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leonid [27]
2 years ago
14

Certain substances have a degreasing effect. Which of the substances below has a degreasing effect in solution?

Chemistry
1 answer:
Eduardwww [97]2 years ago
3 0

Answer:

Sodium Carbonate

Explanation:

Washing Soda�Sodium Carbonate (Na2CO3) (Na = Sodium, water softening)

Sodium softens water by binding with Calcium and Magnesium forming solid, which can be rinsed off the fabric. �However, if not rinsed thoroughly, the minerals can redeposit onto the fabric and cause problems. �

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B , your products are on the right side of the reaction. The reactants are on the left side
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If a balloon initially has a volume of 4.0 liters and a temperature of 10 degrees Celsius, what will the volume of the balloon b
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The answer is D. 8.1L
7 0
2 years ago
What is the reaction quotient, Q, for this system when [N2] = 2.00 M, [H2] = 2.00 M, and [NH3] = 1.00 M at 472°C?
cupoosta [38]

Answer : The value of reaction quotient, Q is 0.0625.

Solution : Given,

Concentration of N_2 = 2.00 M

Concentration of H_2 = 2.00 M

Concentration of NH_3 = 1.00 M

Reaction quotient : It is defined as a concentration of a chemical species involved in the chemical reaction.

The balanced equilibrium reaction is,

N_2+3H_2\rightleftharpoons 2NH_3

The expression of reaction quotient for this reaction is,

Q=\frac{[Product]^p}{[Reactant]^r}\\Q=\frac{[NH_3]^2}{[N_2]^1[H_2]^3}

Now put all the given values in this expression, we get

Q=\frac{(1.00)^2}{(2.00)^1(2.00)^3}=0.0625

Therefore, the value of reaction quotient, Q is 0.0625.

3 0
3 years ago
Calculate the pka of hypochlorous acid. The ph of a 0.015 m solution of hypochlorous acid has a ph of 4.64.
o-na [289]

Answer:

  • pKa = 7.46

Explanation:

<u>1) Data:</u>

a) Hypochlorous acid = HClO

b) [HClO} = 0.015

c) pH = 4.64

d) pKa = ?

<u>2) Strategy:</u>

With the pH calculate [H₃O⁺], then use the equilibrium equation to calculate the equilibrium constant, Ka, and finally calculate pKa from the definition.

<u>3) Solution:</u>

a) pH

  • pH = - log [H₃O⁺]

  • 4.64 = - log [H₃O⁺]

  • [H_3O^+]= 10^{-4.64} = 2.29.10^{-5}

b) Equilibrium equation: HClO (aq) ⇄ ClO⁻ (aq) + H₃O⁺ (aq)

c) Equilibrium constant: Ka =  [ClO⁻] [H₃O⁺] / [HClO]

d) From the stoichiometry: [CLO⁻] = [H₃O⁺] = 2.29 × 10 ⁻⁵ M

e) By substitution: Ka = (2.29 × 10 ⁻⁵ M)² / 0.015M = 3.50 × 10⁻⁸ M

f) By definition: pKa = - log Ka = - log (3.50 × 10 ⁻⁸) = 7.46

5 0
3 years ago
What is generally the best approach when liquid from a large reagent bottle is needed in an experiment
Paul [167]

Answer:

See the answer below

Explanation:

The best approach would be to <u>pour the liquid from the large reagent bottle into a small-size beaker or reagent bottle first</u>, before measuring the required quantity out into the reaction vessel. This is necessary in order to maintain safety in the laboratory.

Pouring the liquid directly from the large reagent bottle into the measuring cylinder or directly into the reaction bottle can compromise safety in the laboratory. The liquid might splash out and cause harm to the handler or create other harmful circumstances in the laboratory.

6 0
3 years ago
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