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creativ13 [48]
3 years ago
14

(WILL GIVE BRAINLIEST IF YOU ANSWER ALL 4) Chemistry

Chemistry
1 answer:
Kitty [74]3 years ago
5 0

Answer:

See explanations

Explanation:

a. Molarity = moles/Volume in Liters = 5moles/2Liters = 2.5M in NaCl

b. Freezing Pt Depression

     1. Sprinkling salt on icy surfaces

    2. Using antifreeze in automobile cooling systems

    3. <em>Not an application </em>

    4. Using salt to make ice cream

c. pOH = -log[OHˉ] = -log(1x10ˉ¹⁰) = -(-10) = 10 => pH = 14 – pOH = 14 – 10 = 4

d. H₂O + NH₃ => NH₄⁺ + OHˉ => Bronsted Acid is H₂O  (proton donor)

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avanturin [10]

Answer: C the study of living things

Explanation:

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3 years ago
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Which of the following results in the release of nuclear energy?
Pepsi [2]

c.Both the breaking of nuclear bonds and the forming of nuclear bonds.

Explanation:

Nuclear energy is released by the breaking and forming of nuclear bonds. The breaking of nuclear bonds by unstable atoms is known as nuclear fission. The forming of nuclear bonds by combination of light atoms is known as nuclear fusion.

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3 0
3 years ago
Why is pyridine included in the reaction of an acid chloride and an amine or alcohol? Pyridine will deprotonate the amine or alc
Pavel [41]

Answer:

Pyridine will neutralize the acid by-product of the reaction.

Explanation:

In the esterifications, hydrochloric acid is released as a by-product that reacts directly with the pyridine in the medium to give the pyridine hydrochloride.

7 0
3 years ago
How do I calculate the standard enthalpy change for the following reaction at 25 °C ?
ladessa [460]
MgCl2(s) + H2O(l) → MgO(s) + 2 HCl(g) 

Using the standard enthalpies of formation given in the source below: 

(−601.24 kJ) + (2 x −92.30 kJ) − (−641.8 kJ) − (−285.8 kJ) = +141.76 kJ 
So: 

MgCl2(s) + H2O(l) → MgO(s) + 2 HCl(g), ΔH = +141.76 kJ
4 0
4 years ago
The K w for water at 0 ∘ C is 0.12 × 10 − 14 . Calculate the pH of a neutral aqueous solution at 0 ∘ C.
dusya [7]

Answer:

pH = 7.46

Explanation:

2H₂O  ⇄  H₃O⁺  .  OH⁻          Kw = [H₃O⁺] . [OH⁻]

[H₃O⁺] = [OH⁻]

√0.12×10⁻¹⁴ = [H₃O⁺] → 3.46×10⁻⁸  M

- log  [H₃O⁺] = pH

- log 3.46×10⁻⁸ = pH → 7.46

6 0
3 years ago
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