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Ghella [55]
3 years ago
11

Why is oil unable to dissolve well in water? Water is polar and oil is nonpolar, which means they have no attraction for each ot

her. Oil molecules are too large to fit between the closely spaced water molecules in the liquid state. Water has a greater density than oil, which means that they are not compatible for mixing together. The cohesive properties of water molecules make it difficult for substances like oil to dissolve.
Chemistry
2 answers:
timofeeve [1]3 years ago
5 0

Answer:

Water is polar, oil is non polar, which means they have no attraction for each other

Explanation:

Marat540 [252]3 years ago
3 0
Because water is polar and oil is nonpolar, their molecules are not attracted to each other. The molecules of a polar solvent like water are attracted to other polar molecules, such as those of sugar. This explains why sugar has such a high solubility in water. Ionic compounds, such as sodium chloride, are also highly soluble in water. Because water molecules are polar, they interact with the sodium and chloride ions. In general, polar solvents dissolve polar solutes, and nonpolar solvents dissolve nonpolar solutes. This concept is often expressed as “Like dissolves like.” So many substances dissolve in water that it is sometimes called the universal solvent. Water is considered to be essential for life because it can carry just about anything the body needs to take in or needs to get rid of.
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39. Period is the force of attraction between objects resulting in objects pulling towards each
storchak [24]

Answer:

True

Explanation:

if there was nothing to avoid of objects pulling towards one another many results may happen

8 0
2 years ago
CO2(g)+CCl4(g)⇌2COCl2(g) Calculate ΔG for this reaction at 25 ∘C under these conditions: PCO2PCCl4PCOCl2===0.140 atm0.185 atm0.7
padilas [110]

<u>Answer:</u> The \Delta G for the reaction is 54.425 kJ/mol

<u>Explanation:</u>

For the given balanced chemical equation:

CO_2(g)+CCl_4(g)\rightleftharpoons 2COCl_2(g)

We are given:

\Delta G^o_f_{CO_2}=-394.4kJ/mol\\\Delta G^o_f_{CCl_4}=-62.3kJ/mol\\\Delta G^o_f_{COCl_2}=-204.9kJ/mol

To calculate \Delta G^o_{rxn} for the reaction, we use the equation:

\Delta G^o_{rxn}=\sum [n\times \Delta G_f(product)]-\sum [n\times \Delta G_f(reactant)]

For the given equation:

\Delta G^o_{rxn}=[(2\times \Delta G^o_f_{(COCl_2)})]-[(1\times \Delta G^o_f_{(CO_2)})+(1\times \Delta G^o_f_{(CCl_4)})]

Putting values in above equation, we get:

\Delta G^o_{rxn}=[(2\times (-204.9))-((1\times (-394.4))+(1\times (-62.3)))]\\\Delta G^o_{rxn}=46.9kJ=46900J

Conversion factor used = 1 kJ = 1000 J

The expression of K_p for the given reaction:

K_p=\frac{(p_{COCl_2})^2}{p_{CO_2}\times p_{CCl_4}}

We are given:

p_{COCl_2}=0.735atm\\p_{CO_2}=0.140atm\\p_{CCl_4}=0.185atm

Putting values in above equation, we get:

K_p=\frac{(0.735)^2}{0.410\times 0.185}\\\\K_p=20.85

To calculate the gibbs free energy of the reaction, we use the equation:

\Delta G=\Delta G^o+RT\ln K_p

where,

\Delta G = Gibbs' free energy of the reaction = ?

\Delta G^o = Standard gibbs' free energy change of the reaction = 46900 J

R = Gas constant = 8.314J/K mol

T = Temperature = 25^oC=[25+273]K=298K

K_p = equilibrium constant in terms of partial pressure = 20.85

Putting values in above equation, we get:

\Delta G=46900J+(8.314J/K.mol\times 298K\times \ln(20.85))\\\\\Delta G=54425.26J/mol=54.425kJ/mol

Hence, the \Delta G for the reaction is 54.425 kJ/mol

7 0
3 years ago
Calculate the density for a rectangular block using the following measurements: Length = 10 cm, Width = 1.1 cm, Height = 15 cm,
Ksivusya [100]

Answer:

0.40 g/cm3

Explanation:

density = mass / volume.

mass = 65.2 grams

volume = 10*1.1*15=165 cm3

so density = 65.2/165=0.40 g/cm3

3 0
2 years ago
Pls help this my final 20 points!!!! Be right
Anastaziya [24]

Answer:

I think it's B

Explanation:

I hope it helps

5 0
2 years ago
8. Which gas contributes the largest part of air? *
timurjin [86]

Answer:

Nitrogen

Explanation:

Nitrogen present 78% in the earth's atmosphere

5 0
3 years ago
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