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gtnhenbr [62]
1 year ago
5

When a substance is dissolved in water, the properties of the solution will be different from those of the water?

Chemistry
1 answer:
Flauer [41]1 year ago
8 0

Answer:

yes

Explanation:

boiling point, freezing point, ph , density  are just a few that would change

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The partial pressure of CO2 gas in a bottle of carbonated water is 4.60 atm at 25 ºC. How much CO2 gas (in g) will be released f
frutty [35]

If the partial pressure of CO₂ in a bottle of carbonated water decreases from 4.60 atm to 1.28 atm, the mass of CO₂ released is 0.265 g.

The partial pressure of CO₂ gas in a bottle of carbonated water is 4.60 atm at 25 ºC. We can calculate the concentration of CO₂ using Henry's law.

C = k \times P = \frac{1.65 \times 10^{-3} M }{atm}  \times 4.60 atm = 7.59 \times 10^{-3} M

We can calculate the mass of CO₂ in 1.1 L considering its molar mass is 44.01 g/mol.

\frac{7.59 \times 10^{-3} mol}{L} \times  1.1 L \times \frac{44.01 g}{mol}  = 0.367 g

Now, we will repeat the same procedure for a partial pressure of 1.28 atm.

C = k \times P = \frac{1.65 \times 10^{-3} M }{atm}  \times 1.28 atm = 2.11 \times 10^{-3} M

\frac{2.11 \times 10^{-3} mol}{L} \times  1.1 L \times \frac{44.01 g}{mol}  = 0.102 g

The mass of CO₂ released will be equal to the difference in the masses at the different pressures.

m = 0.367 g - 0.102 g = 0.265 g

If the partial pressure of CO₂ in a bottle of carbonated water decreases from 4.60 atm to 1.28 atm, the mass of CO₂ released is 0.265 g.

Learn more: brainly.com/question/18987224

<em>The partial pressure of CO₂ gas in a bottle of carbonated water is 4.60 atm at 25 ºC. How much CO₂ gas (in g) will be released from 1.1 L of the carbonated water when the partial pressure of CO2 is lowered to 1.28 atm? At 25 ºC, the Henry’s law constant for CO₂ dissolved in water is 1.65 x 10⁻³ M/atm, and the density of water is 1.0 g/cm³.</em>

5 0
2 years ago
A 46.9 gram sample of a substance has a volume of about 3.5 centimeters3. It is solid at a room temperature of 23ºC. Out of the
horrorfan [7]

Answer : (C) Hafnium is the most likely identity of the given substance.

Solution :  Given,

Mass of given substance (m) = 46.9 g

Volume of given substance (V) = 3.5 Cm^{3}

First, find the Density of given substance.

Formula used :    

Density=\frac{\text{Mass of given substance}}{\text{Voume of given substance}}

Now,put all the values in this formula, we get

Density=\frac{46.9 g}{3.5 Cm^{3} } = 13.4 g/Cm^{3}

So, we conclude that the density of given substance (13.4 g/Cm^{3}) is approximately equal to the density of Mercury and Hafnium (13.53 and 13.31 g/Cm^{3} respectively).

According to the question the substance is solid at room temperature but Mercury is liquid at room temperature. So, Mercury is not identical to the given substance.

Another element i.e, Hafnium is the element whose density is approximately equal to the given substance and also solid at room temperature. And we know that the melting point of solid is high.

So, Hafnium is the most likely element which is the identity of the given substance.

3 0
3 years ago
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I need help whats 1.09 g/mL to lbs/gal
storchak [24]
9.096491 lbs/gal hope it helps
3 0
3 years ago
Is there such thing as a pure mixture?
prisoha [69]
Pure substances are further broken down into elements and compounds. Mixture are physically combined structures that can be separated into their original components. A chemical substance is composed of one type of atom or molecule.
3 0
3 years ago
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I will give brainlyst for this so pls help
omeli [17]

Answer:

Answer -b

Explanation:

Brainliest pls

8 0
2 years ago
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