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Triss [41]
3 years ago
10

The reaction A + 2B → products was found to follow the rate law: rate = k[A] 2[B]. Predict by what factor the rate of reaction w

ill increase when the concentration of A is doubled, the concentration of B is tripled, and the temperature remains constant
Chemistry
1 answer:
Alexus [3.1K]3 years ago
5 0

Answer:

By a factor of 12

Explanation:

For the reaction;

A + 2B → products

The rate law is;

rate = k[A]²[B]

As you can see, the rate is proportional to the square of the concentration of  A  and the of the concentration of  B .

Let's say initially, [A] = x, [B] = y

The rate law in this case is equal to;

rate1 = k. x².y

Now you double the concentration of A and triple the concentration of B.

[A] = 2x, [B] = 3y

The new rate law is given as;

rate2 = k . (2x)². (3y)

rate2 = k . 4x² . 3y

rate2 = 12 k . x² . y

Comparing rate 2 and rate 1, the ratio is given as; rate 2/ rate 1 = 12

Therefore the rate has increased by a factor of 12.

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mario62 [17]

Answer:

  • <u><em>a. C₂H₄</em></u>

Explanation:

At constant pressure and temperature, the mole ratio of the gases is equal to their volume ratio (a consequence of Avogadro's law).

Hence, the <em>complete combustion reaction</em> that has a ratio of 100 ml of gaseous hydrocarbon to 300 ml of oxygen, is that whose mole ratio is 1 mol hydrocarbon : 3 mol of oxygen.

Then, you must write the balanced chemical equations for the complete combustion of the four hydrocarbons in the list of choices, and conclude which has such mole ratio (1 mol hydrocarbon : 3 mol oxygen).

A complete combustion reaction of a hydrocarbon is the reaction with oxygen that produces CO₂ and H₂O, along with the release of heat and light.

<u>a. C₂H₄:</u>

  • C₂H₄ (g) + 3O₂ (g) → 2CO₂(g)  + 2H₂O (g)

Precisely, for this reaction the mole ratio is 1 mol C₂H₄: 2 mol O₂, hence, this is the right choice.

The following analysis just shows that the other options are not right.

<u>b. C₂H₂:</u>

  • 2C₂H₂ (g) + 5O₂ (g) → 4CO₂(g)  + 2H₂O (g)

The mole ratio for this reaction is 2 mol C₂H₂ :5 mol O₂.

<u>с. С₃Н₈</u>

  • C₃H₈ (g) + 5O₂ (g) → 3CO₂(g)  + 4H₂O (g)

The mole ratio is 1 mol C₃H₈ : 5 mol O₂

<u>d. C₂H₆</u>

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The mole ratio is 2 mol C₂H₆ : 7 mol O₂

7 0
3 years ago
Use the periodic table to determine the electron configuration for iodine (i). express your answer in condensed form.
Dovator [93]
Iodine electron configuration is:

1S^2 2S^2 2P^6 3S^2 3P^6 4S^2 3d^10 4P^6 5S^2 4d^10  5P^5
when Krypton is the noble gas in the row above iodine in the periodic table,
we can change 1S^2  2S^2 2P^6 3S^2 3P^6 4S^2 3d^10 4P^6 by the symbol
[Kr] of Krypton.

So we can write the electron configuration of Iodine:
[Kr] 5S^2 4d^10 5P^5

8 0
3 years ago
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vovikov84 [41]
<span>My only guess is obtain a metal and heat it in a boiling water bath (of known temperature) this will be your initial temperature. Now obtain a calorimeter cup with water of known temperature as well. Place the metal into the calorimeter cup and record the temperature after 5 minutes. You now have delta T, mass of the metal, and Q. Solve for C.

Hope this helps xox :)</span>
3 0
3 years ago
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To what volume in millimeters must 50.0 mL of 18.0 M H2SO4 be diluted to obtain 4.35 M H2SO4?
poizon [28]
We know that to relate solutions of with the factors of molarity and volume, we can use the equation: M_{1}  V_{1} = M_{2}  V_{2}

**NOTE: The volume as indicated in this question is defined in L, not mL, so that conversion must be made. However it is 1000 mL = 1 L.

So now we can assign values to these variables. Let us say that the 18 M H_{2}  SO_{4} is the left side of the equation. Then we have:

(18 M)(0.050 L)=(4.35M) V_{2}

We can then solve for V_{2}:

V_{2}= \frac{(18M)(0.05L)}{4.35M} and V_{2} =0.21 L or 210 mL

We now know that the total amount of volume of the 4.35 M solution will be 210 mL. This is assuming that the entirety of the 50 mL of 18 M is used and the rest (160 mL) of water is then added.


7 0
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. How many ethyl groups are in the molecule 3, 3-diethylpentane?
Mamont248 [21]

Answer:

2 groups

Explanation:

The molecule 3, 3-diethylpentane has 2 ethyl groups because the prefix di- means that there are 2 homogeneous substituent groups present in the molecule.

6 0
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