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Serga [27]
3 years ago
10

A chemist mixes 1.00 g CuCl2 with an excess of (NH4)2HPO4 in dilute aqueous solution . He measures the evolution of 670 J of hea

t as the two substances react to give Cu3(PO4)2(s). Compute the ΔH that would result from the reaction of 1.00 mo! CuCl2 with an excess of (NH4)2HPO4.
Chemistry
1 answer:
kaheart [24]3 years ago
4 0

Answer:

\Delta H will be 90054 J

Explanation:

Number of moles = (mass)/(molar mass)

Molar mass of CuCl_{2} = 134.45 g/mol

So, 1.00 g of CuCl_{2} = \frac{1.00}{134.45}mol of CuCl_{2} = 0.00744 mol of CuCl_{2}

0.00744 mol of CuCl_{2} produces 670 J of heat

So, 1 mol of CuCl_{2} produces \frac{670}{0.00744}J of heat or 90054 J of heat

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A 35.66g sample of copper is heated using 600j of energy. if the original temperature of the copper is 85C what is its final tem
ira [324]

This problem is providing the mass, energy, initial temperature and specific heat of a sample of copper that is required to calculate the final temperature.

Thus, we recall the general heat equation:

Q=mC(T_f-T_i)\\

Which has to be solved for the final temperature, T_f as follows:

T_f=T_i+\frac{Q}{mC}

Finally, we plug in the numbers to obtain:

T_f=85\°C+\frac{600J}{35.66g*0.38\frac{J}{g\°C} } \\\\T_f=129.3\°C

However, this result is not given in the choices.

Learn more:

  • brainly.com/question/14383794
8 0
2 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
Calculate the volume of water which was liberated during the decomposition of hydrogen peroxide.
nikitadnepr [17]

The answer is :

The normal equation for the decomposition of hydrogen peroxide indicates that 2 vol. peroxide vapor should give rise to 2 vol. water vapor and 1 vol. oxygen.

What is hydrogen peroxide ?

  • Hydrogen peroxide is water (H2O) with an extra oxygen molecule (H2O2).
  • The extra oxygen molecule oxidizes, which is how peroxide gets its power says Dr. Beers.
  • This oxidation kills germs and bleaches color from porous surfaces like fabrics.
  • Hydrogen peroxide is found in biological systems including the human body.
  • Enzymes that use or decompose hydrogen peroxide are classified as peroxidases.

To learn more about Hydrogen peroxide visit: brainly.com/question/18709693?

#SPJ4

4 0
1 year ago
Name three ways in which unicellular organisms can move. Describe one of them.
Viefleur [7K]

Answer:

Unicellular organisms achieve locomotion using cilia and flagella. By creating currents in the surrounding environment, cilia and flagella can move the cell in one direction or another.

Explanation:

3 0
2 years ago
Read 2 more answers
How many atoms of nitrogen are present in 2.49 moles of nitrogen trifluoride ? atoms of nitrogen?
stepladder [879]
<span>Nitrogen trifluoride - NF3.
1 mol NF3 contains 1 mol atoms of Nitrogen
2.49 mol NF3 contains 2.49 mol atoms of Nitrogen

1 mol   ----  6.02 *10²³ atoms
2.49 mol ----- 2.49*6.02*10²³ = 15.0*10²³ atoms of N</span>
6 0
2 years ago
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