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ANTONII [103]
3 years ago
7

When a 2.5 mol of sugar (C12H22O11) are added to a certain amount of water the boiling point is raised by 1 Celsius degree. If 2

.5 mol of aluminum nitrate is added to the same amount of water, by how much will the boiling point be changed? Use 3 – 4 sentences to explain your answer.
Please Help!
Will give most brainly
Chemistry
2 answers:
Mademuasel [1]3 years ago
6 0

Answer is: boiling point will be changed by 4°C.

Chemical dissociation of aluminium nitrate in water: Al(NO₃)₃ → Al³⁺(aq) + 3NO⁻(aq).

Change in boiling point: ΔT =i · Kb · b.

Kb - molal boiling point elevation constant of water is 0.512°C/m, this the same for both solution.

b -  molality, moles of solute per kilogram of solvent., this is also same for both solution, because ther is same amount of substance.

i - Van't Hoff factor.

Van't Hoff factor for sugar solution is 1, because sugar do not dissociate on ions.

Van't Hoff factor for aluminium nitrate solution is approximately 4, because it dissociates on four ions (one aluminium cation and three nitrate anions). So ΔT is four times bigger.

nlexa [21]3 years ago
4 0

The answer should be 4°C....

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Alexandra [31]

Answer:

The appropriate answer is "9.225 g".

Explanation:

Given:

Required level,

= 63 ppm

Initial concentration,

= 22 ppm

Now,

The amount of free SO₂ will be:

= Required \ level -Initial \ concentration

= 63-22

= 41 \ ppm

The amount of free SO₂ to be added will be:

= 41\times 225

= 9225 \ mg

∵ 1000 mg = 1 g

So,

= 9225\times \frac{1}{1000}

= 9.225

Thus,

"9.225 g" should be added.

3 0
3 years ago
Balance the following equation. Then, given the moles of reactant or product below, determine the corresponding amount in moles
Mrrafil [7]

Answer:

Option a → 4 mol NH₃

Explanation:

This the unbalanced reaction

NH₃  +  O₂  ⟶  N₂  +  H₂O

The balanced reaction:

4NH₃  +  3O₂  →  2N₂  +  6H₂O

4 mol of ammonia

3 mol of oxygen

2 mol of nitrogen

6 mol of water

6 0
3 years ago
The following data were collected for the rate of disappearance of NO in the reaction 2NO(g)+O2(g)→2NO2(g)::
Anit [1.1K]

Answer:

a) The rate law is: v = k[NO]² [O₂]

b) The units are: M⁻² s⁻¹

c) The average value of the constant is: 7.11 x 10³ M⁻² s⁻¹

d) The rate of disappearance of NO is 0.8 M/s

e) The rate of disappearance of O₂ is 0.4 M/s

Explanation:

The experimental rates obtained can be expressed as follows:

v1 = k ([NO]₁)ᵃ ([O₂]₁)ᵇ = 1.41 x 10⁻² M/s

v2 = k ([NO]₂)ᵃ ([O₂]₂)ᵇ = 5.64 x 10⁻² M/s

v3 = k ([NO]₃)ᵃ ([O₂]₃)ᵇ = 1.13 x 10⁻¹ M/s

where:

k = rate constant

[NO]₁ = concentration of NO in experiment 1

[NO]₂ = concentration of NO in experiment 2

[NO]₃ = concentration of NO in experiment 3

[O₂]₁ = concentration of O₂ in experiment 1

[O₂]₂ = concentration of O₂ in experiment 2

[O₂]₃ = concentration of O₂ in experiment 3

a and b = order of the reaction for each reactive respectively.

We can see these equivalences:

[NO]₂ = 2[NO]₁

[O₂]₂ = [O₂]₁

[NO]₃ = [NO]₂

[O₂]₃ = 2[O₂]₂

So, v2 can be written in terms of the concentrations used in experiment 1 replacing [NO]₂ for 2[NO]₁ and [O₂]₂ by [O₂]₁ :

v2 = k (2 [NO]₁)ᵃ ([O₂]₁)ᵇ

If we rationalize v2/v1, we will have:

v2/v1 = k *2ᵃ * ([NO]₁)ᵃ * ([O₂]₁)ᵇ / k * ([NO]₁)ᵃ * ([O₂]₁)ᵇ (the exponent "a" has been distributed)

v2/v1 = 2ᵃ

ln(v2/v1) = a ln2

ln(v2/v1) / ln 2 = a

a = 2

(Please review the logarithmic properties if neccesary)

In the same way, we can find b using the data from experiment 2 and 3 and writting v3 in terms of the concentrations used in experiment 2:

v3/v2 = k ([NO]₂)² * 2ᵇ * ([O₂]₁)ᵇ / k * ([NO]₂)² * ([O₂]₂)ᵇ

v3/v2 = 2ᵇ

ln(v3/v2) = b ln 2

ln(v3/v2) / ln 2 = b

b = 1

Then, the rate law for the reaction is:

<u>v = k[NO]² [O₂]</u>

Since the unit of v is M/s and the product of the concentrations will give a unit of M³, the units of k are:

M/s = k * M³

M/s * M⁻³ = k

<u>M⁻² s⁻¹ = k </u>

To obtain the value of k, we can solve this equation for every experiment:

k = v / [NO]² [O₂]

for experiment 1:

k = 1.41 x 10⁻² M/s / (0.0126 M)² * 0.0125 M = 7.11 x 10³ M⁻² s⁻¹

for experiment 2:

k = 7.11 x 10³ M⁻² s⁻¹

for experiment 3:

k = 7.12 x 10³ M⁻² s⁻¹

The average value of k is then:

(7.11 + 7.11 + 7.12) x 10³ M⁻² s⁻¹ / 3 = <u>7.11 x 10³ M⁻² s⁻¹ </u>

The rate of the reaction when [NO] = 0.0750 M and [O2] =0.0100 M is:

v = k [NO]² [O₂]

The rate of the reaction in terms of the disappearance of NO can be written this way:

v = 1/2(Δ [NO] / Δt) (it is divided by 2 because of the stoichiometric coefficient of NO)

where (Δ [NO] / Δt) is the rate of disappearance of NO.

Then, calculating v with the data provided by the problem:

v = 7.11 x 10³ M⁻² s⁻¹ * (0.0750M)² * 0.0100M = 0.4 M/s

Then, the rate of disappearance of NO will be:

2v = Δ [NO] / Δt = <u>0.8 M/s</u>

The rate of disappearance of O₂ has to be half the rate of disappearance of NO because two moles of NO react with one of O₂. Then Δ [O₂] / Δt = <u>0.4 M/s</u>

With calculations:

v = Δ [O₂] / Δt = 0.4 M/s (since the stoichiometric coefficient is 1, the rate of disappearance of O₂ equals the rate of the reaction).

3 0
3 years ago
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Daniel [21]

Answer:

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hope this helps

8 0
1 year ago
A 15.00 % by mass solution of lactose (C 12H 22O 11, 342.30 g/mol) in water has a density of 1.0602 g/mL at 20°C. What is the mo
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Answer:

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Explanation:

342.3g/mol is présent in 1000

what about in 15??

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