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4 years ago

Which contributes to the dissolution of sugar in water?

Chemistry

2 answers:

AURORKA [14]4 years ago

7 0

Option C: The dissociation of a polar covalent compound in water.

Sugar is made up of sucrose molecules containing polar covalent bonds. They have same type of oxygen-hydrogen covalent bonds that are present in water.

Since, sugar is not an ionic compound, option a and option b are incorrect. Also, due to same type of bonding in water and sugar molecules there will be no such force of attraction between them, this will opt out the option d.

Thus, dissolution of sugar in water is the dissociation of a polar covalent compound in water and option c is correct.

Veronika [31]4 years ago

3 0

I think the correct answer from the choices listed above is option d. It is the strong pull of water molecules on a polar covalent compound that contributes to the dissolution of sugar in water. Hope this answers the question.

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Which of the following choices has the compounds correctly arranged in order of increasing solubility in water? (least soluble t

larisa86 [58]

Answer: Option (d) is the correct answer.

Explanation:

As it is known that like dissolves like. So, water being a polar compound is able to dissolve only polar compounds.

Hence, a compound that is ionic or polar in nature will readily dissolve in water. Whereas non-polar compounds will be insoluble in water.

As $CCl_{4}$ is a non-polar compound. Hence, it is insoluble in water.

On the other hand, $CHCl_{3}$ is a polar compound due to difference in electronegativity of chlorine and carbon atom there will be development of partial charges. Hence, there will be dipole-dipole forces existing between them.

Whereas $NaNO_{3}$ is an ionic compound and it will readily dissociate into ions when dissolved in water. Also, there will be ion-dipole interactions between sodium and nitrate ions.

Hence, $NaNO_{3}$ will readily dissolve in water.

Thus, we can conclude that the compounds correctly arranged in order of increasing solubility in water are $CCl_{4}$ < $CHCl_{3}$ < $NaNO_{3}$ .

6 0

3 years ago

The volume, in liters, occupied by 2.50 moles of N2 gas. calculate at STP

kondaur [170]

Answer: The volume occupied by 2.50 moles of $N_2$ gas at STP is 56.0L

Explanation:

According to ideal gas equation:

$PV=nRT$

P = pressure of gas = 1 atm (at STP)

V = Volume of gas = ?

n = number of moles = 2.50

R = gas constant = $0.0821Latm/Kmol$

T =temperature = $273K$ (at STP)

$V=\frac{nRT}{P}$

$V=\frac{2.50mol\times 0.0821Latm/K mol\times 273K}{1atm}=56.0L$

Thus the volume occupied by 2.50 moles of $N_2$ gas at STP is 56.0L

6 0

3 years ago

How much ice (in grams) would have to melt to lower the temperature of 353 mL of water from 26 ∘C to 6 ∘C? (Assume the density o

Rashid [163]

Answer:

The mass of ice required to melt to lower the temperature of 353 mL of water from 26 ⁰C to 6 ⁰C is 85.4197 kg

Explanation:

Heat gain by ice = Heat lost by water

Thus,

Heat of fusion + $m_{ice}\times C_{ice}\times (T_f-T_i)=-m_{water}\times C_{water}\times (T_f-T_i)$

Where, negative sign signifies heat loss

Or,

Heat of fusion + $m_{ice}\times C_{ice}\times (T_f-T_i)=m_{water}\times C_{water}\times (T_i-T_f)$

Heat of fusion = 334 J/g

Heat of fusion of ice with mass x = 334x J/g

For ice:

Mass = x g

Initial temperature = 0 °C

Final temperature = 6 °C

Specific heat of ice = 1.996 J/g°C

For water:

Volume = 353 mL

$Density (\rho)=\frac{Mass(m)}{Volume(V)}$

Density of water = 1.0 g/mL

So, mass of water = 353 g

Initial temperature = 26 °C

Final temperature = 6 °C

Specific heat of water = 4.186 J/g°C

So,

$334x+x\times 1.996\times (6-0)=353\times 4.186\times (26-6)$

$334x+x\times 11.976=29553.16$

345.976x = 29553.16

x = 85.4197 kg

Thus,

The mass of ice required to melt to lower the temperature of 353 mL of water from 26 ⁰C to 6 ⁰C is 85.4197 kg

7 0

3 years ago

Be sure to answer all parts. Butane gas is compressed and used as a liquid fuel in disposable cigarette lighters and lightweight

Salsk061 [2.6K]

Answer:

The correct answer will be "4.60 g".

Explanation:

The given values are:

Volume of Butane = 7.96 mL

Density = 0.579 g/mL

As we know,

⇒ $Mass \ of \ Butane = Density\times Volume$

On putting the estimated values, we get

⇒ $=0.579\times 7.96$

⇒ $=4.60 \ g$

6 0

3 years ago

A closed vessel system of volume 2.5 L contains a mixture of neon and fluorine. The total pressure is 3.32 atm at 0.0°C. When th

gayaneshka [121]

Answer:

moles Ne = 0.154 mol

moles F₂ = 0.217 mol

Explanation:

Step 1: Data given

Volume of the vessel system = 2.5 L

Total pressure = 3.32 atm at 0.0 °C

The mixture is heated to 15.0 °C

The entropy of the mixture increases by 0.345 J/K

The heat capacity of monoatomic gas = 3/2R and that for a diatomic gas = 5/2R

Step 2: Define the gas

Neon is a monoatomic gas, composed of Ne atoms

⇒ Cv(Ne) ≅ (3/2)R

Fluorine is a diatomic gas, composed of F₂ molecules.

⇒ Cv(F₂) ≅ (5/2)R

Step 3: Calculate moles of gas

p*V = n*R*T

⇒ with p = the total pressure = 3.32 atm

⇒ with V = the total volume = 2.5 L

⇒ with n = the number of moles of gas

⇒ with R = the gas constant = 0.08206 L*atm/K*mol

⇒ with T = the temperature = 273.15 Kelvin

n(total) = p*V/RT = (3.32 atm*2.5 L)/(0.08206 L*atm/mol•K*273.15) = 0.3703 mol

Step 4: Calculate moles of Ne and F2

For one mole heated at constant volume,

∆S = Cv*ln(288.15/273.15) = 0.05346*Cv

⇒ ∆S for 0.3703 mol,

∆S = (0.3703 mol)(0.05346)Cv = 0.345 J/K

⇒ Cv = 17.43 J/mol*K for the Ne/F₂ mixture.

For pure Ne, Cv = (3/2)R = 1.5*8.314 J/mol*K = 12.471 J/mol*K

For pure F₂, Cv = (5/2)R = 2.5 * 8.314 J/mol*K = 20.785 J/mol*K

if X is the mole fraction of Ne, we can find X by:

17.43 J/mol*K = X* 12.471 J/mol*K + (1 – X) * 20.785 J/mol*K

⇒ 20.875 – 8.314 * X = 17.43

X = 0.415 , 1 – X = 0.585

moles Ne = (0.415)(0.3703 mol) = 0.154 mol

moles F₂ = (0.585)(0.3703 mol) = 0.217 mol

4 0

4 years ago