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qaws [65]
3 years ago
8

The s orbital can hold

Chemistry
2 answers:
malfutka [58]3 years ago
7 0

Answer:

B. Two electrons

Explanation:

Since the S orbital is closest to the nucleus and it only has one orbital, it can only hold 2 electrons

Mademuasel [1]3 years ago
5 0
B.2electrons
Because it is nearest to nucleus
So we can fill two electrons only
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Suppose 0.245 g of sodium chloride is dissolved in 50. mL of a 18.0 m M aqueous solution of silver nitrate.
Bezzdna [24]

Answer:

\large \boxed{\text{ 0.066 mol/L}}

Explanation:

We are given the amounts of two reactants, so this is a limiting reactant problem.

1. Assemble all the data in one place, with molar masses above the formulas and other information below them.

Mᵣ:       58.44  

            NaCl + AgNO₃ ⟶ NaNO₃ + AgCl

m/g:     0.245

V/mL:                 50.

c/mmol·mL⁻¹:       0.0180

2. Calculate the moles of each reactant  

\text{Moles of NaCl} = \text{245 mg NaCl} \times \dfrac{\text{1 mmol NaCl}}{\text{58.44 mg NaCl}} = \text{4.192 mmol NaCl}\\\\\text{ Moles of AgNO}_{3}= \text{50. mL AgNO}_{3} \times \dfrac{\text{0.0180 mmol AgNO}_{3}}{\text{1 mL AgNO}_{3}} = \text{0.900 mmol AgNO}_{3}

3. Identify the limiting reactant  

Calculate the moles of AgCl we can obtain from each reactant.

From NaCl:  

The molar ratio of NaCl to AgCl is 1:1.

\text{Moles of AgCl} = \text{4.192 mmol NaCl} \times \dfrac{\text{1 mmol AgCl}}{\text{1 mmol NaCl}} = \text{4.192 mmol AgCl}

From AgNO₃:  

The molar ratio of AgNO₃ to AgCl is 1:1.  

\text{Moles of AgCl} = \text{0.900 mmol AgNO}_{3} \times \dfrac{\text{1 mmol AgCl}}{\text{1 mmol AgNO}_{3}} = \text{0.900 mmol AgCl}

AgNO₃ is the limiting reactant because it gives the smaller amount of AgCl.

4. Calculate the moles of excess reactant

                   Ag⁺(aq)  +  Cl⁻(aq) ⟶ AgCl(s)

 I/mmol:      0.900        4.192            0

C/mmol:    -0.900       -0.900        +0.900

E/mmol:      0                3.292          0.900

So, we end up with 50. mL of a solution containing 3.292 mmol of Cl⁻.

5. Calculate the concentration of Cl⁻

\text{[Cl$^{-}$] } = \dfrac{\text{3.292 mmol}}{\text{50. mL}} = \textbf{0.066 mol/L}\\\text{The concentration of chloride ion is $\large \boxed{\textbf{0.066 mol/L}}$}

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