Answer:
69.7% is percent yield
Explanation:
Based on the reaction:
3Ca(NO3)2(aq) + 2Na3PO4(aq) → Ca3(PO4)2(s) + 6NaNO3(aq)
2 moles of Na3PO4 react producing 6 moles of NaNO3.
As 24.2 moles of Na3PO4 react, theoretical moles of NaNO3 produced are:
24.2 moles Na3PO4 * (6 moles NaNO3 / 2 moles Na3PO4) =
72.6 moles of NaNO3
As there are produced 50.6 moles of NaNO3, percent yield is:
50.6 moles NaNO3 / 72.6 moles NaNO3 =
<h3>69.7% is percent yield</h3>
Answer:
The answer to your question is remplacement double
Explanation:
Data
Lead (II) nitrate = Pb(NO₃)₂
Potassium iodide = KI
Process
1.- Write the balanced chemical reaction
Pb(NO₃)₂ + 2KI ⇒ PbI₂ + 2KNO₃
2.- Conclusion
This is a remplacement double reaction because there are two reactants that interchange cations and the products are a combination of the reactants.
Answer:
Rain forest.
Explanation: Because of how humid it is in a rain forest that humidity will bring a lot of pressure from above.
Answer:
the conversion factor is f= 6 mol of glucose/ mol of CO2
Explanation:
First we need to balance the equation:
C6H12O6(s) + O2(g) → CO2(g) + H2O(l) (unbalanced)
C6H12O6(s) + 6O2(g) → 6CO2(g) + 6H2O(l) (balanced)
the conversion factor that allows to calculate the number of moles of CO2 based on moles of glucose is:
f = stoichiometric coefficient of CO2 in balanced reaction / stoichiometric coefficient of glucose in balanced reaction
f = 6 moles of CO2 / 1 mol of glucose = 6 mol of glucose/ mol of CO2
f = 6 mol of CO2/ mol of glucose
for example, for 2 moles of glucose the number of moles of CO2 produced are
n CO2 = f * n gluc = 6 moles of CO2/mol of glucose * 2 moles of glucose= 12 moles of CO2
Answer:
Answer to you need to make a 6.00 x 10-4 M KSCN solution starting with a 2.00 x 10-3 M KSCN solution. You will be making up the so. ... X 10-3 M KSCN Solution. You Will Be Making Up The Solution In A 25 ML Volumetric Flask And Using 0.5 M HNO3 As The Diluent. What Volume Of 2.00 X 10-3 M KSCN Will You Need?
Explanation:
