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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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2 years ago
What is the total pressure (in atm) inside of a vessel containing N2 exerting a partial pressure of 0.256 atm, He exerting a par
Bezzdna [24]

Answer: Total pressure inside of a vessel is 0.908 atm

Explanation:

According to Dalton's law, the total pressure is the sum of individual partial pressures. exerted by each gas alone.

p_{total}=p_1+p_2+p_3

p_{N_2} = partial pressure of nitrogen = 0.256 atm

p_{He} = partial pressure of helium = 203 mm Hg = 0.267 atm  (760mmHg=1atm)

p_{H_2} = partial pressure of hydrogen =39.0 kPa = 0.385 atm  (1kPa=0.00987 atm)

Thus p_{total}=p_{H_2}+p_{He}+p_{H_2}

p_{total} =0.256atm+0.267atm+0.385atm =0.908atm

Thus total pressure (in atm) inside of a vessel is 0.908

8 0
3 years ago
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How many grams of sodium are in 0.820 moles of na2so4?
s344n2d4d5 [400]
<span>the molar mass of a compound is the sum of the products of the atomic masses by the number of atoms of the element.
molar mass of Na</span>₂SO₄<span> is - 142 g/mol.
1 mol of </span>Na₂SO₄<span> has a mass of 142 g.
In 1 mol of </span>Na₂SO₄<span> the mass of Na is 23 g/mol x 2 = 46 g.
                             
Mass of Na in 1 mol of </span>Na₂SO₄ is - 46 g
                           
mass of Na in 0.820 mol of Na₂SO₄ - 46 g /1 mol x 0.820 mol = 37.72 g.
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5 0
3 years ago
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2 years ago
Urgent!! A chemist measured 5.2 g copper(II) bromide tetrahydrate (CuBr2•4(H2O)). How many moles were measured out? Answer in un
bezimeni [28]
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