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2 years ago

Alcohols, such as 1-propanol, combine homogeneously with water because both molecules are

Chemistry

1 answer:

marta [7]2 years ago

8 0

Answer:

Explanation:

Polar molecules combine with polar molecules ( like combines with like)

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A bedroom has a volume of 118 m3 What is its volume in km3?

ruslelena [56]

I believe the problem is just simply asking for us to convert the value from one unit to the other. This case from m^3 to km^3. From the SI units, we know 1 km is equal to 1000 m. We do as follows:

118 m^3 ( 1 km / 1000 m )^3 = 1.18 x 10^-7 km^3

5 0

3 years ago

Read 2 more answers

the standard change in Gibbs free energy is Δ????°′=7.53 kJ/molΔG°′=7.53 kJ/mol . Calculate Δ????ΔG for this reaction at 298 K29

Vera_Pavlovna [14]

Answer:

ΔG = 16.218 KJ/mol

Explanation:

dihydroxyacetone phosphate ↔ glyceraldehyde-3-phosphate

ΔG = ΔG° - RT Ln Q

∴ ΔG° = 7.53 KJ/mol * ( 1000 J / KJ ) = 7530 J/mol

∴ R = 8.314 J/K.mol

∴ T = 298 K

∴ Q = [glyceraldehyde-3-phosphate] / [dihydroxyacetone phosphate]

⇒ Q = 0.00300 / 0.100 = 0.03

⇒ ΔG = 7530J/mol - (( 8.314 J/K.mol) * ( 298 K ) * Ln ( 0.03 ))

⇒ ΔG = 16217.7496 J/mol ( 16.218 KJ/mol )

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3 years ago

Which laws can be combined to form the ideal gas law?

slava [35]

Option 3- Avogadro's, Charles's and Boyle's

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2 years ago

The estimated heat of vaporization of diethyl ether using the Chen's rule is A. 29.7 KJ/mol B. 33.5 KJ/mol C. 26.4 KJ/mol D. 36.

Brums [2.3K]

Answer:

C. 26.4 kJ/mol

Explanation:

The Chen's rule for the calculation of heat of vaporization is shown below:

$\Delta H_v=RT_b\left [ \frac{3.974\left ( \frac{T_b}{T_c} \right )-3.958+1.555lnP_c}{1.07-\left ( \frac{T_b}{T_c} \right )} \right ]$

Where,

$\Delta H_v$ is the Heat of vaoprization (J/mol)

$T_b$ is the normal boiling point of the gas (K)

$T_c$ is the Critical temperature of the gas (K)

$P_c$ is the Critical pressure of the gas (bar)

R is the gas constant (8.314 J/Kmol)

For diethyl ether:

$T_b=307.4\ K$

$T_c=466.7\ K$

$P_c=36.4\ bar$

Applying the above equation to find heat of vaporization as:

$\Delta H_v=8.314\times307.4 \left [ \frac{3.974\left ( \frac{307.4}{466.7} \right )-3.958+1.555ln36.4}{1.07-\left ( \frac{307.4}{466.7} \right )} \right ]$