Which of the following is a true statement about atomic nuclei?

Chemistry

1 answer:

Free_Kalibri [48]3 years ago

8 0

Answer: they have a net positive charge

Explanation:

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Consider the following intermediate chemical equations. When you form the final chemical equation, what should you do with CO? C

-Dominant- [34]

Answer:

Cancel out CO because it appears as a reactant in one intermediate reaction and a product in the other intermediate reaction.

Explanation:

The CO appears twice hence in he intermediate reaction it only forms path of the enabling reagents and it further reacts to form the final product. Accounting for the CO in the intermediate reaction that undergoes further reaction will impact on the stoichiometry of the reaction.

5 0

3 years ago

Give the spectator ions for the reaction that occurs when aqueous solutions of H 2SO 4 and KOH are mixed.

Sever21 [200]

Answer:

The spectator ions is: $K^+$ and $SO^{2-}_4$

Explanation:

The equation of reaction between H₂ SO₄ and KOH is:

$H_2SO_{4(aq)} + 2KOH _{aq} \to K_2SO_{4(aq)} +2H_2O _{(l)}$

Rewriting this equation as ionic;

$[2H^{+} + SO^{2-}_4 + 2K^+ +2OH^- \to 2K^+ SO_4^{2-} + 2H_2O ]$

Spectators ions are ions present on both sides of the ionic equation by the same quantity but do not take part in the net reaction.

5 0

2 years ago

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

just olya [345]

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$_{2}$}$