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Shalnov [3]
2 years ago
5

Suppose you are working with a NaOH stock solution but you need a solution with a lower concentration for your experiment. Calcu

late the volume of the 1.224 M stock NaOH solution needed to prepare 250.0 mL of 0.1281 M dilute NaOH solution.
Chemistry
1 answer:
Monica [59]2 years ago
7 0

Answer: The volume of the 1.224 M NaOH solution needed is 26.16 mL

Explanation:

In order to prepare the dilute NaOH solution, solvent is added to a given amount of the NaOH stock solution up to a final volume of 250.0 mL.

Since only solvent is added, the amount of the solute, NaOH, in the dilute solution is the same as in the volume taken from the stock solution.

Molarity (<em>M)</em> is calculated from the following equation:

<em>M</em> = <em>n</em> ÷ <em>V</em>

where <em>n</em> is the number of moles of the solute in the solution, and <em>V</em> is the volume of the solution.

Accordingly, the number of moles of the solute is given by

<em>n</em> = <em>M</em> x <em>V</em>

Now, let's designate the stock NaOH solution and the dilute solution as (1) and (2), respectively . The number of moles of NaOH in each of these solutions is:

<em>n </em>(1) = <em>M </em>(1) x <em>V </em>(1)

<em>n </em>(2) = <em>M </em>(2) x <em>V </em>(2)

As the amount of NaOH in the dilute solution is the same as in the volume taken from the stock solution,

<em>n</em> (1) = <em>n</em> (2)

and

<em>M</em> (1) x <em>V</em> (1)<em> </em>= <em>M</em> (2) x <em>V</em> (2)

For the stock solution, <em>M</em> (1) = 1.244 M, and <em>V</em> (1) is the volume needed. For the dilute solution, <em>M</em> (2) = 0,1281 M, and <em>V</em> (2) = 250.0 mL.

The volume of the stock solution needed, <em>V</em> (1), is calculated as follows:

<em>V</em> (1) = <em>M</em> (2) x <em>V</em> (2) ÷ <em>M</em> (1)

<em>V</em> (1) = 0.1281 M x 250.0 mL ÷ 1.224 M

<em>V </em>(1) = 26.16 mL

The volume of the 1.224 M NaOH solution needed is 26.16 mL.

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ludmilkaskok [199]

0.3268 moles of PC15 can be produced from 58.0 g of Cl₂ (and excess

P4)

<h3>How to calculate moles?</h3>

The balanced chemical equation is

P_{4}  + 10Cl_{2}  = 4PCl_{5}

The mass of clorine is m(Cl_{2}) = 58.0 g

The amount of clorine is n(Cl_{2}) = m(Cl_{2})/M(Cl_{2}) = 58/70.906 = 0.817 mol

The stoichiometric reaction,shows that

10 moles of Cl_{2} yield 4 moles of PCl_{5};

0.817 of Cl_{2} yield x moles of PCl_{5}

n(PCl_{5}) = 4*0.817/10 = 0.3268 mol

To know more about stoichiometric reaction, refer:

brainly.com/question/14935523

#SPJ9

3 0
1 year ago
7. Suppose each of these isotopes emits a beta particle. Give the iso-
Volgvan

In a beta emission, the mass number of the daughter nucleus remains unchanged while the atomic number of the daughter nucleus increases by one unit. The following are isotopes produced when the following undergo beta emission;

1) potassium-42  ------> Ca - 42

2) iodine-131 ------------> Xe - 131

3)  iron-52 ---------------> Co - 52

4)  sodium-24 -----------> Mg -24

The daughter nucleus formed after beta emission is found one place after its parent in the periodic table.

Regarding the stability of the daughter nuclei, a nucleus is unstable if the neutron-proton ratio is less than 1 or greater than 1.5.

Hence, the following daughter nuclei are stable; Ca - 42, Xe - 131, Mg -24.

Learn more: brainly.com/question/1371390

7 0
2 years ago
Can someone help me please
lesantik [10]
It is an endothermic reaction because the products hAve more heat than the reactions so it was a gain of heat which makes the enthalpy Change positive !
4 0
3 years ago
Equal molar quantities of Ca2 and EDTA (H4Y) are added to make a 0.010 M solution of CaY2- at pH 10. The formation constant for
abruzzese [7]

Answer:

the concentration of free Ca2⁺ in this solution is 7.559 × 10⁻⁷

Explanation:

Given the data in the question;

Ca^{2+ + y^{4- ⇄  CaY^{2-

Formation constant Kf

Kf = CaY^{2- / ( [Ca^{2+][y^{4-] ) = 5.0 × 10¹⁰

Now,

[y^{4-] = \alpha _4CH_4Y; ∝₄ = 0.35

so the equilibrium is;

Ca^{2+ + H_4Y ⇄  CaY^{2- + 4H⁺

Given that; CH_4Y = Ca^{2+     { 1 mol Ca^{2+  reacts with 1 mol H_4Y  }

so at equilibrium, CH_4Y = Ca^{2+ = x

∴

Ca^{2+ + y^{4- ⇄  CaY^{2-

x        + x         0.010-x

since Kf is high, them x will be small so, 0.010-x is approximately 0.010

so;

Kf = CaY^{2- / ( [Ca^{2+][y^{4-] ) =  CaY^{2- / ( [Ca^{2+][\alpha _4CH_4Y] )  = 5.0 × 10¹⁰

⇒ CaY^{2- / ( [Ca^{2+][\alpha _4CH_4Y] )  = 5.0 × 10¹⁰

⇒ 0.010 / ( [x][ 0.35 × x] )  = 5.0 × 10¹⁰

⇒ 0.010 / 0.35x²  = 5.0 × 10¹⁰

⇒ x² = 0.010 / ( 0.35 × 5.0 × 10¹⁰ )

⇒ x² = 0.010 / 1.75 × 10¹⁰

⇒ x² = 0.010 / 1.75 × 10¹⁰

⇒ x² = 5.7142857 × 10⁻¹³

⇒ x = √(5.7142857 × 10⁻¹³)

⇒ x = 7.559 × 10⁻⁷

Therefore, the concentration of free Ca2⁺ in this solution is 7.559 × 10⁻⁷

8 0
2 years ago
Which of the following diagrams shows a pure substance that is composed of only individual atoms and not molecules?
Damm [24]

Answer:

Ar

Explanation:

the last one

8 0
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