Welcome to Westonci.ca, your go-to destination for finding answers to all your questions. Join our expert community today! Discover solutions to your questions from experienced professionals across multiple fields on our comprehensive Q&A platform. Get immediate and reliable solutions to your questions from a community of experienced professionals on our platform.
Sagot :
for resistance we have R=ρ l/a
thus for conductance we have K=σ a/l
conductance,K=1/R
conductivity,σ =1/ρ
σ = .80 Ω-1 cm-1
l =9 cm
a = 3 cm²
K=.80 ×3/9
=0.26 Ω-1
thus for conductance we have K=σ a/l
conductance,K=1/R
conductivity,σ =1/ρ
σ = .80 Ω-1 cm-1
l =9 cm
a = 3 cm²
K=.80 ×3/9
=0.26 Ω-1
Answer: [tex]0.266 ohm^{-1}[/tex]
Explanation: Specific conductance or conductivity is the conductance for unit volume of a solution.
[tex]\kappa=C\times \frac{l}{a}[/tex]
where,
[tex]\kappa[/tex] =Specific conductance or Conductivity=[tex]0.80ohm^{-1}cm^{-1}[/tex]
C = Conductance = ?
l = length= 9.0 cm
a = area = 3.0[tex]cm^2[/tex]
[tex]\frac{length}{area}[/tex]= cell constant =[tex]\frac{9.0cm}{3.0cm^2}=3cm^{-1}[/tex]
[tex]0.80ohm^{-1}cm^{-1}=C\times 3cm^{-1}[/tex]
[tex]C=0.266 ohm^{-1}[/tex]
Thank you for your visit. We are dedicated to helping you find the information you need, whenever you need it. We appreciate your visit. Our platform is always here to offer accurate and reliable answers. Return anytime. Thank you for choosing Westonci.ca as your information source. We look forward to your next visit.
