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34kurt
3 years ago
7

A compound is analyzed and found to contain 22.10%Al, 25.40%P, and 52.50%O. What is the empirical formula of the compound?

Chemistry
1 answer:
Sindrei [870]3 years ago
5 0

A compound is analyzed and found to contain 22.10%Al, 25.40%P, and 52.50%O

Let the total mass of compound = 100g

The mass of Aluminum = 22.10 g

Moles of Al = mass / atomic mass = 22.10 /23= 0.96

The mass of P = 25.40 g

Moles of P = 25.40 / 31 = 0.819

The mass of oxygen  = 52.50 g

Moles of oxygen = 52.50 / 16= 3.28

The mole ratio of the two elements will be

Al = 0.96/0.819 = 1.17

P = 0.819/0.819 = 1

O = 3.28 / 0.819 = 4

The whole number ratio will be 1 : 1 : 4

So formula will be AlPO4


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\qquad\qquad\huge\underline{{\sf Answer}}

Here we go ~

1 mole of \sf{ COCl_2} has 6.022 × 10²³ molecules of the given compound.

So, 0.78 mole of \sf{COCl_2} will have ~

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5 0
2 years ago
How many grams of CuF2 are needed to make a 2.8 M solution
11111nata11111 [884]
Molarity is given as,

                              Molarity  =  Moles / Volume of Solution  ----- (1)

Also, Moles is given as,

                              Moles  =  Mass / M.mass

Substituting value of moles in eq. 1,

                              Molarity  =  Mass / M.mass × Volume

Solving for Mass,

                              Mass  =  Molarity × M.mass × Volume  ---- (2)

Data Given;

                  Molarity  =  2.8 mol.L⁻¹

                  M.mass  =  101.5 g.mol⁻¹

                  Volume  =  1 L (I have assumed it because it is not given)

Putting values in eq. 2,

                              Mass  =  2.8 mol.L⁻¹ × 101.5 g.mol⁻¹ × 1 L

                              Mass  =  284.2 g of CuF₂
5 0
3 years ago
How does melting order relate to melting point?
KatRina [158]

Answer:

I think A.

Explanation:I say A because of the substance melting the quicking does have the highest melting point because its the highest.

5 0
3 years ago
Calculate the equilibrium constant k for the isomerization of glucose-1-phosphate to fructose-6-phosphate at 298 k. express your
k0ka [10]
We cannot solve this problem without using empirical data. These reactions have already been experimented by scientists. The standard Gibb's free energy, ΔG°, (occurring in standard temperature of 298 Kelvin) are already reported in various literature. These are the known ΔG° for the appropriate reactions.

<span>glucose-1-phosphate⟶glucose-6-phosphate          ΔG∘=−7.28 kJ/mol
fructose-6-phosphate⟶glucose-6-phosphate          ΔG∘=−1.67 kJ/mol
</span>
Therefore, the reaction is a two-step process wherein glucose-6-phosphate is the intermediate product.

glucose-1-phosphate⟶glucose-6-phosphate⟶fructose-6-phosphate 

In this case, you simply add the ΔG°. However, since we need the reverse of the second reaction to end up with the terminal product, fructose-6-phosphate, you'll have to take the opposite sign of ΔG°.

ΔG°,total = −7.28 kJ/mol  + 1.67 kJ/mol = -5.61 kJ/mol

Then, the equation to relate ΔG° to the equilibrium constant K is

ΔG° = -RTlnK, where R is the gas constant equal to 0.008317 kJ/mol-K.
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lnK = 2.2635
K = e^2.2635
K = 9.62


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Answer:

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Explanation:

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