E.m.f of the following cell at 298K in V is x×10-2.
Zn|Zn2+(0.1M)||Ag+(0.01M)|Ag
The value of x is _________. (Rounded off to the nearest integer)
Given:EZn2+/Zn°=-0.76;EAg+/Ag°=+0.80V;2.303RTF=0.059
Step1:CalculationofE°cellE°cell=E°Ag/Ag+-E°Zn2+/Zn=0.80-(-0.76)=1.56VStep2:CalculationofEmfofcelli.e.EcellUsingNernstEquation,Ecell=E°cell-0.0592log[Zn2+][Ag+]2=1.56-0.0592log[0.1][0.01]2{Substitutingthegivenvalues}=1.56-0.0592log3=1.56-0.0885=1.4715=147.15×10-2≃147×10-2=x×10-2Hence,thevalueofxis147.15V≃147V.
Study the following pattern and write the next steps for given pattern:1×9+2=1112×9+3=111123×9+4=11111234×9+5=11111___+____=__________+____=_______
To draw a histogram to represent the following frequency distribution:
Class interval
Frequency
5-1010-1515-2525-4545-75
61210815
The adjusted frequency for the class 25-45 is
Study the pattern and answer the following:
15×15=22525×25=62535×35=122545×45=2025
55×55=....2565×65=....2575×75=....2585×85=....2595×95=....25105×105=....25
Study the following pattern and write the next two steps for given pattern:43×9=43(10-1)=430-4343×99=43×(100-1)=4300-4343×999=43×(1000-1)=43000-43___×____=__×____=_____-_________×____=__×____=_____-______
Multiply
42×9______________________