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nadezda [96]
3 years ago
9

33.56 g of fructose (C6H,206) and 18.88 g of water are mixed to obtain a 40.00 ml solution a. What is this solution's density? b

. What is the mole fraction of fructose in this solution? c. What is the solution's average molar mass? d. What is the specific molar volume of the solution?
Chemistry
1 answer:
Darina [25.2K]3 years ago
5 0

Explanation:

Mass of fructose = 33.56 g

Mass of water =  18.88  g

Total mass of the solution =  Mass of fructose + Mass of water = M

M = 33.56 g + 18.88  g =52.44 g

Volume of the solution = V = 40.00 mL

Density =\frac{Mass}{Volume}

a) Density of the solution:

\frac{M}{V}=\frac{52.44 g}{40.00 mL}=1.311 g/mL

b) Molar mass of fructose = 180.16 g/mol

Moles of fructose = n_1=\frac{ 33.56 g}{180.16 g/mol}=0.1863 mol

Molar mass of water = 18.02 g/mol

Moles of water= n_2=\frac{ 18.88 g}{18.02 g/mol}=1.0477 mol

Mole fraction of fructose in this solution:\chi_1

\chi_1=\frac{n_1}{n_1+n_2}=\frac{0.1863 mol}{0.1863 mol+1.0477 mol}

\chi_1=0.1510

Mole fraction of water = \chi_2=1-\chi_1=0.8490

c) Average molar mass of of the solution:

=\chi_1\times 180.16 g/mol+\chi_2\times 18.02 g/mol

=0.1510\times 180.16 g/mol+0.8490\times 18.02 g/mol=42.50 g/mol

d) Mass of 1 mole of solution = 42.50 g/mol

Density of the solution = 1.311 g/mL

d) Specific molar volume of the solution:

\frac{\text{Average molar mass}}{\text{Density of the mass}}

=\frac{42.50 g/mol}{1.311 g/mL}=32.42 mL/mol

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3 years ago
Given that the initial rate constant is 0.0110s−1 at an initial temperature of 21 ∘C , what would the rate constant be at a temp
gulaghasi [49]

The rate constant is mathematically given as

K2=2.67sec^{-1}

<h3>What is the Arrhenius equation?</h3>

The rate constant for a particular reaction may be calculated with the use of the Arrhenius equation. This constant can be stated in terms of two distinct temperatures, T1 and T2, as follows:

ln(\frac{K2}{K1})= (\frac{Ea}{R})*(\frac{1}{T1}-\frac{1}{T2})

Therefore

KT1= 0.0110^{-1}

T1= 21+273.15

T1= 294.15K

T2= 200  

T2=200+273.15

T2= 473.15K

Ea= 35.5 Kj/Mol

Hence, in  j/mol R Ea is

Ea=35.5*1000 j/mol R

ln(\frac{K2}{0.0110})= (\frac{35.5*1000}{8.314})*(\frac{1}{294.15}-\frac{1}{473.15}\\\\ln(\frac{K2}{0.0110})=5.492

K2/0.0110 =e^(5.492)

K2/0.0110 =242.74

K2= 242.74*0.0110

K2=2.67sec^{-1}

In conclusion, rate constant

K2=2.67sec^{-1}

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5 0
2 years ago
What is equal to the number of particles in 36 grams of carbon-12?
HACTEHA [7]

Answer:

3 moles

Explanation:

To solve this problem we will  use the Avogadro numbers.

The number 6.022×10²³ is called Avogadro number and it is the number of atoms, ions or molecules in one mole of substance.  According to this,

1.008 g of hydrogen = 1 mole =  6.022×10²³  atoms.

18 g water = 1 mole = 6.022×10²³ molecules

we are given 36 g of C-12. So,

12 g of C-12 = 1 mole

24 g of C-12 =  2 mole

36 g of C-12 = 3 mole

So 3 moles of C-12 equals to the number of particles in 36 g of C-12.

4 0
3 years ago
Someone please help i don’t have much time left
REY [17]

Answer: Energy of reactants = 30, Energy of products = 10

Exothermic

Activation energy for forward reaction is 10.

Explanation:

Exothermic reactions are defined as the reactions in which energy of the product is lesser than the energy of the reactants. The total energy is released in the form of heat and \Delta H for the reaction comes out to be negative.

Energy of reactants = 30

Energy of products = 10

Thus as energy of the product < energy of the reactant, the reaction is exothermic.

Activation energy (E_a) is the extra energy that must be supplied to reactants in order to cross the energy barrier and thus convert to products.

E_a for forward reaction is (40-30) = 10.

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7 0
4 years ago
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