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omeli [17]
3 years ago
12

Determine the number of moles in 4.21 x 10^23 molecules of CaCI2?????

Chemistry
1 answer:
Alla [95]3 years ago
8 0

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

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Answer: 2.3 moles

Explanation:

Recall that based on Avogadro's law, 1 mole of any substance has 6.02 x 10^23 atoms

So if 1 mole of Aluminum = 6.02 x 10^23 atoms

Then, Z moles = 1.4 x 10^24 atoms

To get the value of Z, we cross multiply:

1 mole x 1.4 x 10^24 atoms = Z x (6.02 x 10^23 atoms)

1.4 x 10^24 atoms = Z x (6.02 x 10^23)

Hence, Z = (1.4 x 10^24 atoms) ➗ (6.02 x 10^23 atoms)

Z =2.3 moles

Thus, there are 2.3 moles in 1.4 x 10^24 atoms of aluminum.

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For the reaction A+B+C→D+E, the initial reaction rate was measured for various initial concentrations of reactants. The followin
erastovalidia [21]

Answer : The initial rate for a reaction will be 3.4\times 10^{-3}Ms^{-1}

Explanation :

Rate law is defined as the expression which expresses the rate of the reaction in terms of molar concentration of the reactants with each term raised to the power their stoichiometric coefficient of that reactant in the balanced chemical equation.

For the given chemical equation:

A+B+C\rightarrow D+E

Rate law expression for the reaction:

\text{Rate}=k[A]^a[B]^b[C]^c

where,

a = order with respect to A

b = order with respect to B

c = order with respect to C

Expression for rate law for first observation:

6.0\times 10^{-5}=k(0.20)^a(0.20)^b(0.20)^c ....(1)

Expression for rate law for second observation:

1.8\times 10^{-4}=k(0.20)^a(0.20)^b(0.60)^c ....(2)

Expression for rate law for third observation:

2.4\times 10^{-4}=k(0.40)^a(0.20)^b(0.20)^c ....(3)

Expression for rate law for fourth observation:

2.4\times 10^{-4}=k(0.40)^a(0.40)^b(0.20)^c ....(4)

Dividing 1 from 2, we get:

\frac{1.8\times 10^{-4}}{6.0\times 10^{-5}}=\frac{k(0.20)^a(0.20)^b(0.60)^c}{k(0.20)^a(0.20)^b(0.20)^c}\\\\3=3^c\\c=1

Dividing 1 from 3, we get:

\frac{2.4\times 10^{-4}}{6.0\times 10^{-5}}=\frac{k(0.40)^a(0.20)^b(0.20)^c}{k(0.20)^a(0.20)^b(0.20)^c}\\\\4=2^a\\a=2

Dividing 3 from 4, we get:

\frac{2.4\times 10^{-4}}{2.4\times 10^{-4}}=\frac{k(0.40)^a(0.40)^b(0.20)^c}{k(0.40)^a(0.20)^b(0.20)^c}\\\\1=2^b\\b=0

Thus, the rate law becomes:

\text{Rate}=k[A]^2[B]^0[C]^1

Now, calculating the value of 'k' by using any expression.

Putting values in equation 1, we get:

6.0\times 10^{-5}=k(0.20)^2(0.20)^0(0.20)^1

k=7.5\times 10^{-3}M^{-2}s^{-1}

Now we have to calculate the initial rate for a reaction that starts with 0.75 M of reagent A and 0.90 M of reagents B and C.

\text{Rate}=k[A]^2[B]^0[C]^1

\text{Rate}=(7.5\times 10^{-3})\times (0.75)^2(0.90)^0(0.90)^1

\text{Rate}=3.4\times 10^{-3}Ms^{-1}

Therefore, the initial rate for a reaction will be 3.4\times 10^{-3}Ms^{-1}

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