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3 years ago

if it takes 54 mL of 0.1 NaOH to neutralize 125 mL of an HCL solution, what is the concentration of HCL?

Chemistry

1 answer:

alexira [117]3 years ago

6 0

Answer:

0.0432 M

Explanation:

We are given;

Volume of NaOH as 54 mL

Molarity of NaOH as 0.1 M

Volume of HCl as 125 mL

We are required to determine the concentration of HCl

Step 1; We write a balanced equation for the reaction between NaOH and HCl

The balanced equation is given by;

NaOH(aq) + HCl(aq) → NaCl(aq) + H₂O(l)

Step 2: Determine the number of moles of NaOH

Moles = Volume × molarity

Therefore;

Moles of NaOH = 0.054 L × 0.1 M

= 0.0054 Moles

Step 3: We use the mole ratio to determine the moles of HCl

From the equation;

1 mole of NaOH reacts with 1 mole of HCl

Therefore;

Moles of NaOH = Moles of HCl

Thus; moles of HCl = 0.0054 moles

Step 4: Determine the concentration of HCl

We know that;

Molarity = Moles ÷ Volume

Therefore;

Molarity of HCl = 0.0054 moles ÷ 0.125 L

= 0.0432 M

Therefore, the concentration of HCl is 0.0432 M

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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 $\large \boxed{\textbf{0.066 mol/L}}$}$

8 0

3 years ago

A student isolated 25 g of a compound following a procedure that would theoratically yield 81 g. Wht was his percentt yield?

Sati [7]

<u>Answer:</u> The percent yield of the compound is 30.86 %.

<u>Explanation:</u>

To calculate the percentage yield of a compound, we use the equation:

$\%\text{ yield}=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100$

Experimental yield of compound = 25 g

Theoretical yield of compound = 81 g

Putting values in above equation, we get:

$\%\text{ yield of compound}=\frac{25g}{81g}\times 100\\\\\% \text{yield of compound}=30.86\%$

Hence, the percent yield of the compound is 30.86 %.

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Answer:

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Explanation:

cause that's the only one that makes sense because it goes: cell, tissues,organ,organ system , organism

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3. tissues - would be made of one kind of cell

4- an organ system - meets all the needs of a organism

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