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makvit [3.9K]
3 years ago
10

How many grams of calcium phosphate are theoretically produced if we start with 552.2 grams of Ca(NO3)2 and 285.4 grams of Li3PO

4? Reaction: 3Ca(NO3)2 + 2Li3PO4 → 6LiNO3 +Ca3(PO4)2
Chemistry
1 answer:
Sedbober [7]3 years ago
7 0
<span>molecular weight of Ca(NO3)2 --> 164 Ca=40 N=14 O=16 molecular weight of Li3PO4 --> 115.8 molecular weight of LiNO3 --> 68.9 Ca3(PO4)2 produced ---> 310.2 g/mol molecular weight Hence the answer would be 310.2 g/mol of calcium phosphate will be produced</span>
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Determine the change in boiling point for 397.7 g of carbon disulfide (Kb = 2.34°C kg/mol) if 35.0 g of a nonvolatile, nonionizi
uysha [10]

Answer: The change in boiling point for 397.7 g of carbon disulfide (Kb = 2.34°C kg/mol) if 35.0 g of a nonvolatile, nonionizing compound is dissolved in it is 2.9^0C

Explanation:

Elevation in boiling point:

T_b-T^o_b=i\times k_b\times \frac{w_2\times 1000}{M_2\times w_1}

where,

T_b = boiling point of solution = ?

T^o_b = boiling point of pure carbon disulfide= 46.2^oC

k_b = boiling point constant  =2.34^0Ckg/mol

m = molality

i = Van't Hoff factor = 1 (for non-electrolyte)

w_2 = mass of solute = 35.0 g

w_1 = mass of solvent (carbon disulphide) = 397.7 g

M_2 = molar mass of solute = 70.0 g/mol

Now put all the given values in the above formula, we get:

(T_b-46.2)^oC=1\times (2.34^oC/m)\times \frac{(35.0g)\times 1000}{70.0\times (397.7g)}

T_b=49.1^0C

Therefore, the change in boiling point is (49.1-46.2)^oC=2.9^0C

5 0
3 years ago
If oxygen is removed from a sample of air as iron rust, what happens to the particle pressure of oxygen in the air?
klasskru [66]

Answer : It increases

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So once iron rusts, there is oxygen, just not in air; it's in the iron oxide.


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A solution is prepared by dissolving 38.6 g sucrose (C12H22O11) in 495 g of water. Determine the mole fraction of sucrose if the
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Answer:

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Let's convert the mass to moles (mass / molar mass)

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Total moles = 0.113 m + 27.5 m = 27.0613 moles

Mole fraction of sucrose = Moles of sucrose / Total moles

0.113 m / 27.0613 moles = 4.09×10⁻³

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