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Dennis_Churaev [7]
3 years ago
7

5. Calculate the time required to deposit 0.37g of Silver from a silverNitrate solution using a current of 2.5A( Charge of Ag in

AgNO3 is + 1)​
Chemistry
1 answer:
Monica [59]3 years ago
8 0

Time required : 132.2 s

<h3>Further explanation</h3>

Given

mass of Silver = 0.37 g

current = 2.5 A

Required

time do deposit

Solution

Faraday's Law :

\tt W=\dfrac{e.i.t}{96500}

e = ar/valence (valence Ag = 1)

i = current

t = time = s

\tt 0.37~g=\dfrac{108/1\times 2.5~A\times t }{96500}\\\\t=132.2~s

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Nitric acid is produced commercially by the Ostwald process, represented by the following equations. 4 NH3(g) + 5 O2(g) 4 NO(g)
Marrrta [24]

Answer:

The answer to your question is: 1538095.2 kg of NH3

Explanation:

MW HNO3 = 63 kg

MW NO2 = 46 kg

                         3 NO2(g) + H2O(l)--- 2 HNO3(aq) + NO(g)

                            3(46) kg--------------   2(63) kg  

                                  x     ---------------  7600000 kg

           x = 7600000 x 138/126 = 8323809.5 kg og NO2

MW NO = 30            

                         2 NO(g) + O2(g)---2 NO2(g)

                       2(30) ------------------2(46)

                        x        ---------------- 8323809.5 kg  

             x = 8323809.5 x 60/92 = 5428571.4 kg of NO

MW NH3 = 17 kg

                      4 NH3(g) + 5 O2(g) 4 NO(g) + 6 H2O(g)

                       4(17) -------------------- 4(30)

                         x ----------------------- 5428571.4

x = 5428571.4 x 34 / 120

x = 1538095.2 kg of NH3

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