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Goryan [66]
3 years ago
12

Part 1: What is the final volume in milliliters when 0.730 L of a 44.8 % (m/v) solution is diluted to 23.3 % (m/v)?

Chemistry
1 answer:
Andre45 [30]3 years ago
6 0

part 1 : the final volume : 1.404 L

part 2 : the initial concentration : 4.06 M

<h3>Further explanation </h3>

Dilution is the process of adding a solvent to get a more dilute solution.

The moles(n) before and after dilution are the same.

Can be formulated :

M₁V₁=M₂V₂

M₁ = Molarity of the solution before dilution  

V₁ = volume of the solution before dilution  

M₂ = Molarity of the solution after dilution  

V₂ = Molarity volume of the solution after dilution

part 1 :

M₁=44.8%

V₁=0.73 L

M₂=23.3%

\tt V_2=\dfrac{M_1.V_1}{M_2}\\\\V_2=\dfrac{44.8\times 0.73}{23.3}\\\\V_2=1.404~L

part 2 :

V₁=739 ml=0.739 L

V₂=1.5 L

M₂=2

\tt M_1=\dfrac{M_2.V_2}{V_1}\\\\M_1=\dfrac{2\times 1.5}{0.739}\\\\M_1=4.06

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allochka39001 [22]

Answer : The theoretical yield of water formed from the reaction is 0.54 grams.

Solution : Given,

Mass of HCl = 1.1 g

Mass of NaOH = 2.1 g

Molar mass of HCl = 36.5 g/mole

Molar mass of NaOH = 40 g/mole

Molar mass of H_2O = 18 g/mole

First we have to calculate the moles of HCl and NaOH.

\text{ Moles of }HCl=\frac{\text{ Mass of }HCl}{\text{ Molar mass of }HCl}=\frac{1.1g}{36.5g/mole}=0.030moles

\text{ Moles of }NaOH=\frac{\text{ Mass of }NaOH}{\text{ Molar mass of }NaOH}=\frac{2.1g}{40g/mole}=0.525moles

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

HCl+NaOH\rightarrow NaCl+H_2O

From the balanced reaction we conclude that

As, 1 mole of HCl react with 1 mole of NaOH

So, 0.030 mole of HCl react with 0.030 mole of NaOH

From this we conclude that, NaOH is an excess reagent because the given moles are greater than the required moles and HCl is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of H_2O

From the reaction, we conclude that

As, 1 mole of HCl react to give 1 mole of H_2O

So, 0.030 moles of HCl react to give 0.030 moles of H_2O

Now we have to calculate the mass of H_2O

\text{ Mass of }H_2O=\text{ Moles of }H_2O\times \text{ Molar mass of }H_2O

\text{ Mass of }H_2O=(0.030moles)\times (18g/mole)=0.54g

Therefore, the theoretical yield of water formed from the reaction is 0.54 grams.

6 0
3 years ago
STATES THAT FOR AN ATOM IN THE GROUND STATE, THE NUMBER OF UNPAIRED ELECTRONS IS THE MAXIMUM POSSIBLE AND THESE UNPAIRED ELECTRO
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Answer:

Hund's rule

Explanation:

Hund's rule  states that for an atom in the ground state, the number of unpaired electrons is the maximum possible and these unpaired electrons have the same spin.

I will clearly illustrate this using the ground state of the nitrogen atom. Nitrogen atom contains seven electrons. Its electronic configuration is; 1s2 2s2 2px1 2py1 2pz1.

Notice that the number of unpaired electrons in this configuration is maximum and they are all expected to possess the same spin in accordance with Hund's rule .

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A way of describing a chemical change that indicates what substances were present before the change and what substances are pres
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Answer:

\\ \rm\dashrightarrow Chemical\: Equation

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7 0
2 years ago
17. Physical properties of a material include all but ________________.
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Answer:

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8 0
3 years ago
When aqueous solutions of K3PO4 and Ba(NO3)2 are combined, Ba3(PO4)2 precipitates. Calculate the mass, in grams, of the Ba3(PO4)
Firdavs [7]

Answer:

Mass of Ba₃(PO₄)₂ = 0.0361 g

Explanation:

Given data:

Volume of Ba(NO₃)₂ = 1.2 mL (1.2 × 10⁻³ L )

Molarity of Ba(NO₃)₂ = 0.152 M

Volume of K₃PO₄ = 4.2 mL (4.2 × 10⁻³ L)

Molarity of K₃PO₄ =  0.604 M

Mass of Ba₃(PO₄)₂ produced = ?

Solution:

Chemical equation:

3Ba(NO₃)₂  + 2K₃PO₄  → Ba₃(PO₄)₂  + 6KNO₃

Number of moles of Ba(NO₃)₂ = Molarity × Volume in litter

Number of moles of Ba(NO₃)₂ = 0.152 M × 1.2 × 10⁻³ L

Number of moles of Ba(NO₃)₂ = 0.182 × 10⁻³ mol

Number of moles of K₃PO₄ = Molarity × Volume in litter

Number of moles of K₃PO₄ = 0.604 M × 4.2 × 10⁻³ L

Number of moles of K₃PO₄ = 2.537 × 10⁻³ mol

Now we will compare the moles of Ba₃(PO₄)₂ with K₃PO₄ and Ba(NO₃)₂ .

              Ba(NO₃)₂        :         Ba₃(PO₄)₂

                   3                :               1

              0.182 × 10⁻³    :              1/3 ×0.182 × 10⁻³ = 0.060 × 10⁻³ mol

                K₃PO₄           :          Ba₃(PO₄)₂

                   2                 :                1

              2.537 × 10⁻³     :               1/2 ×  2.537 × 10⁻³= 1.269 × 10⁻³ mol

The number of moles of Ba₃(PO₄)₂ produced by  Ba(NO₃)₂  are less it will limiting reactant.

Mass of Ba₃(PO₄)₂ = moles × molar mass

Mass of Ba₃(PO₄)₂ = 0.060 × 10⁻³ mol × 601.93 g/mol

Mass of Ba₃(PO₄)₂ = 36.12 × 10⁻³ g

Mass of Ba₃(PO₄)₂ = 0.0361 g

6 0
3 years ago
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