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densk [106]
3 years ago
14

How to balance lead and silver acetate yields lead (||) acetate and silver

Chemistry
1 answer:
arsen [322]3 years ago
3 0
The given elements put into an equation using their symbols  are as follows:
Pb + AgC_{2}H_{3}O_{2} = Pb(II){(C_{2}H_{3}O_{2})_2} + Ag

Since there are 2 Pb on the right side of the equation, you would change the coefficient of Pb on the left side to 2:
2Pb + AgC_{2}H_{3}O_{2} = Pb(II){(C_{2}H_{3}O_{2})_2} + Ag

Since there are 2 Acetate on the right side of the equation, you would change the coefficient of Silver Acetate on the left side to 2:
2Pb + 2AgC_{2}H_{3}O_{2} = Pb(II){(C_{2}H_{3}O_{2})_2} + Ag

Now there are 2 Silver on the left side, so you change the coefficient of Silver on the right side to 2:
2Pb + 2AgC_{2}H_{3}O_{2} = Pb(II){(C_{2}H_{3}O_{2})_2} + 2Ag

That is your final equation

The coefficients are 2 + 2 = 1 + 2
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Answer:

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4 0
2 years ago
How long does it take for a 12.62g sample of ammonia to heat from 209K to 367K if heated at a constant rate of 6.0kj/min? The me
Georgia [21]
First, consider the steps to heat the sample from 209 K to 367K.

1) Heating in liquid state from 209 K to 239.82 K

2) Vaporaizing at 239.82 K

3) Heating in gaseous state from 239.82 K to 367 K.


Second, calculate the amount of heat required for each step.

1) Liquid heating

Ammonia = NH3 => molar mass = 14.0 g/mol + 3*1g/mol = 17g/mol

=> number of moles = 12.62 g / 17 g/mol = 0.742 mol

Heat1 = #moles * heat capacity * ΔT

Heat1 = 0.742 mol * 80.8 J/mol*K * (239.82K - 209K) = 1,847.77 J

2) Vaporization

Heat2 = # moles * H vap

Heat2 = 0.742 mol * 23.33 kJ/mol = 17.31 kJ = 17310 J

3) Vapor heating

Heat3 = #moles * heat capacity * ΔT

Heat3 = 0.742 mol * 35.06 J / (mol*K) * (367K - 239.82K) = 3,308.53 J

Third, add up the heats for every steps:

Total heat = 1,847.77 J + 17,310 J + 3,308.53 J = 22,466.3 J

Fourth, divide the total heat by the heat rate:

Time = 22,466.3 J / (6000.0 J/min) = 3.7 min

Answer: 3.7 min


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In the step gradient separation, four separate fractions were collected. How were these related to the polarities of the column
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hope this helps</span>
7 0
3 years ago
Read 2 more answers
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