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aleksandr82 [10.1K]
3 years ago
5

If there are 7 moles of copper dissolved in 15 moles of zinc to make 22 moles of a brass solution. What is the mole fraction of

copper?
Chemistry
1 answer:
iragen [17]3 years ago
6 0
Mole fraction of copper is 7/22
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Determine the number of moles in 4.21 x 10^23 molecules of CaCI2?????
Alla [95]

Explanation:

Given parameters:

Number of molecules = 4.21 x 10²³ molecules

Unknown parameters:

Number of moles

Solution:

A mole can be defined as the amount of a substance that contains the avogadro's number of particles i.e 6.02 x 10²³

  To find the number of moles:

          Number of moles = \frac{number of molecules }{[tex]6.02 x 10^{23}}[/tex]

  Number of moles =  \frac{4.21 x 10[tex]^{23} }{6.02 x 10^{23}}[/tex]

Number of moles of CaCl₂ = 0.699moles

Learn more;

mole calculation brainly.com/question/13064292

#learnwithBrainly

8 0
3 years ago
How many atoms does 2.0 moles of He represent
Rudiy27

We use the following formula to calculate the number of atoms:

n (mol) = N(number of atoms) / NA

N(He) = n(mol) · NA

N(He) = 2,0 moles · 6.02·1023 = 12.04·1023 atoms

6 0
4 years ago
What is the limiting reactant for the following balanced equation when 9 moles of AlF3 are mixed with 12 miles of O2?
tamaranim1 [39]
<h2>Answer:AlF_{3} </h2>

Explanation:

The chemical equation of the reaction that occurs when AlF_{3} reacts with O_{2} is

4AlF_{3}+3O_{2}→2Al_{2}O_{3}+6F_{2}

4 moles of AlF_{3} requires 3 moles of O_{2}.

1 mole of AlF_{3} requires \frac{3}{4} moles of O_{2}.

Given that we have 9 moles of AlF_{3}.

9 moles of AlF_{3} requires \frac{3}{4}\times 9=6.75 moles of O_{2}.

But we have 12 moles of O_{2}.

So,AlF_{3}  will be consumed first.

So,AlF_{3}  is the limiting reagent.

3 0
3 years ago
1 mol super cooled liquid water transformed to solid ice at -10 oC under 1 atm pressure.
Arada [10]

Answer:

Explanation:

Given that:

number of moles of super cooled liquid water = 1

Melting enthalpy of ice = 6020 J/mol

Freezing point =0 °C = (0 + 273 K)= 273 K

The decrease in entropy of the system during freezing for 1 mol (i.e during transformation from liquid water to solid ice )  = - 6020 J/mol × 1 mol /273 K = -22.051 J/K

Entropy change during further cooling from 0 °C (273 K) to -10 °C (263 K)

\Delta \ S = \int\limits^{T_2}_{T_1}\dfrac{nC_p(s)dT}{T}

\Delta \ S = {nC_p(s)In \dfrac{T_2}{T_1}

\Delta \ S = {(1*37.7)In \dfrac{263}{273}

Δ S = -1.4 J/K

Total entropy change of the system = - 22.05 J/K - 1.4 J/K = - 23.45 J/K

Entropy change of universe = entropy change of the system+ entropy change of the surrounding

According to the second law of thermodynamics

Entropy change of universe  >0

SO,

Entropy change of the system + entropy change in the surrounding > 0

Entropy change in the surrounding > - entropy change of the system

Entropy change in the surrounding > - (- 23.53 J/K)

Entropy change in the surrounding > 23.53 J/K

b) Make some comments on entropy changes from the obtained data.

From the data obtained; we will realize that the entropy of the system decreases as cooling takes place when water is be convert to ice , randomness of these molecules reduces and as cooling proceeds , hence, entropy reduces more as well and the liberated heat will go into the surrounding due to this entropy of the surrounding increasing.

4 0
3 years ago
Look at the picture and answer.
Llana [10]

Answer:

In order to balance the chemical equation, you need to make sure the number of atoms of each element on the reactants side is equal to the number of atoms of each element on the product side. In order make both sides equal, you will need to multiply the number of atoms in each element until both sides are equal.

Hope this helped you.

6 0
3 years ago
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