How has the work of chemists affected the environment?
2 answers:
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1) Balance the chemical equation.
2) List the known and unknown quantities.
Reactant 1: Ca(NO3)2.
Amount of substance: 3.40 mol.
Reactant 2: Li3PO4.
Amount of substance: 2.40 mol.
Product: Ca3(PO4)2
Mass: unknown.
3) Which is the limiting reactant?
<em>3.1-How many moles of Li3PO4 do we need to use all of the Ca(NO3)2?</em>
The molar ratio between Li3PO4 and Ca(NO3)2 is 2 mol Li3PO4: 3 mol Ca(NO3)2.
<em>We need 2.2667 mol Li3PO4 and we have 2.40 mol Li3PO4. We have enough Li3PO4. </em>This is the excess reactant.
<em>3.2-How many moles of Ca(NO3)2 do we need to use all of the Li3PO4?</em>
The molar ratio between Li3PO4 and Ca(NO3)2 is 2 mol Li3PO4: 3 mol Ca(NO3)2.
<em>We need 3.60 mol Ca(NO3)2 and we have 3.40 mol Ca(NO3)2. We do not have enough Ca(NO3)2. </em>This is the limiting reactant.
4) Moles of Ca3(PO4)2 produced from the limiting reactant.
We have 3.40 mol Ca(NO3)2 of the limiting reactant.
The molar ratio between Ca(NO3)2 and Ca3(PO4)2 is 3 mol Ca(NO3)2: 1 mol Ca3(PO4)2.
5) Mass of Ca3(PO4)2 produced.
The molar mass of Ca3(PO4)2 is 310.1767 g/mol.
<em>The mass of Ca3(PO4)2 produced is</em> 351 g Ca3(PO4)2.
Option D.
.
Answer:
Explanation:
Given that:
Half life = 14.0 days
Where, k is rate constant
So,
The rate constant, k = 0.04951 days⁻¹
Initial concentration [A₀] = 60.0 g
Time = 46.7 hrs
Considering, 1 hr = 0.041667 days
So, time = 1.9458 days
Using integrated rate law for first order kinetics as:
Where,
is the concentration at time t
So,