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kvv77 [185]
2 years ago
11

11. The density of liquid chloroform is 1.48 g/mL. What is its density in units of lb/in?

Chemistry
1 answer:
omeli [17]2 years ago
6 0

Answer:

5.35\times10^-2\ lb/in^3

Explanation:

Given density of chloroform is 1.48 g/mL

We need to find the density in lb/in^3

Also, we have given 2.54 cm = 1 in and 2.205 lb = 1 kg

We know 1ml=1cm^3

Also, 2.54 cm = 1 in

2.54^3\ cm^3=1\ in^3\\16.38\ cm^3=1\ in^3\\16.38\ cm^3=16.38\ ml=1\ in^3

And 1000 g = 1 kg

1\ g=\frac{1kg}{1000}\\\\1\ kg=2.205\ lb\\\\1\ g=\frac{2.205\ lb}{1000}=2.205\times10^{-3}\ lb

Now we will plug

1\ ml=\frac{1}{16.38}\ in^3 \\\\And\ 1\ g=2.205\times10^-3\ lb

So,

1.48\ \frac{g}{ml}=1.48\times 2.205\times10^{-3}\frac{lb}{\frac{in^3}{16.38} }\\ \\=1.48\times2.205\times16.38\times10^{-3}\frac{lb}{in^3}\\ =5.35\times10^{-2}\frac{lb}{in^3}

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Lady_Fox [76]

Complete question:

Consider the reaction.

At equilibrium at 600 K, the concentrations are as follows.

2HF -----> H₂ + F₂

[HF] = 5.82 x 10-2 M

[H2] = 8.4 x 10-3 M

[F2] = 8.4 x 10-3 M

What is the value of Keq for the reaction expressed in scientific notation?

2.1 x 10-2

2.1 x 102

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Answer:

2.1 × 10^-2

Explanation:

Kequilibrum(Keq) = product/reactant

Equation for the reaction :

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Therefore,

Keq = [H2][F2] / [HF]^2

Keq = [8.4 x 10-3][8.4 x 10-3] / [5.82 x 10-2]^2

Keq = [70.56 × 10^(-3 + - 3)]/[33.8724 × 10^(-2×2)]

Keq = [70.56 × 10^-6] / [33.8724 × 10^-4]

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7 0
2 years ago
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Compound w , c6h13cl, undergoes base-promoted e2 elimination to give a single c6h12 alkene, y. compound x, c6h13br, undergoes a
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If the mass percentage composition of a compound is 72.1% Mn and 27.9% O, its empirical formula is
mojhsa [17]

Answer:

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Explanation:

Empirical formula: This is the formula that shows the ratio of elements

present in a  

compound.

   

How to determine Empirical formula

1. First arrange the symbols of the elements present in the compound

alphabetically to  determine the real empirical formula. Although, there

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2. Divide the percentage composition by the mass number.

3. Then divide through by the smallest number.

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                                                                              Mn                         O    

                         

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Divide by mass number                                       54.94                     16  

                                 

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Divide by the smallest number                         1.31                      1.31                          

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The resulting ratio is 1:1

 

Hence the Empirical formula is MnO, Manganese oxide

8 0
2 years ago
The value of ka for nitrous acid (hno2) at 25 ∘c is 4.5×10−4. what is the value of δg at 25 ∘c when [h+] = 5.9×10−2m , [no2-] =
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To get the value of ΔG we need to get first the value of ΔG°:

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when ΔG° = -19 KJ

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ΔG = -19 KJ - 0.00831 * 298K* ㏑(0.21/5.9x10^-2*6.3 x10^-4 )

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8 0
2 years ago
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