An incandescent bulb becomes hotter than a fluorescent bulb when turned on because in a regular incandescent bulb, there is tungsten wire where electricity is converts into heat. A regular incandescent light bulb requires 4 times more energy than a fluorescent bulb in order to produce the same amount of light. The conversion is such that for a 75-watt bulb, temperature get raised to approximately 2000 K. For such a high temperature, the radiating energy from the wire have some visible light. In such bulbs, 90% of the electricity get consumed in producing heat and only 10% produces light thus, they are not much efficient source of light.
On the other hand, fluorescent bulbs produce light with less amount of heat. In them, 40% of electricity is consumed in producing light and 60% in heat which is very less as compared to heat produced by a incandescent bulb. This is because when it get turned on, mercury atoms inside the bulb collides with electrons and produce UV light which is then converted into visible light using thin layer of phosphor power present inside the bulb. This produces low amount of heat thus, the bulb stays cooler, the bigger size of bulb also helps in dispersing heat.
Therefore, a fluorescent light bulb is not as hot as an incandescent light bulb.
Answer:
K I will attempt
Explanation:
a)

b)
1 : 2 : 2 (I don't know if this is what the question wants but it is what I would answer)
c)
Hydrogen because it requires 2 moles of H2 to react with 1 mole of O2
d)
24 moles of water. Look at stoichiometric coefficient. 2:2 means 24 moles you get 24 moles
e)
Oxygen. 2 < 5/2. Remember, 1 mole of O2 requires 2 moles of H2. But 5/2 is still greater than 2
f)
First, let's find out how many moles of water we can get. Since O2 is the limiting reactant, and O2:H2O ratio is 1:2, we will get 4 moles of H2O. Then, we can multiply 4 by Avogadro's number which is
to get the number of molecules. We get: 2.41 * 10^24 molecules of water.
Answer:
187.34 atm
Explanation:
From the question,
PV = nRT.................. Equation 1
Where P = Pressure, V = Volume, n = number of mole, R = molar gas constant, T = Temperature.
make P the subject of the equation
P = nRT/V.............. Equation 2
n = mass(m)/molar mass(m')
n = m/m'............... Equation 3
Substitute equation 3 into equation 2
P = (m/m')RT/V............ Equation 4
Given: m = 46 g, T = 25°C = (25+273) = 298 K, V = 3.00 L
Constant: m' = 2 g/mol, R = 0.082 atmL/K.mol
Substitute these values into equation 4
P = (46/2)(0.082×298)/3
P = (23×0.082×298)/3
P = 187.34 atm
Answer:

Explanation:
To convert from moles to grams, the molar mass must be used.
1. Find Molar Mass
The compound is iron (III) chloride: FeCl₃
First, find the molar masses of the individual elements in the compound: iron (Fe) and chlorine (Cl).
There are 3 atoms of chlorine, denoted by the subscript after Cl. Multiply the molar mass of chlorine by 3 and add iron's molar mass.
- FeCl₃: 3(35.45 g/mol)+(55.84 g/mol)=162.19 g/mol
This number tells us the grams of FeCl₃ in 1 mole.
2. Calculate Moles
Use the number as a ratio.

Multiply by the given number of grams, 345.0.

Flip the fraction so the grams of FeCl₃ will cancel.



Divide.

3. Round
The original measurement of grams, 345.0, has 4 significant figures. We must round our answer to 4 sig figs.
For the answer we calculated, that is the thousandth place.
The 1 in the ten thousandth place tells us to leave the 7 in the thousandth place.

There are about <u>2.127 mole</u>s of iron (III) chloride in 345.0 grams.