Δmc
2
For one reaction:
Mass Defect =Δm
=2(m
H
)−m
He
−m
n
=2(2.015)−3.017−1.009
=0.004 amu
1 amu=931.5 MeV/c
2
Hence,
E=0.004×931.5 MeV=3.724 MeV
E=3.726×1.6×10
−13
J=5.96×10
−13
J
For 1 kg of Deuterium available,
moles=
2g
1000g
=500
N=500N
A
=3.01×10
26
Energy released =
2
N
×5.95×10
−13
J
=8.95×10
13
One example of matter could be <em>Light.</em>
The properties of nonmetals- nonductile(cannot be turned into wire), dull color, high solubility, poor conductors, brittle(difficult to break) and have a wide range of colors. Compared to the properties of a metal- polished color, conducters of heat and electricity, cannot be dissolved, malleable(can bend). Would you like me to use certain elements to show this?
There's a lot of things built from petroleum, such as toothpaste.
Here's also another list.Gasoline 46%
Heating Oil / Diesel Fuel 20%
Jet Fuel ( kerosene) 8%
Propane / Propylene 7%
NGL / LRG 6%
Still Gas 4%
Petrochemical Feedstocks 2%
Petroleum Coke 2%
Residual / Heavy Fuel Oil 2%
Asphalt / Road Oil 2%
Lubricants 1%
Miscellaneous Products / Special Naphthas 0.4%
Other Liquids 1%
Aviation Gasoline 0.1%
Waxes 0.04%
Kerosene 0.02%
<span>H2O2
First, let's determine how many moles of hydrogen and oxygen atoms we have. Start by looking up the atomic weights of those elements:
Atomic weight hydrogen = 1.00794
Atomic weight oxygen = 15.999
Moles hydrogen = 1.33 g / 1.00794 g/mol = 1.319522987 mol
Moles oxygen = 21.3 g / 15.999 g/mol = 1.331333208 mol
We now have a ratio of 1.319522987 : 1.331333208 and we want a ratio of small integers that is close. Start by dividing all the numbers in the ratio by the smallest value, giving:
1 : 1.008950371
This ratio is acceptably close to 1:1 so I assume the formula is of the form HnOn where n is a small integer. Let's initially assume that n is 1, so the mass would be
1.00794 + 15.999 = 17.00694
Obviously 17 is far smaller than 34.1. So let's divide 34.1 by 17.00694 and see what n should be:
34.1 / 17.00694 = 2.005063815
So the formula we want is H2O2, which is hydrogen peroxide.</span>