Answer:
Final [B] = 1.665 M
Explanation:
3A + 4B → C + 2D
Average rection rate = 3[A]/Δt = 4[B]/Δt = [C]/Δt = 2[D]/Δt
0.05600 M/s = 4 [B]/ 2.50 s
[B] = 0.035 M (concentration of B consumed)
Final [B] = initial [B] - consumed [B]
Final [B] = 1.700 M - 0.035 M
Final [B] = 1.665 M
Answer:
b. 2.28 M
Explanation:
The reaction of neutralization of NaOH with H2SO4 is:
2NaOH + H2SO4 → Na2SO4 + 2H2O
<em>Where 2 moles of NaOH react per mole of H2SO4</em>
<em />
To solve the concentration of NaOH we need to find the moles of H2SO4. Using the chemical equation we can find the moles of NaOH that react and with the volume the molar concentration as follows:
<em>Moles H2SO4:</em>
45.7mL = 0.0457L * (0.500mol/L) = 0.02285 moles H2SO4
<em>Moles NaOH:</em>
0.02285 moles H2SO4 * (2moles NaOH / 1 mol H2SO4) = 0.0457moles NaOH
<em>Molarity NaOH:</em>
0.0457moles NaOH / 0.020L =
2.28M
Right option:
<h3>b. 2.28 M</h3>
Answer : The pH of the solution is, 9.63
Explanation : Given,
The dissociation constant for HCN = 
First we have to calculate the moles of HCN and NaCN.
and,
The balanced chemical reaction is:
Initial moles 0.1116 0.0461 0.08978
At eqm. (0.1116-0.0461) 0 (0.08978+0.0461)
0.0655 0.1359
Now we have to calculate the pH of the solution.
Using Henderson Hesselbach equation :
![pH=pK_a+\log \frac{[Salt]}{[Acid]}](https://tex.z-dn.net/?f=pH%3DpK_a%2B%5Clog%20%5Cfrac%7B%5BSalt%5D%7D%7B%5BAcid%5D%7D)
Now put all the given values in this expression, we get:
Therefore, the pH of the solution is, 9.63
A word equation is a chemical reaction described using words.
A common example is the act of photosynthesis - the process plants use to make glucose (sugar) to use as 'food'.
Plants convert water and carbon dioxide into oxygen and glucose.
A word equation to express this is:
Water + Carbon Dioxide → Glucose + Oxygen
The other type of equation is a symbol equation - this uses the symbols of the elements instead of the common names:
H₂O + CO₂ → C₆H₁₂O₆ + O₂
There is also a balanced version:
6H₂O + 6CO₂ → C₆H₁₂O₆ + 6O₂
<em>If you want information on the balanced symbol equations, feel free to PM me.</em>
