Answer: Total pressure inside of a vessel is 0.908 atm
Explanation:
According to Dalton's law, the total pressure is the sum of individual partial pressures. exerted by each gas alone.
= partial pressure of nitrogen = 0.256 atm
= partial pressure of helium = 203 mm Hg = 0.267 atm (760mmHg=1atm)
= partial pressure of hydrogen =39.0 kPa = 0.385 atm (1kPa=0.00987 atm)
Thus 
=0.256atm+0.267atm+0.385atm =0.908atm
Thus total pressure (in atm) inside of a vessel is 0.908
There appears to be no table shown, but I can still answer. Aluminum has 3 valence electrons
The range is negative numbers.
The interval for the range is .
***You might want to look at your functions again because I don't see a choice that matches.
Step-by-step explanation:
Given functions:
We are asked to find the range of .
I'm also going to look at the domain just to see if this possibly might change my range .
is the inner function. So we will consider the domain of that function first.
You only have to worry about division by zero for the function .
Since we are dividing by , we don't want to be zero.
So far the domain is all real numbers except .
Now let's move out.
exists for all numbers, . So we didn't want to include from before.
Now let's put it together:
So the domain is still all real numbers except at since we cannot divide by 0 and is 0 when .
with .
is positive for all numbers except .
So is negative for all numbers since negative divided by positive is negative.
So the range is only negative numbers.
Let's also look at the inverse:
Multiply both sides by :
Divide both sides by :
Take the square root of both sides:
.
So can't be 0 and it also can't be positive because the inside of the square root will be negative (since negative divided by positive results in negative).
Answer:
- Mass of monobasic sodium phosphate = 1.857 g
- Mass of dibasic sodium phosphate = 1.352 g
Explanation:
<u>The equilibrium that takes place is:</u>
H₂PO₄⁻ ↔ HPO₄⁻² + H⁺ pka= 7.21 (we know this from literature)
To solve this problem we use the Henderson–Hasselbalch (<em>H-H</em>) equation:
pH = pka + ![log\frac{[A^{-} ]}{[HA]}](https://tex.z-dn.net/?f=log%5Cfrac%7B%5BA%5E%7B-%7D%20%5D%7D%7B%5BHA%5D%7D)
In this case [A⁻] is [HPO₄⁻²], [HA] is [H₂PO₄⁻], pH=7.0, and pka = 7.21
If we use put data in the <em>H-H </em>equation, and solve for [HPO₄⁻²], we're left with:
![7.0=7.21+log\frac{[HPO4^{-2} ]}{[H2PO4^{-} ]}\\ -0.21=log\frac{[HPO4^{-2} ]}{[H2PO4^{-} ]}\\\\10^{-0.21} =\frac{[HPO4^{-2} ]}{[H2PO4^{-} ]}\\0.616 * [H2PO4^{-}] = [HPO4^{-2}]](https://tex.z-dn.net/?f=7.0%3D7.21%2Blog%5Cfrac%7B%5BHPO4%5E%7B-2%7D%20%5D%7D%7B%5BH2PO4%5E%7B-%7D%20%5D%7D%5C%5C%20-0.21%3Dlog%5Cfrac%7B%5BHPO4%5E%7B-2%7D%20%5D%7D%7B%5BH2PO4%5E%7B-%7D%20%5D%7D%5C%5C%5C%5C10%5E%7B-0.21%7D%20%3D%5Cfrac%7B%5BHPO4%5E%7B-2%7D%20%5D%7D%7B%5BH2PO4%5E%7B-%7D%20%5D%7D%5C%5C0.616%20%2A%20%5BH2PO4%5E%7B-%7D%5D%20%3D%20%5BHPO4%5E%7B-2%7D%5D)
From the problem, we know that [HPO₄⁻²] + [H₂PO₄⁻] = 0.1 M
We replace the value of [HPO₄⁻²] in this equation:
0.616 * [H₂PO₄⁻] + [H₂PO₄⁻] = 0.1 M
1.616 * [H₂PO₄⁻] = 0.1 M
[H₂PO₄⁻] = 0.0619 M
With the value of [H₂PO₄⁻] we can calculate [HPO₄⁻²]:
[HPO₄⁻²] + 0.0619 M = 0.1 M
[HPO₄⁻²] = 0.0381 M
With the concentrations, the volume and the molecular weights, we can calculate the masses:
- Molecular weight of monobasic sodium phosphate (NaH₂PO₄)= 120 g/mol.
- Molecular weight of dibasic sodium phosphate (Na₂HPO₄)= 142 g/mol.
- mass of NaH₂PO₄ = 0.0619 M * 0.250 L * 120 g/mol = 1.857 g
- mass of Na₂HPO₄ = 0.0381 M * 0.250 L * 142 g/mol = 1.352 g
