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Nat2105 [25]
2 years ago
9

Elements in the same period share the same number of:

Chemistry
1 answer:
Anuta_ua [19.1K]2 years ago
7 0

Answer:       electron on the shell  of the atom

Explanation:  that what I trying to remember

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Smhsmz bomb b V mzvznshx
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3 years ago
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150 mL of 0.25 mol/L magnesium chloride solution and 150 mL of 0.35 mol/L silver nitrate solution are mixed together. After reac
Murljashka [212]

Answer:

0.175\; \rm mol \cdot L^{-1}.

Explanation:

Magnesium chloride and silver nitrate reacts at a 2:1 ratio:

\rm MgCl_2\, (aq) + 2\, AgNO_3\, (aq) \to Mg(NO_3)_2 \, (aq) + 2\, AgCl\, (s).

In reality, the nitrate ion from silver nitrate did not take part in this reaction at all. Consider the ionic equation for this very reaction:

\begin{aligned}& \rm Mg^{2+} + 2\, Cl^{-} + 2\, Ag^{+} + 2\, {NO_3}^{-} \\&\to  \rm Mg^{2+} + 2\, {NO_3}^{-} + 2\, AgCl\, (s)\end{aligned}.

The precipitate silver chloride \rm AgCl is insoluble in water and barely ionizes. Hence, \rm AgCl\! isn't rewritten as ions.

Net ionic equation:

\begin{aligned}& \rm Ag^{+} + Cl^{-} \to AgCl\, (s)\end{aligned}.

Calculate the initial quantity of nitrate ions in the mixture.

\begin{aligned}n(\text{initial}) &= c(\text{initial}) \cdot V(\text{initial}) \\ &= 0.25\; \rm mol \cdot L^{-1} \times 0.150\; \rm L \\ &= 0.0375\; \rm mol \end{aligned}.

Since nitrate ions \rm {NO_3}^{-} do not take part in any reaction in this mixture, the quantity of this ion would stay the same.

n(\text{final}) = n(\text{initial}) = 0.0375\; \rm mol.

However, the volume of the new solution is twice that of the original nitrate solution. Hence, the concentration of nitrate ions in the new solution would be (1/2) of the concentration in the original solution.

\begin{aligned} c(\text{final}) &= \frac{n(\text{final})}{V(\text{final})} \\ &= \frac{0.0375\; \rm mol}{0.300\; \rm L} = 0.175\; \rm mol \cdot L^{-1}\end{aligned}.

6 0
3 years ago
4 moles of hydrogen react with 5 moles of oxygen to form water, identify the excess reagent?
Fynjy0 [20]

Answer:

Excess Reagent = oxygen

Explanation:

Limiting reagent: The substance that is totally consumed when the reaction is completed.

Excess reagent: The substance left after the limiting reagent is consumed completely

The balanced chemical equation for formation of water is as follow:

2H_{2}+O_{2}\rightarrow 2H_{2}O

This means when 2 moles of hydrogen reacts with 1 mole of oxygen, 2 moles of water is produced.

Hence the ratio in which hydrogen and oxygen gas reacts is 2:1

Now if 2 mole hydrogen require 1 mole of oxygen ,then 4 mole hydrogen need 2 mole of oxygen.

H_{2}:O_{2}=2:1

or

H_{2}:O_{2}=4:2

Here 5 mole of oxygen is reacting but only 2 mole is required .

Oxygen is in excess.

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