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Salsk061 [2.6K]
2 years ago
13

Using this equation, m1v2=m2v2 , calculate the diluted molarity of 100 mL of a 0.5 M solution when 50 mL of

Chemistry
1 answer:
geniusboy [140]2 years ago
8 0

The molarity of the diluted solution is 0.33 M

From the question given above, the following data were obtained:

Molarity of stock solution (M₁) = 0. 5 M

Volume of stock solution (V₁) = 100 mL

Volume of diluted solution (V₂) = 100 + 50 = 150 mL

<h3>Molarity of diluted solution (M₂) =? </h3>

The molarity of the diluted solution can be obtained by using the dilution formula as illustrated below:

<h3>M₁V₁ = M₂V₂</h3>

0.5 × 100 = M₂ × 150

50 = M₂ × 150

Divide both side by 150

M₂ = 50 / 150

<h3>M₂ = 0.33 M</h3>

Therefore, the molarity of the diluted solution is 0.33 M

Learn more: brainly.com/question/24625656

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Loy and Sam have a dog that has a spotted coat. The dog is also excellent at performing tricks.
Wewaii [24]

Answer: the correct option is A ( The spotted coat is an INHERITED TRAIT, and the ability to perform tricks is a LEARNED TRAIT.

Explanation:

Dog is one of the first domesticated animal by man because it has the ability to perceive information and retain it as knowledge for applying to solve problems.

Therefore, the ability to perform tricks is a LEARNED TRAIT. This is so because they have acquired the ability to understand and communicate with humans, and they are uniquely attuned to human behaviors.

INHERITED TRAIT is a trait that can be passed down from one generation to another which are found in genes within the strands of dogs DNA. Some of the inheritable traits includes the following:

- coat color,

- ear type, and.

- tail style.

Therefore the spotted coat is an INHERITED TRAIT

4 0
3 years ago
How many moles of 02 can be prepared from 6.75 moles of KCIO3 ?
Varvara68 [4.7K]

The equation for this question could be 2KClO_{3}→2KCl+3O_{2} .

so for 6.75 moles of KClO_{3}  *3 moles of O_{2}/2 moles of KClO_{3}

= 10.125

5 0
3 years ago
A 27.9 mL sample of 0.289 M dimethylamine, (CH3)2NH, is titrated with 0.286 M hydrobromic acid.
sesenic [268]

Answer:

(1) Before the addition of any HBr, the pH is 12.02

(2) After adding 12.0 mL of HBr, the pH is 10.86

(3) At the titration midpoint, the pH is 10.73

(4) At the equivalence point, the pH is 5.79

(5) After adding 45.1 mL of HBr, the pH is 1.18

Explanation:

First of all, we have a weak base:

  • 0 mL of HBr is added

(CH₃)₂NH  + H₂O  ⇄  (CH₃)₂NH₂⁺  +  OH⁻            Kb = 5.4×10⁻⁴

0.289 - x                             x                x

Kb = x² / 0.289-x

Kb . 0.289 - Kbx - x²

1.56×10⁻⁴ - 5.4×10⁻⁴x - x²

After the quadratic equation is solved x = 0.01222 → [OH⁻]

- log  [OH⁻] = pOH → 1.91

pH = 12.02   (14 - pOH)

  • After adding 12 mL of HBr

We determine the mmoles of H⁺, we add:

0.286 M . 12 mL = 3.432 mmol

We determine the mmoles of base⁻, we have

27.9 mL . 0.289 M = 8.0631 mmol

When the base, react to the protons, we have the protonated base plus water (neutralization reaction)

(CH₃)₂NH     +      H₃O⁺        ⇄  (CH₃)₂NH₂⁺  +  H₂O

8.0631 mm       3.432 mm                 -

4.6311 mm                                  3.432 mm

We substract to the dimethylamine mmoles, the protons which are the same amount of protonated base.

[(CH₃)₂NH] → 4.6311 mm / Total volume (27.9 mL + 12 mL) = 0.116 M

[(CH₃)₂NH₂⁺] → 3.432 mm / 39.9 mL = 0.0860 M

We have just made a buffer.

pH = pKa + log (CH₃)₂NH  / (CH₃)₂NH₂⁺

pH = 10.73 + log (0.116/0.0860) = 10.86

  • Equivalence point

mmoles of base = mmoles of acid

Let's find out the volume

0.289 M . 27.9 mL = 0.286 M . volume

volume in Eq. point = 28.2 mL

(CH₃)₂NH     +      H₃O⁺        ⇄  (CH₃)₂NH₂⁺  +  H₂O

8.0631 mm       8.0631mm               -

                                                8.0631 mm

We do not have base and protons, we only have the conjugate acid

We calculate the new concentration:

mmoles of conjugated acid / Total volume (initial + eq. point)

[(CH₃)₂NH₂⁺] = 8.0631 mm /(27.9 mL + 28.2 mL)  = 0.144 M

(CH₃)₂NH₂⁺   +  H₂O   ⇄   (CH₃)₂NH  +  H₃O⁻       Ka = 1.85×10⁻¹¹

 0.144 - x                                  x               x

[H₃O⁺] = √ (Ka . 0.144) →  1.63×10⁻⁶ M  

pH = - log [H₃O⁺] = 5.79

  • Titration midpoint (28.2 mL/2)

This is the point where we add, the half of acid. (14.1 mL)

This is still a buffer area.

mmoles of H₃O⁺ = 4.0326 mmol (0.286M . 14.1mL)

mmoles of base = 8.0631 mmol - 4.0326 mmol

[(CH₃)₂NH] = 4.0305 mm / (27.9 mL + 14.1 mL) = 0.096 M

[(CH₃)₂NH₂⁺] = 4.0326 mm (27.9 mL + 14.1 mL) = 0.096 M

pH = pKa + log (0.096M / 0.096 M)

pH = 10.73 + log 1 =  10.73

Both concentrations are the same, so pH = pKa. This is the  maximum buffering capacity.

  • When we add 45.1 mL of HBr

mmoles of acid = 45.1 mL . 0.286 M = 12.8986 mmol

mmoles of base = 8.0631 mmoles

This is an excess of H⁺, so, the new [H⁺] = 12.8986 - 8.0631 / Total vol.

(CH₃)₂NH     +      H₃O⁺        ⇄  (CH₃)₂NH₂⁺  +  H₂O

8.0631 mm     12.8986 mm             -

       -               4.8355 mm                        

[H₃O⁺] = 4.8355 mm / (27.9 ml + 45.1 ml)

[H₃O⁺] = 4.8355 mm / 73 mL → 0.0662 M

- log [H₃O⁺] = pH

- log 0.0662 = 1.18 → pH

7 0
3 years ago
If the power source in a motor was to be replaced with an electrical load, and the armature was turned by hand, then what device
timofeeve [1]

Answer:

This question doesnt make much sense but ill try to answer it.

A brushed DC electric motor is an internally commutated electric motor designed to be run from a direct current power source. Brushed motors were the first commercially important application of electricity. If the shaft of a DC motor is turned by an external force, the motor will act as a generator. DC Motor rotation does have to do with the voltage polarity and the direction of the current flow.

4 0
3 years ago
Can the process of rusting be called combustion? ​
gregori [183]

Answer:

no the answer is oxidation

8 0
2 years ago
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