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Rudiy27
3 years ago
12

The following initial rate data are for the reaction of hypochlorite ion with iodide ion in 1M aqueous hydroxide solution: OCI+r

→ or +CI
Experiment [OCI] M I(-M) Rate (M/s)2
1 3.48 x 10-3 5.05 x 10-3 1.34 x 10-3
2 3.48 x 10-3 1.01 x 10-2 2.68 x 10-3
3 6.97 x 10-3 5.05 x 10-3 2.68 x 10-3
4 6.97 x 10-3 1.01 x 10-2 5.36 x 10-3
What is the rate law for this reaction?
a. Rate = [1-1
b. Rate = K[OCI-11-2
c. Rate = K[OCI,20-1
d. Rate = K[OCH
e. Rate = K[OC1-20-1
Chemistry
1 answer:
Vinil7 [7]3 years ago
5 0

Answer:

Rate = k [OCl] [I]

Explanation:

OCI+r → or +CI

Experiment [OCI] M I(-M) Rate (M/s)2

1 3.48 x 10-3 5.05 x 10-3 1.34 x 10-3

2 3.48 x 10-3 1.01 x 10-2 2.68 x 10-3

3 6.97 x 10-3 5.05 x 10-3 2.68 x 10-3

4 6.97 x 10-3 1.01 x 10-2 5.36 x 10-3

The table above able shows how the rate of the reaction is affected by changes in concentrations of the reactants.

In experiments 1 and 3, the conc of iodine is constant, however the rate is doubled and so is the conc of OCl. This means that the reaction is in first order with OCl.

In experiments 3 and 4, the conc of OCl is constant, however the rate is doubled and so is the conc of lodine. This means that the reaction is in first order with I.

The rate law is given as;

Rate = k [OCl] [I]

You might be interested in
100.0 mL of Ca(OH)2 solution is titrated with 5.00 x 10–2 M HBr. It requires 36.5 mL of the acid solution for neutralization. Wh
miskamm [114]

Answer:

The number of moles HBr = 0.001825

The concentration of Ca(OH)2 = 0.009125 M

Explanation:

Step 1: Data given

Volume of the Ca(OH)2 = 100.0 mL = 0.100 L

Molarity of HBr = 5.00 * 10^-2 M

Volume of HBR = 36.5 mL = 0.0365 L

Step 2: The balanced equation

Ca(OH)2 + 2HBr → CaBr2 + 2H2O

Step 3: Calculate molarity of Ca(OH) 2

b*Va* Ca = a * Vb*Cb

⇒with b = the coefficient of HBr = 2

⇒with Va = the volume of Ca(OH)2 = 0.100 L

⇒with ca = the concentration of Ca(OH)2 = TO BE DETERMINED

⇒with a = the coefficient of Ca(OH)2 = 1

⇒with Vb = the volume of HBr = 0.0365 L

⇒with Cb = the concentration of HBr = 5.00 * 10^-2 = 0.05 M

2 * 0.100 * Ca = 1 * 0.0365 * 0.05

Ca = (0.0365*0.05) / 0.200

Ca = 0.009125 M

Step 4: Calculate moles HBr

Moles HBr = concentration HBr * volume HBr

Moles HBr = 0.05 M * 0.0365 L

Moles HBr = 0.001825 moles

3 0
3 years ago
Potassium is isotopic and has RAM of 39.5 work out the percentage abundance of each isotope in a given sample of potassium which
Troyanec [42]

Yo sup??

Let the percentage of K-39 be x

then the percentage of K-40 is 100-(x+0.01)

We know that the net weight should be 39.5. Therefore we can say

(39*x+40*(100-(x+0.01))+38*0.01)/100=39.5  

(since we are taking it in percent)

39*x+40*(100-(x+0.01))+38*0.01=3950

39x+4000-40x-0.4+0.38=3950

2x=49.98

x=24.99

=25 (approx)

Therefore K-39 is 25% in nature and K-40 is 75% in nature.

Hope this helps.

