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3 years ago

5

The nucleus of an atom contains which subatomic particles? A) neutrons only B) protons and neutrons C) neutrons and electrons D)

protons and electrons

2 answers:

kvv77 [185]3 years ago

8 0

The nucleus of an atom contains B) protons and neutrons.

The electrons are <em>outside</em> the nucleus.

Airida [17]3 years ago

3 0

Protons and neutrons

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are either positively charged or negatively charged species in which the number of protons and electrons are not equal. ________

fredd [130]

Answer:

1. Ions are either negatively or positively charged species in which the number of electrons and protons are not equal.

2. The chemical bond that arises due to the sharing of electrons is termed a covalent bond.

3. The positively charged ions are called cations, which comprise more protons than electrons.

4. An example of a polyatomic anion is the hydroxide anion.

5. The system of assigning an unambiguous name to a compound is called nomenclature.

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2 years ago

Tell uses of nitrogen

Aleksandr [31]

Answer:

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Preservation of Food. Nitrogen gas is used to help with food preservation by preventing oxidative damage leading to food spoiling.
Pharmaceuticals Industry.
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Explanation:

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3 years ago

What percentage of the earth is water?

OleMash [197]

Answer:

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3 years ago

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What are glaze defects or flaws

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2 years ago

in a simulation mercury removal from industrial wastewater, 0.020 L of 0.10 M sodium sulfide reacts with 0.050 L of 0.010 M merc

Margarita [4]

Answer: 0.1161 grams of mercury(II) sulfide) form.

Explanation:

To calculate the number of moles for given molarity, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}\times 1000}{\text{Volume of solution (in L)}}$ .....(1)

a) Molarity of $Na_2S$ solution = 0.10 M

Volume of solution = 0.020 L

Putting values in equation 1, we get:

$0.10M=\frac{\text{Moles of }Na_2S}{0.020L}\\\\\text{Moles of Na_2S}={0.10mol/L\times 0.020}=0.002mol$

$\text {Moles of}Na_2S=0.10M\times 0.020L=0.002mol$

b) Molarity of $Hg(NO_3)_2$ solution = 0.010 M

Volume of solution = 0.050 L

Putting values in equation 1, we get:

$0.010M=\frac{\text{Moles of }Hg(NO_3)_2}{0.050L}\\\\\text{Moles of }Hg(NO_3)_2={0.010mol/L\times 0.050}=0.0005mol$

$Na_2S+Hg(NO_3)_2\rightarrow HgS+2NaNO_3$

According to stoichiometry :

1 mole of $Hg(NO_3)_2$ reacts with 1 mole of $Na_2S$

Thus 0.0005 moles of $HgNO_3$ reacts with= $\frac{1}{1}\times 0.0005=0.0005$ moles of $Hg(NO_3)_2$

Thus $Hg(NO_3)_2$ is the limiting reagent and $Na_2S$ is the excess reagent.

According to stoichiometry :

1 mole of $Hg(NO_3)_2$ forms= 1 mole of $Hg_2S$

Thus 0.0005 moles of $Hg(NO_3)_2$ forms= $\frac{1}{1}\times 0.0005=0.0005$ moles of $Hg_2S$

mass of $H_2S=moles\times {\text {Molar mass}}=0.0005mol\times 232.2g/mol=0.1161g$

Thus 0.1161 grams of mercury(II) sulfide) form.

5 0

3 years ago