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vaieri [72.5K]
3 years ago
14

Which of the following has no electrons? Use a periodic table. Question 1 options: H He H– He 2+ Save

Chemistry
1 answer:
Elodia [21]3 years ago
6 0

An electron is a negatively charged subatomic particle present in the space outside the nucleus of an atom. The loss of electron from an atom results in the formation of cation whereas gaining of electron by an atom results in the formation of anion. The cation possesses positive charge due to loss of electron and anion possesses negative charge due to gain of electron.

The neutral atom has no charge on it.

For given atomic symbols:

  • H

The atomic number of hydrogen is 1 and the given symbol has no charge that means it is in its neutral state. So, the number of electrons in H is 1.

  • He

The atomic number of helium is 2 and the given symbol has no charge that means it is in its neutral state. So, the number of electrons in He is 2.

  • H^{-}

The atomic number of hydrogen is 1 and the given symbol has a negative charge that represents a gain of electron. So, the number of electrons in H^{-} is 2.

  • He^{2+}

The atomic number of helium is 2 and the given symbol has two positive charge that represents loss of two electrons. So, the number of electrons in He^{2+} is 0.

Hence, He^{2+} has no electrons.

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N2(g) + 3H2(g) → 2NH3(g) How many grams of N2 are required to produce 240.0g NH3?
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Answer:

\large \boxed{\text{197.4 g}}

Explanation:

We will need a chemical equation with masses and molar masses, so, let's gather all the information in one place.

Mᵣ:     28.01               17.03

            N₂ + 3H₂ ⟶ 2NH₃

m/g:                          240.0

(a) Moles of NH₃

\text{Moles of NH}_{3} = \text{240.0 g NH}_{3}\times \dfrac{\text{1 mol NH}_{3}}{\text{17.03 g NH}_{3}}= \text{14.09 mol NH}_{3}

(b) Moles of N₂

\text{Moles of N$_{2}$} = \text{14.09 mol NH}_{3} \times \dfrac{\text{1 mol N$_{2}$}}{\text{2 mol NH}_{3}} = \text{7.046 mol N$_{2}$}

(c) Mass of N₂

\text{Mass of N$_{2}$} =\text{7.046 mol N$_{2}$} \times \dfrac{\text{28.01 g N$_{2}$}}{\text{1 mol N$_{2}$}} = \textbf{197.4 g N$_{2}$}\\\\\text{The reaction requires $\large \boxed{\textbf{197.4 g}}$ of N$_{2}$}

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If 40.0 mL of a calcium nitrate solution reacts with excess potassium carbonate to yield 0.524 grams of a precipitate, what is t
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Answer : The molarity of calcium ion on the original solution is, 0.131 M

Explanation :

The balanced chemical reaction is:

Ca(NO_3)_2+K_2CO_3\rightarrow CaCO_3+3KNO_3

When calcium nitrate react with potassium carbonate to give calcium carbonate as a precipitate and potassium nitrate.

First we have to calculate the moles of CaCO_3

\text{Moles of }CaCO_3=\frac{\text{Mass of }CaCO_3}{\text{Molar mass of }CaCO_3}

Given:

Mass of CaCO_3 = 0.524 g

Molar mass of CaCO_3 = 100 g/mol

\text{Moles of }CaCO_3=\frac{0.524}{100g/mol}=0.00524mol

Now we have to calculate the concentration of CaCO_3

\text{Concentration of }CaCO_3=\frac{\text{Moles of }CaCO_3}{\text{Volume of solution}}=\frac{0.00524mol}{0.040L}=0.131M

Now we have to calculate the concentration of calcium ion.

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Thus, the concentration or molarity of calcium ion on the original solution is, 0.131 M

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