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Answer : The voltage applied by the batteries is, 6.0 V
Solution : Given,
Resistance of flashlight = 2.4 ohm
Current in the circuit = 2.5 Ampere
Formula used :
where,
V = applied voltage
I = current in the circuit
R = resistance of light
Now put all the given values in the above formula, we get
Therefore, the voltage applied by the batteries is, 6.0 V
5.732 grams of AgCl is formed when 0.200 L of 0.200 M AGNO3 reacts with an excess of CaCl2.
Explanation:
The balanced equation:
2 AgNO3(aq) + CaCl2(aq) -----> 2 AgCl(s) + Ca(NO3)2(aq)
data given:
volume of AgNO3 = 0.2 L
molarity of AgNO3 = 0.200 M
atomic weight of AgCl= 143.32 gram/mole
from the formula, number of moles can be calculated
Molarity =
number of moles of AgNO3 = 0.04
From the reaction:
2 moles of AgNO3 reacts to form 2 moles of AgCl
0.04 moles of AgNO3 reacts to form x mole of AgCl
=
= 0.04 moles of AgCl is formed
mass of AgCl formed is calculated by multiplying number of moles with atomic mass of AgCl
mass of AgCl = 0.04 x 143.32
= 5.732 grams of AgCl is formed.