What is the electron configuration of P+?
1 answer:
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1. S
+ Sr(OH)
⇒ SrSO4 + 2 H2O is the balanced reaction.
2. 0.034 liters of KI will be required completely react with 2.43 g of Cu(NO3)2.
3. 0.55 liters is the volume in L of a 0.724 M Nal solution contains0.405 mol of NaI.
4. 33.3 ml many mL of 0.300 M NaF would be required to make a 0.0400 M solution of NaF when diluted to 250.0 mL with water
Explanation:
Balance chemical reaction of neutralization:
1. S
+ Sr(OH)
⇒ SrSO4 + 2 H2O
2. Data given:
balance chemical reaction:
2Cu(NO₃)₂(aq) + 4KI(aq) → 2CuI(aq) + I₂(s) + 4KNO₃(aq)
molarity of KI = 0.209 M
mass of Cu(NO₃)₂ = 2.43 grams
number of moles of Cu(NO₃)₂ will be calculated as:
number of moles =
atomic mass of Cu(NO₃)₂ = 187.56 grams/mole
putting the values in the equation,
number of moles=
= 0.0129 moles
2 moles of Cu(NO₃)₂ will react with 4 moles of KI
0.0129 moles will react with x moles of KI
=
x = 0.0258
atomic mass of KI = 166.00 grams/mole
mass = 166.00 x 0.0258
= 4.28 grams or ml is the final volume of KI
so molarity =
so molarity of KI is 0.0258 M, volume is 1 litre.
Using the formula
Minitial x Vinitial = M final Vfinal
V initial =
=
= 34.67 ml 0.034 liters of KI will be required.
3) Data given:
molarity of NaI = 0.724
number of moles of NaI =?
Volume in litres =?
formula used:
molarity =
volume in litres =
= 0.55 liters is the volume
4) Data given:
Initial molarity = 0.3 M
initial volume = ?
final molarity = 0.04 M
final volume diluted by = 250 ml
formula used:
M initial X Vinitial = Mfinal X V final
putting the values in the equation:
Vinitial =
= 33.3 ml of 0.3 M solution will be required.
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.