20 ohms in parallel with 16 ohm= 8.89
20x16/20+16. Product over sum
A diagram of a closed circuit with power source on the left labeled 12 V, a resistor on the top labeled 10 Ohms, a resistor on t
he right labeled 8.0 Ohms and a resistor on the bottom labeled 6.0 Ohms.
Use the circuit diagram to determine the voltage drop across the 8 Ω resistor.
Answer is 4
Use the circuit diagram to determine the voltage drop across the 8 Ω resistor.
Answer is 4
2 answers:
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20 ohms in parallel with 16 ohm= 8.89
20x16/20+16. Product over sum
Answer:
so rate constant is 4.00 x 10^-4
Explanation:
Given data
first-order reactions
85% of a sample
changes to propene t = 79.0 min
to find out
rate constant
solution
we know that
first order reaction are
ln [A]/[A]0 = -kt
here [A]0 = 1 and (85%) = 0.85 has change to propene
so that [A] = 1 - 0.85 = 0.15.
that why
[A] / [A]0= 0.15 / 1
[A] / [A]0 = 0.15
here t = (79) × (60s/min) = 4740 s
so
k = - {ln[A]/[A]0} / t
k = -ln 0.15 / 4740
k = 4.00 x 10^-4
so rate constant is 4.00 x 10^-4