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aliina [53]
3 years ago
10

Consider the data [X] [Y] [Z] initial rate M M M M · s −1 Exp 1 0.30 0.20 0.35 0.210 Exp 2 0.60 0.10 0.70 0.420 Exp 3 0.60 0.20

0.70 0.420 Exp 4 0.60 0.40 0.35 0.105 What is a correct rate law for the reaction?
Chemistry
1 answer:
sergij07 [2.7K]3 years ago
3 0

Answer:

Rate = k [X]⁻¹ [Z]²

Explanation:

[X] [Y] [Z] initial rate M M M M · s −1

Exp 1 0.30 0.20 0.35      0.210

Exp 2 0.60 0.10 0.70      0.420

Exp 3 0.60 0.20 0.70     0.420

Exp 4 0.60 0.40 0.35      0.105

In Experiment 2 and 3 where the concentrations of Y and Z were constant, doubling the concentration of Y had no effect on the rate of the reaction. This means, that the rate of the reaction is zero order with respect to Y.

In experiment 3 and 4, dividing the concentration of Z by 2, causes the rate of the reaction to decrease by 4. This means the rate of the reaction is second order with respect to Z.

In experiment 1 and 4, doubling the concentration of X, causes the rate of the reaction to decrease by half. This means that X has an order of -1 with respect to the rate of the reaction.

The rate expression is given as;

Rate = k [X]⁻¹[Y]⁰[Z]²

Rate = k [X]⁻¹ [Z]²

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1. What mass of Mgo will I make from 48 g of Mg?<br> 2Mg + O2 + 2MgO
statuscvo [17]

Answer:

80.6 g

Explanation:

MgO=24.3+16=40.3 g

48.6/24.3= 2 moles

64/2x16= 2 moles

2x40.3= 80.6g

8 0
2 years ago
What volumes of 0.200 M HCOOH and 2.00 M NaOH would make 500. mL of a buffer with the same pH as one made from 475 mL of 0.200 M
Fed [463]

<u>36 ml of NaOh and</u><u> 464 ml</u><u> of </u><u>HCOOH</u><u> would be enough to form 500 ml of a buffer with the same pH as the buffer made with </u><u>benzoic acid </u><u>and NaOH.</u>

What is benzoic acid found in?

  • Some natural sources of benzoic acid include: ​Fruits:​ Apricots, prunes, berries, cranberries, peaches, kiwi, bananas, watermelon, pineapple, oranges.
  • ​Spices:​ Cinnamon, cloves, allspice, cayenne pepper, mustard seeds, thyme, turmeric, coriander.

Calculate the amount of moles in NaOH and benzoic acid. This calculation is done by multiplying molarity by volume.

Amount of moles of NaOH -2 × 0.025 =  0.05 mol

Amount of moles of benzoic acid 2 × 0.475 = 0.095 mol

In this case, we can calculate the pH produced by the buffer of these two reagents, as follows

pH = pKa + log\frac{base}{acid}

4.2 + log\frac{0.05}{0.045} = 4.245

We must repeat this calculation, with the values shown for HCOOH and NaOH. In this case, we can calculate as follows

pH = pKa + log\frac{base}{acid}

4.245 = 3.75 + log\frac{base}{acid}

log\frac{base}{acid} = 0.5

\frac{base}{acid} = 3.162

Now we must solve the equation above. This will be done using the following values

\frac{2(0.5 - x)}{0.2x - 2(0.5 - x)} = 0.464 L

With these values, we can calculate the volumes of NaOH and HCOOH needed to make the buffer.

NaOH volume

( 0.5 - 0.464)L

0.036L .................... 36ml

HCOOH volume

500 - 36 = 464mL

Learn more about benzoic acid

brainly.com/question/24052816

#SPJ4

3 0
1 year ago
If the solution containing the metal complex ion is made acidic, why will the silver chloride precipitate
Ne4ueva [31]

Answer,

For example, silver ion can be precipitated with hydrochloric acid to yield solid silver chloride. Because many cations will not react with hydrochloric acid in this way, this simple reaction can be used to separate ions that form insoluble chlorides from those that do not.

7 0
2 years ago
It is possible to have more than the maximum amount of solute dissolved in a solvent.
vladimir1956 [14]

B. False

When the maximum amount of solute has been dissolved in a given amount of solvent, we say that the solution is saturated with solute.

3 0
3 years ago
Read 2 more answers
A detailed explanation, one paragraph of the colligative property being discussed and why that property changes the way that it
levacccp [35]
<span>
You can do it on the icing of roads, reverse osmosis for desalination of water, dissolved CO2 in soda cans, osmotic pressure involving blood vessels and IV solutions, etc.</span>
6 0
3 years ago
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