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julsineya [31]
3 years ago
12

A solution is prepared by dissolving 91.7 g fructose in 545 g of water. Determine the mole fraction of fructose if the final vol

ume of the solution is 576 mL.
Chemistry
1 answer:
oee [108]3 years ago
5 0

Answer:

Mole fraction = 0,0166

Explanation:

Mole fraction is defined as mole of a compound per total moles of the mixture. In the solution, the solute is fructose and the solvent is water. That means you need to find moles of fructose and moles of water.

The molecular mass of fructose is 180,16g/mol and mass of water is 18,02 g/mol. Using these values:

91,7g fructose × (1mol / 180,16g) = <em>0,509 moles of fructose</em>

545g water × (1mol / 18,02g) = <em>30,24 moles of water</em>

Thus, mole fraction of fructose is:

\frac{0,509 moles}{0,509mol + 30,24mol} = 0,0166

<em>Mole fraction = 0,0166</em>

I hope it helps!

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sergejj [24]

Answer:

The products are carbon dioxide and water

Explanation:

Step 1: Data given

Combustion = a reaction in which a substance reacts with oxygen gas, releasing energy in the form of light and heat. Combustion reactions must involve  O2  as one reactant.

Step 2: The complete combustion of C3H7OH:

For the combustion of 1-propanol, we need O2.

The products of this combustion are CO2 and H2O.

C3H7OH + O2→ CO2 + H2O

On the left side we have 3x C (in c3H7OH), on the right side we have 1x C (in CO2). To balance the amount of C, we have to multiply CO2 on the right side by 3

C3H7OH + O2→ 3CO2 + H2O

On the left side we have 8x H (in C3H7OH) and 2x on the right side (in H2O). To balance the amount of H, we have to multiply H2O, on the right side by 4.

C3H7OH + O2→ 3CO2 + 4H2O

On the left side we have 3x O (1x in C3H7OH and 2x in O2), on the right side we have 10x O (6x in CO2 and 4x in H2O).

To balance the amount of O on both sides, we have to multiply C3H7OH by 2, multiply O2 by 9. Then we have to multiply 3CO2 by 2 and 4H2O by 2. Now the equation is balanced.

2C3H7OH + 9O2→ 6CO2 + 8H2O

For 2 moles propanol, we need 9 moles of O2 to produce 6 moles of CO2 and 8 moles Of H2O

The products are carbon dioxide and water

7 0
3 years ago
Which buffer would be better able to hold a steady pH on the addition of strong acid, buffer 1 or buffer 2? Explain. Buffer 1: a
Talja [164]

Answer:

Buffer 1.

Explanation:

Ammonia is a weak base. It acts like a Bronsted-Lowry Base when it reacts with hydrogen ions.

\rm NH_3\; (aq) + H^{+}\; (aq) \to {NH_4}^{+}\; (aq).

\rm NH_3 gains one hydrogen ion to produce the ammonium ion \rm {NH_4}^{+}. In other words, \rm {NH_4}^{+} is the conjugate acid of the weak base \rm NH_3.

Both buffer 1 and 2 include

  • the weak base ammonia \rm NH_3, and
  • the conjugate acid of the weak base \rm {NH_4}^{+}.

The ammonia \rm NH_3 in the solution will react with hydrogen ions as they are added to the solution:

\rm NH_3\; (aq) + H^{+}\; (aq) \to {NH_4}^{+}\; (aq).

There are more \rm NH_3 in the buffer 1 than in buffer 2. It will take more strong acid to react with the majority of \rm NH_3 in the solution. Conversely, the pH of buffer 1 will be more steady than that in buffer 2 when the same amount of acid has been added.

5 0
3 years ago
If the concentration of Mg2+ in the solution were 0.039 M, what minimum [OH−] triggers precipitation of the Mg2+ ion? (Ksp=2.06×
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Answer:

2.30 × 10⁻⁶ M

Explanation:

Step 1: Given data

Concentration of Mg²⁺ ([Mg²⁺]): 0.039 M

Solubility product constant of Mg(OH)₂ (Ksp): 2.06 × 10⁻¹³

Step 2: Write the reaction for the solution of Mg(OH)₂

Mg(OH)₂(s) ⇄ Mg²⁺(aq) + 2 OH⁻(aq)

Step 3: Calculate the minimum [OH⁻] required to trigger the precipitation of Mg²⁺ as Mg(OH)₂

We will use the following expression.

Ksp = 2.06 × 10⁻¹³ = [Mg²⁺] × [OH⁻]²

[OH⁻] = 2.30 × 10⁻⁶ M

3 0
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An interplanetary probe returns to Earth with soil samples. The atoms in the sample are sorted by their number of protons. A new
Leya [2.2K]

Answer is: the average atomic mass is 232.

ω₁ = 20% ÷ 100%.

ω₁ = 0.20.

ω₂ = 80% ÷ 100%.

ω₂ = 0.80.

Ar₁ = 120 (number of protons) + 120 (number of neutrons).

Ar₁ = 240.

Ar₂ = 120 + 110 .

Ar₂ = 230.

Average atomic mass of atoms of bolognium =  

Ar₁ · ω₁ + Ar₂ · ω₂.  

Average atomic mass of atoms of bolognium =  240 · 0.2 + 230 · 0.8.  

Average atomic mass of atoms of bolognium = 48 + 184.  

Average atomic mass of atoms of bolognium = 232.

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