Answer:
4.00 is the pH of the mixture
Explanation:
The ethyl amine reacts with HNO3 as follows:
C2H5NH2 + HNO3 → C2H5NH3⁺ + NO3⁻
To solve this question we need to find the moles of ethyl amine and the moles of HNO3:
<em>Moles C2H5NH2:</em>
0.0500L * (0.100mol/L) = 0.00500 moles ethyl amine
<em>Moles HNO3:</em>
0.201L * (0.025mol/L) = 0.005025 moles HNO3
That means HNO3 is in excess. The moles in excess are:
0.005025 moles HNO3 - 0.00500 moles ethyl amine =
2.5x10⁻⁵ moles HNO₃
In 50 + 201mL = 251mL = 0.251L:
2.5x10⁻⁵ moles HNO₃ / 0.251L = 9.96x10⁻⁵M = [H+]
As pH = -log [H+]
pH = -log 9.96x10⁻⁵M
pH = 4.00 is the pH of the mixture
29.26 gram
Explanation:
No of molecules = no of mole × Avogadro's number
No of molecules = mass in gram/ molar mass × Avogadros number
Mass in gram =No of molecules×molar mass / Avogadros number
Mass in gram = 3.012 ×10^23 × 58.5/6.02×10^23
29.26 gram
Answer:
44.28 grams.
Explanation:
Let us write the balanced reaction:
As per balanced equation, six moles of fluorine gas will give four moles of PF₃.
The mass of PF₃ required = 120 g
The molar mass of PF₃ = 88g/mol
Moles of PF₃ required =
The moles of fluorine gas required = 
the mass of fluorine gas required = moles X molar mass = 0.91x38 = 34.58g
Now this much mass will be required if the reaction is of 100% yield
But as given that the yield of reaction is only 78.1%
The mass of fluorine required = 
Answer is: 20 ions, left side.
Unbalanced half reaction: C₅O₅²⁻(g) → CO₃²⁻(aq).
1) There are 5 carbon atoms on the left side of half reaction and 1 carbon atom on right, so first add coefficient 5 in fron of carbon dioxide to balance carbon atoms: C₅O₅²⁻(g) → 5CO₃²⁻(aq).
2) Because there are more oxygen atoms on the right, add OH⁻ ions on the left side of half reaction and water on the right: OH⁻(aq) + C₅O₅²⁻(g) → 5CO₃²⁻(aq) + H₂O(l).
3) Balance oxygen (25 atoms on boths side) and hydrogen (20 atoms on both side of half reaction) atoms:
Balanced half reaction: 20OH⁻(aq) + C₅O₅²⁻(g) → 5CO₃²⁻(aq) + 10H₂O(l).