3 0
3 years ago
Please somebody give me the answers
RSB [31]

Answer:

1. 9.4 grams of methane produce<u> 25.85</u> grams of CO2

2.Grams of water produced = <u>11.81 grams</u>

3.Mass of Methane produced by 10.1 gram of O2 = <u>2.52 grams</u>

4.Amount of methane consumed = <u>46.9 grams</u>

5. Grams of Co2 produced =<u> 8.32 grams</u>

<u></u>

Explanation:

Molar masses :

Methane = CH4 = mass of C + 4x (mass of H)

CH4 = 12 +4(1) = 16 grams

<u>1 mole of CH4 = 16 gram</u>

Oxygen O2 = 2 x (mass of O) = 2x(16) = 32 gram (1 mole of O2 =32 gram)

Carbon Dioxide =CO2 = mass of C + 2(mass of O)

= 12 + 2(16)

= 44 grams <u>(1 mole of CO2 = 44 gram )</u>

Water = H2O = 18 grams ( 1 mole of H2O = 18 gram)

1 mole of each molecule is equal to their molar masses

The balanced equation is :

1CH_{4}(g)+2O_{2}\rightarrow 1CO_{2}+2H_{2}O(l)

According to Stoichiometry :

1 mole of CH4 = 2 Mole of O2 = 1 mole of CO2 = 2 mole of H2O

1. From the equation ,

1 mole of methane produce  =1 mole of CO2

16 gram of methane = 44 gram of CO2

1 gram of methane =

\frac{44}{16} gram of CO2

9.4 gram of CH4 =

\frac{44}{16}\times 9.4 gram of CO2

= 25 .85 gram of CO2

2.

2 mole of O2 produces = 2 mole of H2O(water)

1 mole of O2 produces = 1 mole of H2O

32 gram of O2 = 18 gram of water

1 gram of O2 =

\frac{18}{32}

21 gram of O2 =

\frac{18}{32}\times 21

11.81 gram of water

3. 1 mole of CH4 = 2 mole of O2

16 gram of CH4 = 2(32)  = 64 grams of O2

64 gram of O2 needs = 16 grams of CH4

1 gram of O2 needs =

\frac{16}{64}

10.1 gram of O =

\frac{16}{64}\times 10.1 of CH4

= 2.52 gram

4.

1 mole of CO2 is produced from = 1 mole of CH4

44 gram of CO2 is produced from 16 gram of CH4

1 gram CO2 =

\frac{16}{44} gram of CH4

129 gram of CO2 =

\frac{16}{44}\times 129 gram of CH4

= 46.90 grams

5.

2 mole of O2  produce = 1 mole of CO2

2x 32 gram of O2 = 44 gram of CO2

1 gram of O2 =

\frac{44}{64} of CO2

12.1 gram of O2 produce=

\frac{44}{64}\times 12.1 of CO2

= 8.318 gram

Note : Write the quantity give on left side of "="

write the substance asked on right side of "="

8 0
3 years ago
Name each of the following organic molecules. 
Setler [38]
A. 2-Fluoro-3-Chlorohexane
B. 1-Bromo-2-Chloro-3-Fluorocyclopentane
C. 4-ethyl-5-methylhexane
D. 2,4,5-trimethylheptane
7 0
2 years ago
When carbon is burned in air, it reacts with oxygen to form carbon dioxide. When 27.6 g of carbon were burned in the presence of
In-s [12.5K]

Answer:

The answer to your question is: 101.2 g of CO2

Explanation:

C = 27.6 g

O₂ = 86.5 g   remained 12.9 g

O₂ that reacted = 86.5 - 12.9 = 73.6 g

                     C     + O₂      ⇒        CO₂      The equation is balanced

                    27.6    73.6                 ?

MW               12        32                  44

Rule of three

                        12 g of C------------------  44 g  CO2

                       27.6 g C  ------------------    x

                  x = 27.6(44)/12 = 101.2 g of CO2

                       32 g of O2 ---------------    44 g of CO2

                        73.6 g of O2 ------------      x

                  x = 73.6(44)/32 = 101.2 g of CO2

6 0
3 years ago
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