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Alenkinab [10]
4 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]4 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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What is the volume of 0.640 grams of Oz gas at Standard Temperature and Pressure (STP)?
Ilia_Sergeevich [38]

Answer: The volume of 0.640 grams of O_{2} gas at Standard Temperature and Pressure (STP) is 0.449 L.

Explanation:

Given: Mass of O_{2} gas = 0.640 g

Pressure = 1.0 atm

Temperature = 273 K

As number of moles is the mass of substance divided by its molar mass.

So, moles of O_{2} (molar mass = 32.0 g/mol) is as follows.

No. of moles = \frac{mass}{molar mass}\\= \frac{0.640 g}{32.0 g/mol}\\= 0.02 mol

Now, ideal gas equation is used to calculate the volume as follows.

PV = nRT

where,

P = pressure

V = volume

n = no. of moles

R = gas constant = 0.0821 L atm/mol K

T = temperature

Substitute the values into above formula as follows.

PV = nRT\\1.0 atm \times V = 0.02 mol \times 0.0821 L atm/mol K \times 273 K\\V = 0.449 L

Thus, we can conclude that the volume of 0.640 grams of O_{2} gas at Standard Temperature and Pressure (STP) is 0.449 L.

6 0
3 years ago
Which of the following questions about the flu is a scientific question?
qwelly [4]
D. Is a fly shot effective in preventing the flu?

Answer A touches on the topics of economics and industrial feasibility of the flu shot.
B is more of an ethical / situational question.
C is a social question.
D is the only question which focuses on the science of microbiology and medicine.
5 0
3 years ago
Read 2 more answers
What is the solution to the problem expressed to the correct number of significant figures
quester [9]

I don't know what the problem is, but here are some rues to help you out:

  1. All non-zero figures are significant
  2. When a zero falls between non-zero digits, that zero is significant.
  3. When a zero falls after a decimal point, that zero is significant.
  4. When multiplying and dividing significant figures, the answer is limited to the number of sig figs equal to the least number of sig figs in the problem.
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4 0
3 years ago
Be sure to answer all parts. The standard enthalpy of formation and the standard entropy of gaseous benzene are 82.93 kJ/mol and
skelet666 [1.2K]

Answer : The values of \Delta H^o,\Delta S^o\text{ and }\Delta G^o are 33.89kJ,95.94J/K\text{ and }5.299kJ/mol respectively.

Explanation :

The given balanced chemical reaction is,

C_6H_6(l)\rightarrow C_6H_6(g)

First we have to calculate the enthalpy of reaction (\Delta H^o).

\Delta H^o=H_f_{product}-H_f_{reactant}

\Delta H^o=[n_{C_6H_6(g)}\times \Delta H_f^0_{(C_6H_6(g))}]-[n_{C_6H_6(l)}\times \Delta H_f^0_{(C_6H_6(l))}]

where,

\Delta H^o = enthalpy of reaction = ?

n = number of moles

\Delta H_f^0_{(C_6H_6(g))} = standard enthalpy of formation  of gaseous benzene = 82.93 kJ/mol

\Delta H_f^0_{(C_6H_6(l))} = standard enthalpy of formation  of liquid benzene = 49.04 kJ/mol

Now put all the given values in this expression, we get:

\Delta H^o=[1mole\times (82.93kJ/mol)]-[1mole\times (49.04J/mol)]

\Delta H^o=33.89kJ/mol=33890J/mol

Now we have to calculate the entropy of reaction (\Delta S^o).

\Delta S^o=S_f_{product}-S_f_{reactant}

\Delta S^o=[n_{C_6H_6(g)}\times \Delta S^0_{(C_6H_6(g))}]-[n_{C_6H_6(l)}\times \Delta S^0_{(C_6H_6(l))}]

where,

\Delta S^o = entropy of reaction = ?

n = number of moles

\Delta S^0_{(C_6H_6(g))} = standard entropy of formation  of gaseous benzene = 269.2 J/K.mol

\Delta S^0_{(C_6H_6(l))} = standard entropy of formation  of liquid benzene = 173.26 J/K.mol

Now put all the given values in this expression, we get:

\Delta S^o=[1mole\times (269.2J/K.mol)]-[1mole\times (173.26J/K.mol)]

\Delta S^o=95.94J/K.mol

Now we have to calculate the Gibbs free energy of reaction (\Delta G^o).

As we know that,

\Delta G^o=\Delta H^o-T\Delta S^o

At room temperature, the temperature is 25^oC\text{ or }298K.

\Delta G^o=(33890J)-(298K\times 95.94J/K)

\Delta G^o=5299.88J/mol=5.299kJ/mol

Therefore, the values of \Delta H^o,\Delta S^o\text{ and }\Delta G^o are 33.89kJ,95.94J/K\text{ and }5.299kJ/mol respectively.

6 0
3 years ago
Read 2 more answers
2. Potassium is a metal with one valence electron. With which other elements does it form bonds? a elements with one electron b.
Anika [276]

Answer:

Atoms with one or two valence electrons more than a closed shell are highly reactive ... Hydrogen can form bonds to many other elements such as nitrogen NH 3 ... the electron dot structure and iii the chemical K 1 valence electron 1 dot K . B two. ... The alkaline earth metals IIA elements lose two electrons to form a 2 cation.

Explanation:

6 0
3 years ago
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