Answer:4 and 3
Explanation:just took the test
A. SO2Cl2(g) --> SO2(g) + Cl2(g)
<span>1 mole of SOCl2 becomes 1 mole SO2 and 1 mole Cl2 </span>
<span>1 mole --> 2 moles </span>
<span>entropy increases </span>
Answer:
Molarity of the sodium hydroxide solution is 1.443 M/L
Explanation:
Given;
0.60 M concentration of NaOH contains 2.0 L
3.0 M concentration of NaOH contains 495 mL
Molarity is given as concentration of the solute per liters of the solvent.
If the volumes of the two solutions are additive, then;
the total volume of NaOH = 2 L + 0.495 L = 2.495 L
the total concentration of NaOH = 0.6 M + 3.0 M = 3.6 M
Molarity of NaOH solution = 3.6 / 2.495
Molarity of NaOH solution = 1.443 M/L
Therefore, molarity of the sodium hydroxide solution is 1.443 M/L
Answer:
Explanation:
Hello,
For the given chemical reaction:
We first must identify the limiting reactant by computing the reacting moles of Al2S3:
Next, we compute the moles of Al2S3 that are consumed by 2.50 of H2O via the 1:6 mole ratio between them:
Thus, we notice that there are more available Al2S3 than consumed, for that reason it is in excess and water is the limiting, therefore, we can compute the theoretical yield of Al(OH)3 via the 2:1 molar ratio between it and Al2S3 with the limiting amount:
Finally, we compute the percent yield with the obtained 2.10 g:
Best regards.