Which graph shows the relationship between the temperature and volume of agas according to Charles's law?Volume vs. Temperature
of a Gasvolume vs. Temperature of a Gas160140140'E 20o A. 100O B.1005060o 50 100 150 200 250 30000 50 100 150 200 250 300Temperature (C)Temperature (C)Volume vs. Temperature of a GasVolume vs. Temperature of a Gas180180160140140E 120OC. 100O D. 100600 500 100 150 200 250 30080 100 150 200 250 300Temperature (C)Temperature (c)
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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:
1.12M
Explanation:
Given parameters:
Volume of solution = 2.5L
Mass of Calcium phosphate = 600g
Unknown:
Concentration = ?
Solution:
Concentration is the number of moles of solute in a particular solution.
Now, we find the number of moles of the calcium phosphate from the given mass;
Formula of calcium phosphate = Ca₃PO₄
molar mass = 3(40) + 31 + 4(16) = 215g/mol
Number of moles of Ca₃PO₄ = = 2.79moles
Now;
Concentration =
Concentration = = 1.12M