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lana66690 [7]
3 years ago
7

- The subatomic particle possessing the largest

Chemistry
1 answer:
Ierofanga [76]3 years ago
6 0

Answer:

B. electrons possess the largest charge-to-mass ratio among the subatomic particles listed in the four choices.

Explanation:

Consider the mass of each particle. Express the masses in atomic mass units:

  • Protons: approximately 1.007 amu each;
  • Neutrons: approximately 1.009 amu each;
  • Electrons: approximately 0.0005 amu each.

Similarly, consider the charge on each particle. Express the charges in multiples of the fundamental charge:

  • Protons: +1 e;
  • Neutrons: 0;
  • Electrons: -1 e.

Calculate the charge-to-mass ratio for the three species:

  • Protons: approximately \rm 0.99\; e\cdot amu^{-1};
  • Neutrons: 0;
  • Electrons: approximately \rm 1,800\;e \cdot amu^{-1}.

Almost all nuclei contain protons and neutrons. The only exception is the hydrogen-1 nucleus, which contains only one proton and no neutron. The mass of the nucleus is approximately the same as the sum of its components' masses. The extra neutron will only add to the mass of the nucleus (the denominator) without contributing to the charge (the numerator.) As a result, the charge-to-mass ratio of nuclei will be positive but no greater than the charge-to-mass ratio of protons.

Among the particles in the four choices, the charge-to-mass ratio is the greatest for electrons.

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deff fn [24]

Answer:

Explanation:

This is a limiting reactant problem.

Mg(s)

+

2HCl(aq)

→

MgCl

2

(

aq

)

+ H

2

(

g

)

Determine Moles of Magnesium

Divide the given mass of magnesium by its molar mass (atomic weight on periodic table in g/mol).

4.86

g Mg

×

1

mol Mg

24.3050

g Mg

=

0.200 mol Mg

Determine Moles of 2M Hydrochloric Acid

Convert

100 cm

3

to

100 mL

and then to

0.1 L

.

1 dm

3

=

1 L

Convert

2.00 mol/dm

3

to

2.00 mol/L

Multiply

0.1

L

times

2.00 mol/L

.

100

cm

3

×

1

mL

1

cm

3

×

1

L

1000

mL

=

0.1 L HCl

2.00 mol/dm

3

=

2.00 mol/L

0.1

L

×

2.00

mol

1

L

=

0.200 mol HCl

Multiply the moles of each reactant times the appropriate mole ratio from the balanced equation. Then multiply times the molar mass of hydrogen gas,

2.01588 g/mol

0.200

mol Mg

×

1

mol H

2

1

mol Mg

×

2.01588

g H

2

1

mol H

2

=

0.403 g H

2

0.200

mol HCl

×

1

mol H

2

2

mol HCl

×

2.01588

g H

2

1

mol H

2

=

0.202 g H

2

The limiting reactant is

HCl

, which will produce

0.202 g H

2

under the stated conditions.

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