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3 years ago

Ca3(PO4)2 + 3 H2SO4 ⟶ 2 H3PO4 + 3 CaSO4

Chemistry

1 answer:

Pachacha [2.7K]3 years ago

4 0

Answer:

The answer to your question is below

Explanation:

Balanced chemical reaction

Ca₃(PO₄)₂ + 3H₂SO₄ ⇒ 2H₃SO₄ + 3CaSO₄

To answer this question just calculate the molar mass of both reactants.

Molar mass of Ca₃(PO₄)₂ = (3 x 40) + (2 x 31) + (8 x 16)

= 120 + 62 + 128

= 310 g

Molar mass of 3H₂SO₄ = 3[(2 x 1) + (1 x 32) + (4 x 16)]

= 3[2 + 32 + 64]

= 3[98]

= 294 g

Conclusion

310 g of Ca₃(PO₄)₂ will react with 294 g of 3H₂SO₄

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How much heat, in kJ, will be absorbed by a 25.0 g piece of aluminum (specific heat = 0.930 J/g･°C) as it changes temperature fr

faust18 [17]

Answer:

quantity of heat=mc*theta

=25*0.930(76-25)

=25*0.930*51

=1185.75J

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3 years ago

What is the theoretical yield of so3 produced by the quantities described in part a? express your answer numerically in grams?

PSYCHO15rus [73]

Sulfur reacts with oxygen to yield SO3 as shown in the equation below;
2S(g)+ 3O2(g) = 2SO3(g)
From part A 7.49 g of S were used.
The atomic mass of sulfur is 32.06 g/mol
Hence, the number of moles of sulfur used
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4 years ago

What I know about the Alveoli and Gas exchange

Reptile [31]

Explanation:

What happens during gas exchange in the alveoli?

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What is the role of alveoli in gas exchange?

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7 0

3 years ago

How many grams will be produced

Anni [7]

Answer:

$\Large \boxed{\sf 34.2 \ g}$

Explanation:

Mole ratio

$9:4$

$1.75:x$

Moles of CO₂

$\displaystyle \frac{4 \times 1.75}{9} =0.78$

Use formula to find mass

$\displaystyle moles=\frac{mass}{M_r}$

Relative molecular mass of CO₂ = 12 + 16 × 2 = 44

$\displaystyle 0.78=\frac{mass}{44}$

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3 years ago

How many moles of carbon, hydrogen, and oxygen are present in a 100-g sample of ascorbic acid?

Y_Kistochka [10]

There are:

3.41 moles of C

4.54 moles of H

3.40 moles of O.

Why?

To solve the problem, the first thing that we need to do is to write the chemical formula of the ascorbic acid.

$C_{6}H_{8}O_{6}$

Now, we know that there are 100 grams of the compound, so, the masses of each element will represent the percent in the compound.

We have that:

$C_{6}=12.0107g*6=72.08g\\\\H_{8}=1.008g*8=8.064g\\\\O_{6}=15.999g*6=95.994g\\\\C_{6}H_{8}O_{6}=72.08g+8.064g+95.994g=176.138g$

To know the percent of each element, we need to to the following:

$C=\frac{72.08g}{176.138g}*100=0.409*100=40.92(percent)\\\\H=\frac{8.064g}{176.138g}*100=4.58(percent)\\\\O=\frac{95.994}{176.138g}*100=54.49(percent)$