A circuit consists of a series combination of 5.50 âkΩ and 5.00 âkΩ resistors connected across a 50.0-V battery having negligi
ble internal resistance. You want to measure the true potential difference (that is, the potential difference without the meter present) across the 5.00 âkΩ resistor using a voltmeter having an internal resistance of 10.0 kΩ.1. What potential difference does the voltmeter measure across the 5.00 âkΩ resistor?2. What is the true potential difference across this resistor when the meter is not present?3. By what percentage is the voltmeter reading in error from the true potential difference?
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Complete question:
Consider the hypothetical reaction 4A + 2B → C + 3D
Over an interval of 4.0 s the average rate of change of the concentration of B was measured to be -0.0760 M/s. What is the final concentration of A at the end of this same interval if its concentration was initially 1.600 M?
Answer:
the final concentration of A is 0.992 M.
Explanation:
Given;
time of reaction, t = 4.0 s
rate of change of the concentration of B = -0.0760 M/s
initial concentration of A = 1.600 M
⇒Determine the rate of change of the concentration of A.
From the given reaction: 4A + 2B → C + 3D
2 moles of B ---------------> 4 moles of A
-0.0760 M/s of B -----------> x
⇒Determine the change in concentration of A after 4s;
ΔA = -0.152 M/s x 4s
ΔA = -0.608 M
⇒ Determine the final concentration of A after 4s
A = A₀ + ΔA
A = 1.6 M + (-0.608 M)
A = 1.6 M - 0.608 M
A = 0.992 M
Therefore, the final concentration of A is 0.992 M.