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Alenkinab [10]
3 years ago
8

Be sure to answer all parts. thallium(i) is oxidized by cerium(iv) as follows: tl+(aq) + 2ce4+(aq) → tl3+(aq) + 2ce3+(aq) the el

ementary steps, in the presence of aqueous mn(ii), are as follows: step 1: ce4+ + mn2+ → ce3+ + mn3+ step 2: ce4+ + mn3+ → ce3+ + mn4+ step 3: tl+ + mn4+ → tl3+ + mn2+ the rate law is given by the equation: rate = k[ce4+][mn2+] identify the catalyst: tl+ tl3+ ce3+ ce4+ mn2+ mn3+ mn4+ identify the intermediates: tl+ tl3+ ce3+ ce4+ mn2+ mn3+ mn4+ identify the rate-determining step
Chemistry
1 answer:
Doss [256]3 years ago
5 0
<span>Best Answer: Mn(ii) is catalyst Stetep-1 is slow step Steps-2,3 are fast steps intermediates are Mn(iii)and Mn(iv) since step -1 is slow rate depends on Ce(iv) and Mn(ii) only not on Tl(i) as it is involved in fast step-3</span>
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Name the following alkyne:
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Answer:

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Explanation:

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Calculate the vapor pressure at 35ºC of a solution made by dissolving 20.2 g of sucrose (C12H22O11)in 60.5 g of water. The vapor
sukhopar [10]

Answer:

P' = 41.4 mmHg → Vapor pressure of solution

Explanation:

ΔP = P° . Xm

ΔP = Vapor pressure of pure solvent (P°) - Vapor pressure of solution (P')

Xm = Mole fraction for solute (Moles of solvent /Total moles)

Firstly we determine the mole fraction of solute.

Moles of solute → Mass . 1 mol / molar mass

20.2 g . 1 mol / 342 g = 0.0590 mol

Moles of solvent → Mass . 1mol / molar mass

60.5 g . 1 mol/ 18 g = 3.36 mol

Total moles = 3.36 mol + 0.0590 mol = 3.419 moles

Xm = 0.0590 mol / 3.419 moles → 0.0172

Let's replace the data in the formula

42.2 mmHg - P' = 42.2 mmHg . 0.0172

P' = - (42.2 mmHg . 0.0172 - 42.2 mmHg)

P' = 41.4 mmHg

5 0
3 years ago
A mutation in a certain type of bees allows for bees to pollinate only one type of flour how could this mutation be both a benef
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3 years ago
A sample of Xe gas is observed to effuse through a pourous barrier in 4.83 minutes. Under the same conditions, the same number o
Solnce55 [7]

Answer:

28.93 g/mol

Explanation:

This is an extension of Graham's Law of Effusion where \frac{R1}{R2} = \sqrt{\frac{M2}{M1} } = \frac{t2}{t1}

We're only talking about molar mass and time (t) here so we'll just concentrate on \sqrt{\frac{M2}{M1} } = \frac{t2}{t1}. Notice how the molar mass and time are on the same position, recall effusion is when gas escapes from a container through a small hole. The time it takes it to leave depends on the molar mass. If the gas is heavy, like Xe, it would take a longer time (4.83 minutes). If it was light it would leave in less time, that gives us somewhat an idea what our element could be, we know that it's atleast an element before Xenon.

Let's plug everything in and solve for M2. I chose M2 to be the unknown here because it's easier to have it basically as a whole number already.

\sqrt{\frac{M2}{131} } = \frac{2.29}{4.83}

The square root is easier to deal with if you take it out in the first step, so let's remove it by squaring each side by 2, the opposite of square root essentially.

(\sqrt{\frac{M2}{131} } )^2= (\frac{2.29}{4.83})^2

{\frac{M2}{131} } = (0.47)^2

{\frac{M2}{131} } = 0.22

M2= 0.22 x 131

M2= 28.93 g/mol

8 0
2 years ago
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