The answer to this question

Determine the freezing point of a water solution of fructose (C6H12O6) made by dissolving 92.0 g of fructose in 202 g of water.

Naya [18.7K]

Answer:

THE FREEZING POINT OF A WATER SOLUTION OF FRUCTOSE MADE BY DISSOLVING 92 g OF FRUCTOSE IN 202g OF WATER IS -4.70 ◦C

Explanation:

To calculate the freezing point of a water solution of fructose,

1. calculate the molar mass of Fructose

( 12 * 6 + 1*12 + 16*6) =72 + 12 +96 = 180 g/mol

2. calculate the number of moles of fructose in the solution

number of moles = mass / molar mass

n = 92 g / 180 g/mol

n = 0.511 moles.

3. calculate the molarity of the solution

molarity = moles / mass of water in kg

molarity = 0.5111 / 202 g /1000 g

molarity = 0.5111 / 0.202

molarity = 2.529 M

4. calculate the change in the freezing point of pure solvent and solution ΔTf

ΔTf = Kf * molarity of the solute

Kf = 1.86 ◦C/m for water

ΔTf = 1.86 * 2.529

ΔTf = 4.70 C

5. the freezing point is therefore

0.00 ◦C - 4.70 ◦C = -4.70 ◦C

4 0

3 years ago

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pplication)Using multiple models simultaneously: This FNT refers to a processinvolving three moles of a diatomic gas (which beha

Fiesta28 [93]

Complete Question

Questions Diagram is attached below

Answer:

* $W=1142.86Joule$

* $Q=997.7J$

* $H=2140.5J$

Explanation:

From the question we are told that:

Temperature $T=337K$

Pressure $P=(60-55)Pa*10^5$

Volume $V=(1.6-1.4)m^3*10^{-3}$

Generally the equation for gas Constant is mathematically given by

$\frac{P_2}{P_1}=\frac{V_1}{V_2}^n$

$\frac{55*10^5}{60*10^5}=\frac{1.4*10^{-3}}{1.6*10^{-3}}^n$

$n=0.65$

Therefore

Work-done

$W=\int{pdv}$

$W=\frac{55*10^5*1.6*10^{-3}*60*10^5*1.4*10^{-3}}{1-0.65}$

$W=1142.86Joule$

Generally the equation for internal energy is mathematically given by

$Q=mC_vdT\\\\Q=\frac{3*1*3.314*16}{1.4-1}$

$Q=997.7J$

Therefore

$H=Q+W$

$H=997.7J-11.42.9$

$H=2140.5J$

5 0

3 years ago

Aluminum metal reacts with bromine, a red-brown liquid with a noxious odor. The reaction is vigorous and produces aluminum bromi

GuDViN [60]

<u>Answer:</u> The mass of bromine reacted is 160.6 grams.

<u>Explanation:</u>

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ .....(1)

Given mass of aluminium = 18.1 g

Molar mass of aluminium = 27 g/mol

Putting values in equation 1, we get:

$\text{Moles of aluminium}=\frac{18.1g}{27g/mol}=0.670mol$

The chemical equation for the reaction of aluminium and bromide follows:

$2Al+3Br_2\rightarrow 2AlBr_3$

By Stoichiometry of the reaction:

2 moles of aluminium reacts with 3 moles of bromine gas

So, 0.670 moles of aluminium will react with = $\frac{3}{2}\times 0.670=1.005mol$ of bromine gas.

Now, calculating the mass of bromine gas, we use equation 1:

Moles of bromine gas = 1.005 moles

Molar mass of bromine gas = 159.81 g/mol

Putting values in equation 1, we get:

$1.005mol=\frac{\text{Mass of bromine}}{159.81g/mol}\\\\\text{Mass of bromine}=(1.005mol\times 159.81g/mol)=160.6g$

Hence, the mass of bromine reacted is 160.6 grams.

5 0

3 years ago

Drag each label to the correct location on the image identify the parts of the energy diagram

diamong [38]

If you would’ve attached a picture I’m sure it would’ve been a lot easier.

4 0

3 years ago

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Is oxygen a transional element?

Ede4ka [16]

Transitional elements are usually described as elements of the d block. oxygen is not in the d block, and hence is not a transitional element.

6 0

4 years ago