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Lelu [443]
2 years ago
12

SOLVE The density of a gas is 0.68 g/mL. What amount of space would 23.8 g of this gas occupy?

Chemistry
1 answer:
Contact [7]2 years ago
6 0

Answer:

<h2>The answer is 35 mL</h2>

Explanation:

Density of a substance can be found by using the formula

Density(\rho) =  \frac{mass}{volume}

From the question we are finding the amount of space the gas will occupy that's the volume of the gas

Making volume the subject we have

volume =  \frac{mass}{Density}

From the question

mass = 23.8 g/mL

Density = 0.68 g/mL

Substitute the values into the above formula and solve

That's

volume =  \frac{23.8}{0.68}

We have the final answer as

<h3>35 mL</h3>

Hope this helps you

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As a hydrated compound is heated, it decreases in
Sergio [31]
A hydrated substance will decrease in mass when heated since the water is being boiled away.  

I hope this helps.  Let me know if anything is unclear.
8 0
3 years ago
1]
balu736 [363]

Explanation:

Given -

  • An organic compound gives H₂ gas with Na
  • On treatment with alkaline iodine it gives yellow ppt.
  • On oxidation with CrO₃/H⁺ forms an aldehyde (C₂H₄O)

To Find -

  • Name the compound and write the reaction involved

Now,

Let A be the organic compound.

Then,

  1. A + Na → + H₂↑
  2. A + I₂ → CHI₃ (yellow ppt.)
  3. A + CrO₃ + H⁺ → C₂H₄O

Now,

Here we see that compound A reacts with chromic oxide (CrO₃) in the presence of acidic medium gives aldehyde.

  • Functional group of aldehyde = —CHO

And It forms only 2 Carbon aldehyde it means, It is Ethanal (CH₃CHO).

Compound A reacts with chromic oxide (CrO₃) in the presence of acidic medium gives ethanal.

It means,

We know that 1° alcohol on oxidation gives aldehyde.

Here it gives 2 Carbon aldehyde.

It means,

Here 2 Carbon and 1° alcohol is used.

Now,

Its cleared that Compound A is Ethanol.

Reaction Involved -

  1. CH₃CH₂OH + Na → CH₃CH₂O⁻Na⁺ + H₂↑
  2. CH₃CH₂OH + I₂ + OH⁻ → CHI₃↓ + HCOO⁻ + HI + H₂O
  3. CH₃CH₂OH + CrO₃ + H⁺ → CH₃CHO

6 0
3 years ago
If the molar absorptivity constant for the red dye solution is 5.56×104 M-1cm-1, calculate the molarity of the red dye solution
Shtirlitz [24]

Explanation:

a) Using Beer-Lambert's law :

Formula used :

A=\epsilon \times c\times l

where,

A = absorbance of solution = 0.945

c = concentration of solution = ?

l = length of the cell = 1.20 cm

\epsilon = molar absorptivity of this solution =5.56\times 10^4 M^{-1} cm^{-1}

0.945=5.56\times 10^4 M^{-1} cm^{-1}\times 1.20 \times c

c=1.4163\times 10^{-5} M=14.16 \mu M

(1\mu M=10^{-6} M)

14.16 μM is the molarity of the red dye solution at the optimal wavelength 519nm and absorbance value 0.945.

b) c=1.4163\times 10^{-5} mol/L

1 L of solution contains 1.4163\times 10^{-5} moles of red dye.

Mass of 1.4163\times 10^{-5} moles of red dye:

1.4163\times 10^{-5}\times 879.86g/mol=0.01246 g

(w/v)\%=\frac{\text{Mass of solute (g)}}{\text{Volume of solvent (mL)}}\times 100

red(w/v)\%=\frac{0.01246 g}{1000 mL}\times 100=0.001246\%

c) In order to dilute red dye solution by 5 times, we will need to add 1 L of water to solution of given concentration.

Concentration of red dye solution = c=1.4163\times 10^{-5} M

Concentration of red solution after dilution = c'

c=c'\times 5

1.4163\times 10^{-5} M=c'\times 5

c'=2.83\times 10^{-6} M

The final concentration of the diluted solution is 2.83\times 10^{-6} M

8 0
3 years ago
When the reaction mixture is worked-up, it is first washed three times with 5% sodium bicarbonate, and then with a saturated nac
Ann [662]

Solution:

After the reaction of mixture is worked-up Washing three times the organic  with sodium carbonate helps to decrease the solubility of the organic layer into the aqueous layer. This allows the organic layer to be separated more easily.

And then the reaction washed by saturated NACL we have The bulk of the water can often be removed by shaking or "washing" the organic layer with saturated aqueous sodium chloride (otherwise known as brine). The salt water works to pull the water from the organic layer to the water layer.

5 0
3 years ago
Calculate the boiling point of a solution of 500.0 g of ethylene glycol (c2h6o2) dissolved in 500.0 g of water. kf = 1.86°c/m a
Bingel [31]

Answer:The boiling point of the solution is 108° C.

Explanation:

Boiling point of pure water=T=100^oC

Boiling point of water after addition of 500 g of ethylene glycol=T_f

Mass of water = 500g = 0.5 kg (1000 g = 1 kg)

\Delta T_f=K_b\times \frac{\text{Mass of ethlyene glycol}}{\text{Molar mass of ethylene glycol}}

\Delta T_f=0.512^oC/m\times \frac{500 g}{62.07 g/mol\times 0.5 kg}

\Delta T_f=8.24 ^oC

\Delta T_f=T_f-T

8.24^oC=T_f-100^oC

T_f=108.24^oC

The boiling point of the solution is 108° C.

7 0
2 years ago
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