During electrolysis of `H_2SO_4`(aq) with high charge density, `H_2S_2O_8` formed as by product. In such electrolysis 22.4L `H_2(g)` and 8.4 L `O_2(g)` liberated at 1 atm and 273 K at electrode. The moles of `H_2S_2O_8` formed is :
a) 0.25
b) 0.5
c) 0.75
d) 1
A
0.25
B
0.5
C
0.75
D
1
Text Solution
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The correct Answer is:
A
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Draw the structures of the following molecule : H_2S_2O_8
Find the number of molecules in 1.8 g of H_2O .
H_2 (g) and O_2 (g), can be produced by the electrolysis of water. What total volume (in L) of O_2 and H_2 are produced at 1 atm and 273K when a current of 30 A is passed through a K_2SO_4 (aq) solution for 193 min?
Decomposition of NH_(4)NO_(2)(aq into N_(2)(g) and 2H_(2)O(l) is first order reaction.
identify the type of reaction in the followng example 2H_2(g) + O_2(g) rarr 2H_2O(l)
In H_2S_2O_8 , the oxidation state of S is +6/+7.
Balance the following equations. (i) H_2S (g) + SO_2 (g) rarr S (s)+ H_2O (l)
19 gm of ice is converted into water at 0^(@)C and 1 atm. The entropies of H_(2)O(s) and H_(2)O(l) are 38.2 and 60 J//"mol " K respectively. The enthalpy change for this conversion is :
The strength of H_(2)O_(2) is expressed in several ways like molarity, normality,% (w/V), volume strength, etc. The strength of "10 V" means 1 volume of H_(2)O_(2) on decomposition gives 10 volumes of oxygen at 1 atm and 273 K or 1 litre of H_(2)O_(2) gives 10 litre of O_(2) at 1 atm and 273 K The decomposition of H_(2)O_(2) is shown as under : H_(2)O_(2)(aq) to H_(2)O(l)+(1)/(2)O_(2)(g) H_(2)O_(2) can acts as oxidising as well as reducing agent. What is the percentage strength (%w/V) of "11.2 V" H_(2)O_(2)
The strength of H_(2)O_(2) is expressed in several ways like molarity, normality,% (w/V), volume strength, etc. The strength of "10 V" means 1 volume of H_(2)O_(2) on decomposition gives 10 volumes of oxygen at 1 atm and 273 K or 1 litre of H_(2)O_(2) gives 10 litre of O_(2) at 1 atm and 273 K The decomposition of H_(2)O_(2) is shown as under : H_(2)O_(2)(aq) to H_(2)O(l)+(1)/(2)O_(2)(g) H_(2)O_(2) can acts as oxidising as well as reducing agent. As oxidizing agent H_(2)O_(2) is converted into H_(2)O and as reducing agent H_(2)O_(2) is converted into O_(2) . For both cases its n-factor is 2. :. "Normality " "of" H_(2)O_(2) " solution " =2xx " molarity of" H_(2)O_(2) solution What is the molarity of "11.2 V" H_(2)O_(2) ?
