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Yakvenalex [24]
3 years ago
6

What is the boiling point of a 1.5 m aqueous solution of fructose? the boiling point elevation constant of water is 0.515°c/m.

assume solute is a nonelectrolyte?
Chemistry
1 answer:
Ronch [10]3 years ago
4 0
Answer is: <span>he boiling point of a 1.5 m aqueous solution of fructose is </span>100.7725°C.
The boiling point elevation is directly proportional to the molality of the solution according to the equation: ΔTb = Kb · b.<span>
ΔTb -  the boiling point elevation.
Kb - the ebullioscopic constant. of water.
b - molality of the solution.
Kb = 0.515</span>°C/m.
b = 1.5 m.
ΔTb = 0.515°C/m · 1.5 m.
ΔTb = 0.7725°C.
Tb(solution) = Tb(water) + ΔTb.
 Tb(solution) = 100°C + 0.7725°C = 100.7725°C.
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This is the equation for the combustion of propane.
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Answer:

B

Explanation:

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reactants -> products

Other phenomena like heat are omitted because are not always present, that is, only compounds are included. Therefore, in this reaction the reactants are C3H8 (propane) and O2 ( oxygen) and the products are CO2 (carbon dioxide) and H2O (water)

8 0
3 years ago
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If you start with 0.30 m mn2 , at what ph will the free mn2 concentration be equal to 4.6 x 10-11 m?
aksik [14]

If you start with 0.30 m Mn₂ , at 12.5 pH, free Mn₂ concentration be equal to 4.6 x 10⁻¹¹ m

Initial molarity of Mn₂ = 0.30 M

Final molarity of Mn₂ = 4.6 x 10⁻¹¹

pH = ?

Ksp [Mn(OH)₂] = 4.6 x 10⁻¹⁴ (standard value)

Write the ionic equation

    Mn(OH)₂   →    Mn⁺² + 2OH⁻

    [Mn⁺²] = 4.6 x 10⁻¹¹

We will calculate the concentration of OH⁻ by using Ksp expression

    Ksp = [Mn⁺²][OH-]²

    [Mn⁺²][OH⁻]² = 4.6 x 10⁻¹⁴

    [OH⁻]² = 4.6 x 10⁻¹⁴ / 4.6 x 10⁻¹¹

    [OH⁻]² = 10⁻³

    [OH⁻] = (10⁻³)¹⁽²

    [OH⁻] = 0.0316 M

Calculate the pOH

    pOH = -log [OH⁻]

    pOH =  -log [0.0316]

    pOH = 1.5

Now calculate pH

   pH = 14 - pOH

   pH = 14 - 1.5

   pH = 12.5

You can also learn about molarity from the following question:

brainly.com/question/14782315

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Add me on discord if u wanna be study buddies or just normal friends lol
Katyanochek1 [597]
Yesssdirrrrrr and yessss
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3 years ago
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Describe the movement of electrons when the atom colors by applying ground state and excitement state?
Sladkaya [172]

Answer:

When the excited electron fall back to the lower energy levels the energy is released in the form of radiations.The characteristics bright colors are due to the these emitted radiations. These emitted radiations can be seen if they are fall in the visible region of spectrum

Explanation:

The electron is jumped into higher level and back into lower level by absorbing and releasing the energy.

The process is called excitation and de-excitation.

Excitation:

When the energy is provided to the atom the electrons by absorbing the energy jump to the higher energy levels. This process is called excitation. The amount of energy absorbed by the electron is exactly equal to the energy difference of orbits.  For example if electron jumped from K to L it must absorbed the energy which is equal the energy difference of these two level. The excited electron thus move back to lower energy level which is K by releasing the energy because electron can not stay longer in higher energy level and comes to ground state.

De-excitation:

When the excited electron fall back to the lower energy levels the energy is released in the form of radiations. this energy is exactly equal to the energy difference between the orbits. The characteristics bright colors are due to the these emitted radiations. These emitted radiations can be seen if they are fall in the visible region of spectrum

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3 years ago
Which of the following ions is least likely to form colored compounds?
andreyandreev [35.5K]

Answer:

i think d maybe correct me if im wrong

Explanation:

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