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Elodia [21]
3 years ago
11

Which of these atoms has the largest number of neutrons in the nucleus?

Chemistry
1 answer:
viktelen [127]3 years ago
8 0

Holonium and Gadolinium has the highest number of neutrons in the nucleus.

Looking at the atoms listed;

Dysprosium has 66 protons

Holonium has 67 protons

Neodymium has 60 protons

Europium has 63 protons

Gadolinium has 64 protons

Then,

Number of neutrons = Mass number - number of protons

For Dysprosium

157 - 66 = 91 neutrons

For Holonium

162 - 67 = 95 neutrons

For Neodymium

149 - 60 = 89 neutrons

For Europium

148 - 63 = 85 neutrons

For Gadolinium

159 - 64 = 95 neutrons

Hence, Holonium and Gadolinium has the highest number of neutrons in the nucleus.

Learn more: brainly.com/question/14156701

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The total energy required for this conversion is equivalent to the sum of the energies that are used. There are three steps:

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The following data were obtained in a kinetics study of the hypothetical reaction A + B + C → products. [A]0 (M) [B]0 (M) [C]0 (
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Answer:

B. First order, Order with respect to C = 1

Explanation:

The given kinetic data is as follows:

A + B + C → Products

     [A]₀     [B]₀    [C]₀       Initial Rate (10⁻³ M/s)

1.   0.4      0.4     0.2       160

2.  0.2      0.4      0.4       80

3.   0.6     0.1       0.2       15

4.   0.2     0.1       0.2        5

5.   0.2     0.2      0.4       20

The rate of the above reaction is given as:

Rate = k[A]^{x}[B]^{y}[C]^{z}

where x, y and z are the order with respect to A, B and C respectively.

k = rate constant

[A], [B], [C] are the concentrations

In the method of initial rates, the given reaction is run multiple times. The order with respect to a particular reactant is deduced by keeping the concentrations of the remaining reactants constant and measuring the rates. The ratio of the rates from the two runs gives the order relative to that reactant.

Order w.r.t A : Use trials 3 and 4

\frac{Rate3}{Rate4}= [\frac{[A(3)]}{[A(4)]}]^{x}

\frac{15}{5}= [\frac{[0.6]}{[0.2]}]^{x}

3 = 3^{x} \\\\x =1

Order w.r.t B : Use trials 2 and 5

\frac{Rate2}{Rate5}= [\frac{[B(2)]}{[B(5)]}]^{y}

\frac{80}{20}= [\frac{[0.4]}{[0.2]}]^{y}

4 = 2^{y} \\\\y =2

Order w.r.t C : Use trials 1 and 2

\frac{Rate1}{Rate2}= [\frac{[A(1)]}{[A(2)]}]^{x}[\frac{[B(1)]}{[B(2)]}]^{y}[\frac{[C(1)]}{[C(2)]}]^{z}

we know that x = 1 and y = 2, substituting the appropriate values in the above equation gives:

\frac{160}{80}= [\frac{[0.4]}{[0.2]}]^{1}[\frac{[0.4]}{[0.4]}]^{2}[\frac{[0.2]}{[0.4]}]^{z}

1 = (0.5)^{z}

z = 1

Therefore, order w.r.t C = 1

8 0
3 years ago
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