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Stells [14]
2 years ago
5

Compound C with molar mass 205gmol-1 , contains 3.758g of carbon , 0.316g of hydrogen and 1.251g of oxygen. Determine the molecu

lar formula.
Chemistry
1 answer:
labwork [276]2 years ago
8 0

Answer:

Molecular formula => C₁₂H₁₂O₃

Explanation:

From the question given above, the following data were obtained:

Molar mass of compound = 205gmol¯¹

Mass of Carbon (C) = 3.758 g

Mass of Hydrogen (H) = 0.316 g

Mass of Oxygen (O) = 1.251 g

Molecular formula =?

We'll begin by determining the empirical formula of the compound. This can be obtained as follow:

C = 3.758 g

H = 0.316 g

O = 1.251 g

Divide by their molar masses

C = 3.758 /12 = 0.313

H = 0.316 /1 = 0.316

O = 1.251 /16 = 0.078

Divide by the smallest.

C = 0.313 / 0.078 = 4

H = 0.316 / 0.078 = 4

O = 0.078 / 0.078 = 1

Thus, the empirical formula of the compound is C₄H₄O

Finally, we shall determine the molecular formula of the compound. This can be obtained as follow:

Molar mass of compound = 205gmol¯¹

Empirical formula => C₄H₄O

Molecular formula =>?

[C₄H₄O]ₙ = 205

[(12×4) + (4×1) +16]n = 205

[48 + 4 +16]n = 205

68n = 205

Divide both side by n

n = 205 / 68

n = 3

Molecular formula => [C₄H₄O]ₙ

Molecular formula => [C₄H₄O]₃

Molecular formula => C₁₂H₁₂O₃

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A buffer solution contains 0.240 M ammonium chloride and 0.499 M ammonia. If 0.0565 moles of perchloric acid are added to 250 mL
snow_lady [41]

Answer:

The pH of the resulting solution is 9.02.

Explanation:

The initial pH of the buffer solution can be found using the Henderson-Hasselbalch equation:

pOH = pKb + log(\frac{[NH_{4}Cl]}{[NH_{3}]})  

pOH = -log(1.78\cdot 10^{-5}) + log(\frac{0.240}{0.499}) = 4.43        

pH = 14 - pOH = 14 - 4.43 = 9.57

Now, the perchloric acid added will react with ammonia:

n_{NH_{3}} = 0.499moles/L*0.250 L - 0.0565 moles = 0.0683 moles

Also, the moles of ammonium chloride will increase in the same quantity according to the following reaction:

NH₃ + H₃O⁺ ⇄ NH₄⁺ + H₂O    

n_{NH_{4}Cl} = 0.240 moles/L*0.250L + 0.0565 moles = 0.1165 moles

Finally, we can calculate the pH of the resulting solution:

pOH = pKb + log(\frac{[NH_{4}Cl]}{[NH_{3}]})  

pOH = -log(1.78\cdot 10^{-5}) + log(\frac{0.1165 moles/0.250 L}{0.0683 moles/0.250 L}) = 4.98  

pH = 14 - 4.98 = 9.02

Therefore, the pH of the resulting solution is 9.02.

I hope it helps you!

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3 years ago
In what orbital would to valence electrons of strontium be found? a.) 5s b.) 4s c.) none d.) 5p​
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Answer:

The correct answer is 5s

Explanation:

Strontium atomic number = 38

Electronic configuration is $1s^{2} 2s^{2} 2p^{6} 3s^{2} 3p^{6} 3d^{10} 4s^{2} 4p^{6} 5s^{2}

As $5s$ orbital is valence orbital in strontium, the outer electrons exists in that orbital only.

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In a nuclear reaction, the energy released is equal to 5.4 x 1015 joules. Calculate the mass lost in this reaction. (1 J = 1 kg
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What volume of a 6.0m hcl solution is required to make 250.0 milliters of 1.5 m hcl
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Answer : The volume for 6.0m HCl solution required = 62.5 ml

Solution : Given,

Initial concentration of HCl solution = 6.0m

Final concentration of HCl solution = 1.5m

Final volume of HCl solution = 250 ml

Initial volume of HCl solution = ?

Formula used for dilution is,

M_1V_1=M_2V_2

where,

M_1 = initial concentration

M_2 = final concentration

V_1 = initial volume

V_2 = final volume

Now put all the given values in above formula, we get the initial volume of HCl solution.

6.0m\times V_1=1.5m\times 250ml

V_1 = 62.5 ml

Therefore, the volume for 6.0m HCl solution required = 62.5 ml


8 0
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