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alukav5142 [94]
3 years ago
9

A.Identify which of the following substances dissolve in water, explaining why for each example: CH4, He, MgSO4, NaHCO3

Chemistry
2 answers:
viva [34]3 years ago
7 0

Answer:

A) MgSO4 and NHCO3 will dissolve in water

B)C12H22O11,Na2SO4 and KCl will dissolve in water

CHCl3,CH4 and C2H6O will dissolve in gasoline

C2H6O will dissolve in both

Explanation:

There is a saying in chemistry that like dissolves like. Polar substances dissolve in polar solvents because they interact effectively with them. Non polar solutes also dissolve in nonpolar solvents.

Ionic substances such as Na2SO4,KCl will dissolve in water.

C12H22O11 contains polar groups which interact with water and lead to its dissolution. C2H6O could interact with water by hydrogen bonding via its OH group and can also dissolve in gasoline because it is an organic substance and volatile. CHCl3,CH4 are purely nonpolar substances which can only dissolve in gasoline.

Mars2501 [29]3 years ago
6 0

A. MgSO4 and NahCO3 are both ionic compounds and a salt so they are soluble

B.

1. Sucrose (C12H22O11) will dissolve in water

2.Na2SO4 will dissolve in water

3. KCl will dissolve in water

4. CHCI3 will dissolve gas

5. Methane (CH4) will dissolve in gas

6. C2H60 will be dissolved in both

The left side of the ethanol molecule has non-polar characteristics because of the C-C and C-H bonds which allows the alcohol to dissolve in a non-polar solvent and on the right side of the ethanol molecule we see an alcohol functional group (OH). This side of the molecule has polar characteristics which allow the molecule to be dissolved in a polar solvent such as water.

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If the vapor pressure of ethanol at 34.7degree C is 100
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Answer:

we will use the Clausius-Clapeyron equation to estimate the vapour pressures of the boiling ethanol at sea level pressure of 760mmHg:

 ln (P2/P1) = \frac{ΔvapH}{R}([tex]\frac{1}{T1}-\frac{1}{T2})

where

P1 and P2 are the vapour pressures at temperatures  T1 and T2

Δ vapH  = the enthalpy of vaporization of the ETHANOL

R  = the Universal Gas Constant

In this problem,

P 1 = 100 mmHg

;  T 1 = 34.7 °C = 307.07 K

P 2 = 760mmHg

T 2 =T⁻²=?

Δ vap H = 38.6 kJ/mol

R = 0.008314 kJ⋅K -1 mol -1

ln ( 760/10)=(0.00325 - T⁻²) (38.6kJ⋅mol-1 /0.008314 )

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