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exis [7]
4 years ago
12

Tarnish on tin is the compound SnO. A tarnished tin plate is placed in an aluminum pan of boiling water. When enough salt is add

ed so that the solution conducts electricity, the tarnish disappears. Imagine that the two halves of this redox reaction were separated and connected with a wire and a salt bridge. Part A Calculate the standard cell potential given the following standard reduction potentials: Al3++3e−→Al;E∘=−1.66 V Sn2++2e−→An;E∘=−0.140 V
Chemistry
1 answer:
natulia [17]4 years ago
5 0

Answer:

1.52V

Explanation:

Oxidation half equation:

2Al(s)−→2Al^3+(aq) + 6e

Reduction half equation

3Sn2^+(aq) + 6e−→3Sn(s)

E°cell= E°cathode - E°anode

E°cathode= −0.140 V

E°anode= −1.66 V

E°cell=-0.140-(-1.66)

E°cell= 1.52V

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Bingel [31]
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4 0
3 years ago
Read 2 more answers
650.J is the same amount of energy as? <br>2720cal<br>1550cal<br>650.cal<br>2.72cal
Oksi-84 [34.3K]

<u>Ans: 650 J = 155 calories</u>


<u>Given:</u>

Energy in joules = 650 J

<u>To determine:</u>

The energy in calories

<u>Explanation:</u>

1 joule = 0.2388 calories

Therefore, 650 joules = 0.2388 calories * 650 J/1 J = 155 calories

4 0
4 years ago
Read 2 more answers
What will the ratio of ions be in any compound formed from a group 1a and a group 7a nonmetals?
julsineya [31]
Group 1a elements (the first column on the left side of the Periodic table) always release one electron to form positive ions with a charge of +1. Group 7a nonmetals (the <em>second to the last </em>column on the right side- the rightmost column are the noble gases) always desire to gain one electron to form negative ions with a charge of -1.
Since their charges are equal and opposite, they will always combine in a 1:1 ratio.

4 0
3 years ago
How many formula units make up 20.6 g of magnesium chloride (MgCl2)?
irina [24]

Answer:The formula of magnesium chloride is MgCl2 . The molar mass of MgCl2 is (24.30 + 2 × 35.45) g/mol=95.20 g/mol .

Explanation:

5 0
3 years ago
The wall of an industrial furnace is constructed from 0.15-m-thick, fireclay brick having a thermal conductivity of LZ W/m.K. Me
aliya0001 [1]

Answer:

The rate of heat loss through the wall is 1700 watts.

Explanation:

The complete statement of the problem is:

<em>The wall of an industrial furnace is constructed from 0.15-m-thick, fireclay brick having a thermal conductivity of 1.7 W/m·K. Measurements made during steady state operation reveal temperatures of 1400 and 1150 K at the inner and outer sur- faces, respectively. What is the rate of heat loss through a wall that is 0.5 m by 1.2 m on a side?</em>

Given that wall of the industrial furnace is under steady conditions of heat transfer and whose configuration is a flat element, we use the equation of conductive heat transfer rate (\dot Q), measured in watts:

\dot Q = \frac{k\cdot w\cdot h}{l}\cdot (T_{i}-T_{o})

Where:

k - Thermal conductivity, measured in watts per meter-Kelvin.

w - Width of the wall, measured in meters.

h - Height of the wall, measured in meters.

l - Thickness of the wall, measured in meters.

T_{i} - Inner surface temperature, measured in Kelvin.

T_{o} - Outer surface temperature, measured in Kelvin.

If we know that k = 1.7 \,\frac{W}{m\cdot K}, w = 1.2\,m, h = 0.5\,m, l = 0.15\,m, T_{i} = 1400\,K and T_{o} = 1150\,K, the steady state heat transfer is:

\dot Q = \left[\frac{\left(1.7\,\frac{W}{m\cdot K} \right)\cdot (1.2\,m)\cdot (0.5\,m)}{0.15\,m} \right]\cdot (1400\,K-1150\,K)

\dot Q = 1700\,W

The rate of heat loss through the wall is 1700 watts.

3 0
3 years ago
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