2. A 6.0 ampere light bulb operates at 120 V. What is the resistance of the bulb?
1 answer:
You might be interested in
The density of the sample : 0.827 g/L
<h3>Further explanation</h3>In general, the gas equation can be written
where
P = pressure, atm , N/m²
V = volume, liter
n = number of moles
R = gas constant = 0.082 l.atm / mol K (P= atm, v= liter),or 8,314 J/mol K (P=Pa or N/m2, v= m³)
T = temperature, Kelvin
n= 1 mol
MW Neon = 20,1797 g/mol
mass of Neon :
The density of the sample :
or We can use the ideal gas formula ta find density :
Answer:
a.36 g of water is produced.
b.64 g of is consumed.
Explanation:
The reaction is ⇒
a.
Given,
Weight of reacted = 4g
Weight of 1 mole of = 2
1 = 2g
Therefore no. of moles of reacted =
= 2 moles;
Also given,
Weight of reacted = 32 g
Weight of 1 mole of = 2
16 = 32 g
Therefore no. of moles of reacted =
= 1
We know that 2 moles of Hydrogen reacts with 1 mole of Oxygen to give 2 moles of water,
As we took 2 moles of Hydrogen and 1 mole of Oxygen,
Directly,from the equation we can tell 2moles of water will be produced.
Therefore no. of moles of produced = 2
Weight of 1 mole of water = = 18
Therefore weight of produced =
= 36gm
b.
Given ,
72 g of is produced.
So,
no. of moles of produced =
= 4 moles
From equation For every 2 moles of water formed , 1 mole of oxygen must be required.
So for producing 4 moles of water,
No. of moles of Oxygen required = 2 moles.
Therefore weight of reacted =
= 64 g
Method 2:
Given,
8 g of has reacted.
So,
no. of moles of reacted =
= 4 moles.
From equation , we know that For every 2 moles of reacted,1 mole of
will react.
Therefore,
No. of moles of that reacts with 4 moles of
=
= 2 moles
Therefore the weight of reacted =
= 64 g