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satela [25.4K]
3 years ago
13

A compound is composed of 85.6% Carbon and the rest is Hydrogen. The molecular mass of the compound is 42.078g/mol. What is the

molecular formula for the compound?
Chemistry
1 answer:
s344n2d4d5 [400]3 years ago
3 0

The molecular formula for the compound is  C_{3} H_{6}

<u>Explanation</u>:

As with all of these problems, we assume 100 g of an unknown compound.

And thus, we determine the elemental composition by the given percentages.

Moles of carbon = 85.64 / 12.011

                            = 7.13 mol.

Moles of hydrogen = 14.36 / 1.00794

                               = 14.25 mol.

There are 2 moles of hydrogen per mole of carbon. And thus the empirical formula is CH_{2}.

And molecular formula = n × (empirical formula)

Thus, 42.08 = n × (12.011 + 2 × 1.00794)

And thus n = 3, and molecular formula = C_{3} H_{6}

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2Na+ 2H20 - &gt; 2NaOH + H2
aalyn [17]

Answer:

3.17mol

Explanation:

moles of Na = m/M ; Where M - molar mass of Na which is 23

moles of Na = 73/23 =3.17mol

but from the eqn n(Na) : n(NaOH) = 2:2 = 1:1

: . moles of NaOH is also 3.17mol

3 0
3 years ago
If the concentration of h uons were to decrease what would happen to the oh ions
stich3 [128]
<span>If the concentration of H⁺ ions will decrease then the concentration of OH⁺ ions will increase.</span>
3 0
4 years ago
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Andrew [12]
Graph would be the answer.

I hope this helps. :)
7 0
4 years ago
What is the pH of a buffer solution upon mixing 15.0 mL of 0.40 M HCl and 20.0 mL of 0.50 M NH? Kb (NH3) = 1.8 x 10 E. 7.00 A. 9
vladimir2022 [97]

<u>Answer:</u> The pH of resulting solution is 9.08

<u>Explanation:</u>

To calculate the number of moles for given molarity, we use the equation:

\text{Molarity of the solution}=\frac{\text{Moles of solute}\times 1000}{\text{Volume of solution (in mL)}}       ........(1)

  • <u>For HCl:</u>

Molarity of HCl = 0.40 M

Volume of solution = 15.0 mL

Putting values in equation 1, we get:

0.40M=\frac{\text{Moles of HCl}\times 1000}{15.0mL}\\\\\text{Moles of HCl}=0.006mol

  • <u>For ammonia:</u>

Molarity of ammonia = 0.50 M

Volume of solution = 20.0 mL

Putting values in equation 1, we get:

0.50M=\frac{\text{Moles of ammonia}\times 1000}{20.0mL}\\\\\text{Moles of ammonia}=0.01mol

The chemical reaction for hydrochloric acid and ammonia follows the equation:

                  HCl+NH_3\rightarrow NH_4Cl

Initial:          0.006      0.01

Final:             -         0.004              0.006

Volume of solution = 15.0 + 20.0 = 35.0 mL = 0.035 L    (Conversion factor:  1 L = 1000 mL)

  • To calculate the pOH of basic buffer, we use the equation given by Henderson Hasselbalch:

pOH=pK_b+\log(\frac{[salt]}{[base]})

pOH=pK_b+\log(\frac{[NH_4Cl]}{[NH_3]})

We are given:

pK_b = negative logarithm of base dissociation constant of ammonia = -\log (1.8\times 10^{-5})=4.74

[NH_4Cl]=\frac{0.006}{0.035}

[NH_3]=\frac{0.004}{0.035}

pOH = ?

Putting values in above equation, we get:

pOH=4.74+\log(\frac{0.006/0.035}{0.004/0.035})\\\\pOH=4.92

To calculate pH of the solution, we use the equation:

pH+pOH=14\\pH=14-4.92=9.08

Hence, the pH of the solution is 9.08

3 0
3 years ago
Calculate the standard potential for a cell that employs the over-all cell reaction: 2Al(s) + 3 I2(s)→2 Al+3 + 6 I- From reducti
Valentin [98]

Answer:

Explanation:

2Al(s) + 3 I₂(s)   →    2 Al⁺³    +     6 I⁻

Aluminium is oxidised and iodine is reduced .

so cell potential = Ereduction - Eoxidation

Al⁺³ + 3e = Al          -  1.66 V

I₂ + 2 e = 2 I⁻             0.54 V

=  .54 - ( - 1.66 )

= 1.66 + .54

= 2.2  V

8 0
3 years ago
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