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Taya2010 [7]
3 years ago
14

13. Write the following balanced reactions using cell notation. Use platinum as an inert electrode, if needed. (a) Mg(s) + Ni2+(

aq) ⟶ Mg2+(aq) + Ni(s) (b) 2Ag+(aq) + Cu(s) ⟶ Cu2+(aq) + 2Ag(s) (c) Mn(s) + Sn(NO3 ) 2 (aq) ⟶ Mn(NO3 ) 2 (aq) + Au(s) (d) 3CuNO3 (aq) + Au(NO3 ) 3 (aq) ⟶ 3Cu(NO3 ) 2 (aq) + Au(s)
Chemistry
1 answer:
sp2606 [1]3 years ago
5 0

Answer:

A) Mg ₍s₎ | Mg²⁺₍aq₎║Ni²⁺₍aq₎ | Ni₍s₎

B)  Cu₍s₎ | Cu²⁺₍aq₎║ Ag⁺  |  Ag⁰₍s₎

C)  Cu₍s₎ | Cu²⁺₍aq₎║ Ag⁺  |  Ag⁰₍s₎

D) Pt₍s₎ | CuNO₃₍ aq₎ , Cu(NO₃)₂₍aq₎ ║ Au(NO₃)₃₍aq₎ | Au₍s₎

Explanation:

A) Mg + Ni²⁺ → Mg²⁺ + Ni

Mg⁰→ Mg²⁺ + 2e⁻    oxidation   anode

Ni²⁺ + 2e⁻ → Ni⁰       reduction   cathode

Mg ₍s₎ | Mg²⁺₍aq₎║Ni²⁺₍aq₎ | Ni₍s₎

anode        salt bridge       cathode

B)   2 Ag⁺₍aq₎ + Cu₍s₎ → Cu²⁺₍aq₎ + 2 Ag₍s₎

2 Ag⁺ + 2e⁻ → 2 Ag⁰   cathode

Cu →  Cu²⁺ + 2e⁻          anode

 Cu₍s₎ | Cu²⁺₍aq₎║ Ag⁺  |  Ag⁰₍s₎

anode       salt bridge       cathode

C) Mn ₍s₎  + Sn(NO₃)₂ ₍aq₎ → Mn (NO₃)₂ ₍aq₎  + Au₍s₎

Mn⁰ → Mn²⁺+ 2e⁻       anode

Sn²⁺ + 2e⁻ → Sn⁰₍s₎      cathode

Mn₍s₎ | Mn(NO₃)₂ ₍aq₎ ║ Sn(NO₃)₂ ₍aq₎ | Sn₍s₎

anode                salt bridge                   cathode

D) 3CuNO₃₍ aq₎ + Au(NO₃)₃₍aq₎  → 3Cu(NO₃)₂₍aq₎  + Au₍s₎

3Cu⁺ → 3Cu²⁺ + 3e⁻  anode

Au³⁺ + 3e⁻ → Au⁰                cathode

Pt₍s₎ | CuNO₃₍ aq₎ , Cu(NO₃)₂₍aq₎ ║ Au(NO₃)₃₍aq₎ | Au₍s₎

       anode                           salt bridge                   cathode

Semi-reactions are noted to determine oxidation and reduction. The oxidation reaction occurs at the anode and the reduction reaction at the cathode. In the case of subsections a, b, c the electrodes may be the solid metal, while in the case d the species are in an accusative medium and require the platinum electrode.

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The decomposition of \rm XY is second order in \rm XY and has a rate constant of6.96Ã10â3M^{-1} \cdot s^{-1} at a certain temper
LUCKY_DIMON [66]

Explanation:

Half life is the amount of time taken by a radioactive material to decay to half of its original value.

