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blagie [28]
3 years ago
7

For the following reaction, provide the missing information

Chemistry
1 answer:
jenyasd209 [6]3 years ago
8 0

Answer:

19. Option B. ⁰₋₁B

20. Option D. ²¹⁰₈₄Po

Explanation:

19. ²²⁸₈₈Ra —> ²²⁸₈₉Ac + ʸₓZ

Thus, we can determine ʸₓZ as follow:

228 = 228 + y

Collect like terms

228 – 228 = y

y = 0

88 = 89 + x

Collect like terms

88 – 89 = x

x = –1

Thus,

ʸ ₓZ => ⁰₋₁Z => ⁰₋₁B

²²⁸₈₈Ra —> ²²⁸₈₉Ac + ʸₓZ

²²⁸₈₈Ra —> ²²⁸₈₉Ac + ⁰₋₁B

20. ᵘᵥX —> ²⁰⁶₈₂Pb + ⁴₂He

Thus, we can determine ᵘᵥX as follow:

u = 206 + 4

u = 210

v = 82 + 2

v = 84

Thus,

ᵘᵥX => ²¹⁰₈₄X => ²¹⁰₈₄Po

ᵘᵥX —> ²⁰⁶₈₂Pb + ⁴₂He

²¹⁰₈₄Po —> ²⁰⁶₈₂Pb + ⁴₂He

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Answer:

(i)  Oxidizing Agent: NO2 / Reducing Agent NH3-

(ii) Oxidizing Agent AgNO3 / Reducing Agent Zn

Explanation:

(i) 8NH3( g) + 6NO2( g) => 7N2( g) + 12H2O( l)

In this reaction, both two reactants contain nitrogen with a different oxidation number and produce only one product which contains nitrogen with a unique oxidation state. So, nitrogen is oxidized and reduced in the same reaction.

Nitrogen Undergoes a change in oxidation state from 4+ in NO2 to 0 in N2. It is reduced because it gains electrons (decrease its oxidation state). NO2 is the oxidizing agent (electron acceptor).

Nitrogen Changes from an oxidation state of 3- in NH3 to 0 in N2. It is oxidized because it loses electrons (increase its oxidation state). NH3 is the reducing agent (electron donor)

(ii) Zn(s) +AgNO3(aq) => Zn(NO3)2(aq) + Ag(s)

Ag changes oxidation state from 1+ to 0 in Ag(s).

Ag is reduced because it gains electrons and for this reason and AgNO3 is the oxidizing agent (electron acceptor)

Zn Changes from an oxidation state of 0 in Zn(s) to 2+ in Zn(NO3)2. It is oxidized and for this reason Zn is the reducing agent (electron donor).

Balanced equation:

Zn(s) +2AgNO3(aq) => Zn(NO3)2(aq) + 2Ag(s)

 

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3 years ago
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3 years ago
You just got home from a run on a hot Atlanta afternoon. You grab a 1.00 liter bottle of water and drink three quarters of it in
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Answer:

¾ litters of water was consumed

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8 0
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5. A sample of benzene (C6H6), weighing 7.05 g underwent combustion in a bomb calorimeter by the following reaction:
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Answer:

A sample of benzene (C6H6), weighing 7.05 g underwent combustion in a bomb calorimeter by the following reaction:

2 C6H6 (l) + 15 O2 (g) → 12 CO2 (g) + 6 H2O (l)

The heat given off was absorbed by 500 g of water and caused the temperature of the water and the calorimeter to rise from 25.00 to 53.13 oC. The heat capacity of water = 4.18 J/g/oC and the heat capacity of the calorimeter = 10.5 kJ/oC. (1) what is the ΔH of the reaction?

Explanation:

The heat energy released by the reaction = heat absorbed by calorimeter + heat absorbed by water

Heat absorbed by water = mass of water x specific heat capacity of water x change in temperature

Heat absorbed by water =  500 g x 4.18 J/g. oC x (53.13-25.00)oC

                                         = 58791.7 J

Heat absorbed by calorimeter = heat capacity of calorimeter x change in temperature

Heat absorbed by calorimeter = 10.5 x 10^3 J /oC  x (53.13-25.00)oC

                                                  =295365 J

Total heat energy absorbed = 58791.7 J + 295365 J  = 354156.7 J

Number of moles of benzene given is:

number of moles = goven mass of benzene /its molar mass

=7.05 g / 78.0 g/mol

=0.0903mol

Hence, the heat released by the reaction is:

= 354156.7 J / 0.0903 mol

=  3922.00 kJ/mol

Answer:

The heat released during the combustion of 7.05g of benzene is 3922.00kJ/mol.                                              

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