Answer: P= 1.64 atm
Explanation: solution attached.
Use Ideal gas law 
PV= nRT
Derive for P
P= nRT/V R= 0.08205 L.atm/mol.K
 Substitute the values.
 
        
             
        
        
        
Answer: The correct answer is a) Acetylcholine.
Explanation:
The acetylcholine is a very important neurotransmitter involved in the mediation of synaptic activity between neurons. Among its most important functions we find: expression of mood states, vasodilation, gastrointestinal motility, bronchoconstriction, circular iris muscle contraction and sweat gland secretion.
Therefore the correct answer is a) Acetylcholine.
 
        
             
        
        
        
When a solid (solute) comes in contact with the liquid (solvent), the solute goes about C) dissolution, in which the solid dissolves into the liquid.
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Answer:
Fe₂O₃
Explanation:
To solve this question we must find the moles of Iron in 1.68g. With the difference of the masses we can find the moles of oxygen. The formula will be obtained with the ratio of both amount of moles:
<em>Moles Fe:</em>
1.68g * (1mol / 56g)  =0.03moles
<em>Moles O:</em>
2.40g-1.68g = 0.72g * (1mol/16g) = 0.045moles
The ratio O/Fe is:
0.045moles / 0.03moles = 1.5 moles. this ratio is obtained if the formula is:
<h3>Fe₂O₃</h3>
 
        
             
        
        
        
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