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jok3333 [9.3K]
3 years ago
6

Increasing the concentration of a reactant shifts the position of chemical equilibrium towards formation of more products. What

effect does adding a reactant have on the rates of the forward and reverse reactions?.
Chemistry
1 answer:
lakkis [162]3 years ago
4 0

Answer:

A.There is no effect on the forward and reverse reactions. For an equilibrium reaction, the forward and reverse rates are always equal.

B. Both the forward and reverse reactions speed up by the same amount.

C. The forward reaction speeds up immediately. As more product is made, the reverse reaction starts to speed up as the forward reaction starts to slow down until they are equal.

D. The forward reaction slows down initially. As the reaction proceeds, the reverse reaction slows down to meet the new forward reaction.

E. The forward reaction speeds up. Eventually, production of the product speeds up the reverse reaction rate to match the new forward rate.

F. It is impossible to say without more specific information.

Explanation:

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Gimme ur bath water plzzzz​
kari74 [83]

Answer:

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6 0
3 years ago
HNO3 + H2O → H3O^+ + NO3<br> which ones are acids and which ones are bases
nlexa [21]

Answer:

  • HNO₃ and  H₃O⁺ are acids
  • H₂O and  NO₃⁻ are bases

Explanation:

The chemical equation is:

  • HNO₃ + H₂O → H₃O⁺ + NO₃⁻

There are several definitions of acid and bases: Arrhenius', Bronsted-Lowry's and Lewis'.

Bronsted-Lowry model defines and <em>acid</em> as a donor of protons, H⁺.

In the given equation HNO₃ is such substance: it releases an donates its hdyrogen to form the H₃O⁺ ion.

On the other hand, a <em>base</em> is a substance that accepts protons.

In the reaction shown, H₂O accepts the proton from HNO₃ to form H₃O⁺.

Thus, H₂O is a base.

In turn, on the reactant sides the substances can be classified as acids or bases.

H₃O⁺ contain an hydrogen that can be donated and form H₂O; thus, it is an acid (the conjugated acid), and NO₃⁻ can accept a proton to form HNO₃; thus it is a base (the conjugated base).

4 0
3 years ago
explain observation made when anhydrous calcium chloride and anhydrous copper (ii) sulphate are separately exposed to the atmosp
Burka [1]

Answer:

Anhydrous calcium chloride  dissolves and becomes liquid

Anhydrous copper (ii) sulphate will produce crystal particles

Explanation:

Anhydrous calcium chloride is deliquescent and hence when it is exposed to air, it absorbs water from air. After absorbing water, it dissolves and after some time a pool of clear liquid appears.

Anhydrous copper (ii) sulphate will form crystal structures  and the following reaction will takes place

CuSO4 + 5 H20 --> CuSO4.5H2O

6 0
3 years ago
Using the volumes of sodium thiosulfate solution you just entered, the mass of bleach sample, and the average molarity of the so
dedylja [7]

Answer:

3.18 (w/w) %

Explanation:

In the problem, you can find mass of NaClO knowing the reaction of NaClO with Na₂S₂O₃ is:

NaClO + 2Na₂S₂O₃ + H₂O → NaCl + Na₂S₄O₆ +2NaOH + NaCl

<em>Where 1 mole of NaClO reacts with 2 moles of Na₂S₂O₃</em>

<em> </em>Moles of thiosulfate in the titration are:

0.0101L ₓ (0.042mol / L) = 4.242x10⁻⁴ moles of Na₂S₂O₃

Thus, moles of NaClO in the initial solution are:

4.242x10⁻⁴ moles of Na₂S₂O₃ ₓ (1mol NaClO / 2 mol Na₂S₂O₃) = 2.121x10⁻⁴ moles NaClO

As molar mass of NaClO is 74.44g/mol, mass of 2.121x10⁻⁴ moles are:

2.121x10⁻⁴ moles ₓ (74.44g / mol) = <em>0.0158g of NaClO</em>

As mass of bleach is 0.496g, mass percent is:

0.0158g NaClO / 0.496g bleach ₓ 100 =

<h3>3.18 (w/w) % </h3>
4 0
3 years ago
What is the part of an experiment that is not being tested and is used for comparison
STALIN [3.7K]
The controlled variable, I think.
7 0
3 years ago
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