what is .......................
Answers
Answer:
When a solution of potassium iodide is added to a solution of lead nitrate taken in a test tube, the precipitation of a yellowish solid is observed. This yellowish solid is lead iodide. Potassium nitrate is formed along with lead iodide. This is a double displacement reaction.
Lead nitrate solution contains particles (called ions) of lead, potassium iodide contains particles (called ions) of iodide. When the solutions mix, the lead particles and iodide particles combine to form a new substance, lead iodide, which is a yellow solid.
When Silver nitrate ) is added to the potassium iodide (KI), we get a negative charged solution because iodide gets adsorbed from the potassium iodide solution and two new products are formed. Where is a colourless solid and KI is a yellow they are combined to produce the white solid.
Potassium iodide and lead(II) nitrate are combined and undergo a double replacement reaction. Potassium iodide reacts with lead(II) nitrate and produces lead(II) iodide and potassium nitrate. Potassium nitrate is water soluble.
Answer:
What is the amount of electrical energy consumed by the bulb.
Formula Used :-
Physics
10 points
\boxed{\bold{\small{Electrical\: energy\: =\: P\: \times t}}}
where,
Power
=
P
•
Time
=
t
•
Solution :-
Given :
Power = 100 W
•
Time = 30 d × 8 h = 240 h
•
According to the question by using the formula we get,
↦ Electrical energy = 100 × 240
↦ Electrical energy = 24000
We know that,
1 kWh = 1000 Watt-Hour
\dfrac{24000}{1000}
=
0
0
0
4
2
,
Then
\dfrac{\cancel{24000}}{\cancel{1000}}
↦ Electrical energy =
➠ Electrical energy = 24 Watt-Hour
energy
electrical
of
amount
The
\therefore
.
Watt-Hour
4
2
is
bulb
the
by
consumed