5E2-%28D-e%29%5E2%5D)
where:
- L is lenght
- D is diameter
- e is thickness
![m_{tot}=\delta _{Al}*\frac{\pi *L}{4}*[D^2-(D-e)^2]+m_{pas-Al}](https://tex.z-dn.net/?f=m_%7Btot%7D%3D%5Cdelta%20_%7BAl%7D%2A%5Cfrac%7B%5Cpi%20%2AL%7D%7B4%7D%2A%5BD%5E2-%28D-e%29%5E2%5D%2Bm_%7Bpas-Al%7D)
For steel (same procedure):
![m_{tot}=\delta _{Steel}*\frac{\pi *L}{4}*[D^2-(D-e)^2]+m_{pas-Steel](https://tex.z-dn.net/?f=m_%7Btot%7D%3D%5Cdelta%20_%7BSteel%7D%2A%5Cfrac%7B%5Cpi%20%2AL%7D%7B4%7D%2A%5BD%5E2-%28D-e%29%5E2%5D%2Bm_%7Bpas-Steel)
Knowing that the total mass the airplane can lift is constant and that aluminum has a lower density than the steel, we can afirm that the plane with aluminium can lift more mass of passangers.
Also you can estimate an average weight of passanger to estimate a number of passangers it can lift.