Half life for second order kinetics is given by:

t_{1/2}=\frac{1}{k\times a_0}

k = rate constant =?

a_0 = initial concentration

a = concentration left after time t

Integrated rate law for second order kinetics is given by:

\frac{1}{a}=kt+\frac{1}{a_0}

a) Initial concentration of XY = a_o=0.100 M

Rate constant of the reaction = k = 6.96\times 10^{-3} M^{-1} s^{-1}

Half life of the reaction is:

t_{1/2}=\frac{1}{6.96\times 10^{-3} M^{-1} s^{-1}\times 0.100 M}

=1,436.78 s

1,436.78 seconds is the half-life for this reaction.

b) Initial concentration of XY = 0.100 M

Final concentration after time t = 12.5% of 0.100 M = 0.0125 M

\frac{1}{0.0125 M}=6.96\times 10^{-3} M^{-1} s^{-1}\times t+\frac{1}{0.100M}

Solving for t;

t = 10,057.47 seconds

In 10,057.47 seconds the concentration of XY will become 12.5% of its initial concentration.

c) Initial concentration of XY = 0.200 M

Final concentration after time t = 12.5% of 0.200 M = 0.025 M

\frac{1}{0.025 M}=6.96\times 10^{-3} M^{-1} s^{-1}\times t+\frac{1}{0.200M}

Solving for t;

t = 5,028.73 seconds

In 5,028.73 seconds the concentration of XY will become 12.5% of its initial concentration.

d) Initial concentration of XY = 0.160 M

Final concentration after time t = 6.20\times 10^{-2} M

\frac{1}{6.20\times 10^{-2} M}=6.96\times 10^{-3} M^{-1} s^{-1}\times t+\frac{1}{0.200M}

Solving for t;

t = 1,419.40 seconds

In 1,419.40 seconds the concentration of XY will become 6.20\times 10^{-2} M.

e)  Initial concentration of SO_2Cl_2= 0.050 M

Final concentration after time t = x

t = 55.0 s

\frac{1}{x}=6.96\times 10^{-3} M^{-1} s^{-1}\times 55.0 s+\frac{1}{0.050 M}

Solving for x;

x = 0.04906 M

The concentration after 55.0 seconds is 0.04906 M.

f) Initial concentration of XY= 0.050 M

Final concentration after time t = x

t = 500 s

\frac{1}{x}=6.96\times 10^{-3} M^{-1} s^{-1}\times 500 s+\frac{1}{0.050 M}

Solving for x;

x = 0.04259 M

The concentration after 500 seconds is 0.0.04259 M.

7 0
3 years ago
What are three sources of information scientist use when they do research
exis [7]

Books, Other scientist's research, the Internet.

6 0
3 years ago
hat is the ultimate source of energy for the Earth? a. photosynthesis b. respiration c. the sun d. nutrient cycles
padilas [110]
The correct answer is C. The sun
6 0
4 years ago
You are examining decomposition of corn leaves following the growing season on a farm. To do this, you place 33 g of corn leaves
m_a_m_a [10]

Answer:

c. 9.94 g

Explanation:

From the question,

Using

mt = m₀e⁻kt.................... Equation 1

Where mt = mass of the leaf remaining in the bag, m₀ = original mass of leave that was placed in the bag, k = decay constant, t = time.

Given: m₀ = 33 g, k = 0.04, t = 30 days.

Substitute into equation 1

mt = 33(e⁻(0.04ˣ30))

mt = 33e⁻¹²/¹⁰

mt = 33/e¹²/¹⁰

mt = 33/3.320

mt = 9.94 g.

Hence the right answer is c. 9.94 g

8 0
4 years ago
1. Calculate how many moles of glycine are in a 130.0-g sample of glycine.2. Calculate the percent nitrogen by mass in glycine.
Alexxx [7]

Answer:

n=1.732mol

\% N=18.7\%

Explanation:

Hello!

In this case, since the molecular formula of glycine is C₂H₅NO₂, we realize that the molar mass is 75.07 g/mol; thus, the moles in 130.0 g of glycine are:

n=130.0g*\frac{1mol}{75.07 g}\\\\ n=1.732mol

Furthermore, we can notice 75.07 grams of glycine contains 14.01 grams of nitrogen; thus, the percent nitrogen turns out:

\% N=\frac{14.01}{75.07}*100\% \\\\\% N=18.7\%

Best regards!

4 0
3 years ago
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