Answer:
1200 Sm^2mol^-1
Explanation:
Given data :
conductivity of water ( kwater ) = 76 mS m^-1 = 0.076 Sm^-1
conductivity of kcl (aq)( Kkcl ) = 1.1639 Sm^-1
Kkcl = 1.1639 - 0.076 = 1.0879 Sm^-1
Resistance = 33.21 Ω
where conductivity can be expressed as = 
hence cell constant = conductivity * Resistance
= 1.0879 * 33.21 = 36.13m^-1
conductivity of CH3COOH ( kCH3COOH ) = 36.13 / 300
= 0.120 Sm^-1
<u>Determine the molar conductivity of acetic acid</u>
= ( kCH3COOH * 1000 ) / C
C = 0.1 mol dm
= (0.120 * 1000) / 0.1 = 1200 Sm^2mol^-1
Answer:
The electric field value is 240 N/C
Explanation:
Given that,
Distance = 5.0 mm
Potential difference = 1.2 V
We need to calculate the electric field value
Using formula of potential difference
Where, E = electric field
V = potential difference
d = distance
Put the value into the formula
Hence, The electric field value is 240 N/C
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Hope that was helpful.
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