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1 year ago

N⁻³ and Na⁺ have same A Atomic no. B Mass No. C No. of electrons D No. of neutrons

Chemistry

1 answer:

gizmo_the_mogwai [7]1 year ago

3 0

Answer:

C.) No. of electrons

Explanation:

A.) is incorrect. The atomic number represents the number of protons in an element. Nitrogen (N) and sodium (Na) always have a differing amount of protons.

B.) is incorrect. The mass number represents the number of protons and neutrons in an element. The number of neutrons and protons are specific to each element (disregarding isotopes). When elements ionize, these amounts are not altered.

C.) is correct. When an element becomes an ion, the number of electrons change. When nitrogen gains 3 electrons and sodium loses 1 electron, they end up having the same number of electrons (10).

D.) is incorrect. When elements ionize, the number of neutrons does not change. The only way two different elements could have the same number of neutrons is if at least one of the elements is an isotope. Isotopes are two or more atoms of the same element that differ in their amounts of neutrons.

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2 years ago

if i add 25 ml of water to 135 ml of a 0.25 M NaOH solution what will the molarity of the diluted solution be

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Answer:

0.21 M. (2 sig. fig.)

Explanation:

The molarity of a solution is the number of moles of the solute in each liter of the solution. The unit for molarity is M. One M equals to one mole per liter.

How many moles of NaOH in the original solution?

$n = c \cdot V$ ,

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$n$ is the number of moles of the solute in the solution.
$c$ is the concentration of the solution. $c = 0.25 \;\text{M} = 0.25\;\text{mol}\cdot\textbf{L}^{-1}$ for the initial solution.
$V$ is the volume of the solution. For the initial solution, $V = 135\;\textbf{mL} = 0.135\;\textbf{L}$ for the initial solution.

$n = c\cdot V = 0.25\;\text{mol}\cdot\textbf{L}^{-1} \times 0.135\;\textbf{L} = 0.03375\;\text{mol}$ .

What's the concentration of the diluted solution?

$\displaystyle c = \frac{n}{V}$ .

$n$ is the number of solute in the solution. Diluting the solution does not influence the value of $n$ . $n = 0.03375\;\text{mol}$ for the diluted solution.
Volume of the diluted solution: $25\;\text{mL} + 135\;\text{mL} = 160\;\textbf{mL} = 0.160\;\textbf{L}$ .

Concentration of the diluted solution:

$\displaystyle c = \frac{n}{V} = \frac{0.03375\;\text{mol}}{0.160\;\textbf{L}} = 0.021\;\text{mol}\cdot\textbf{L}^{-1} = 0.021\;\text{M}$ .

The least significant number in the question comes with 2 sig. fig. Keep more sig. fig. than that in calculations but round the final result to 2 sig. fig. Hence the result: 0.021 M.

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