Question

In a nuclear reaction, the energy released is equal to 8.1 x 1016 joules. Calculate the mass lost in this reaction. (1 J = 1 kg m2/s2)
Answer

7.2 x 1032 kg

2.7 x 108 kg

9.0 x 10-1 kg

7.2 x 10-16 kg

Answer 1

How do I calculate mass lost?

Answer 2

you have E= 5.4 x 10^(15) given, and speed of light 'c'= 3 x 10^8 m/sec.

So, just plug these in "E=(deltaM)C^2, and calculate 'deltaM', which is the mass lost.

P.S. - delta just means 'change' so 'deltaM' means the change in mass nothing else.

Answer 3

Just look at this http://openstudy.com/study#/updates/4fc72c0ce4b022f1e12eac37

Answer 4

prob. B, yes?

Answer 5

please help

Answer 6

I don't understand how to input delta in an equation...how do I do that? how??

Answer 7

because you didn't read anything about mass defect

Answer 8

gaahhhhhh......I don't understand their one paragraph description of the concept.

Answer 9

http://finedrafts.com/files/CUNY/chemistry/General/Brown-Holme/Chemistry_Engineering_Students_2nd_txtbk.pdf
chapter 14

Answer 10

alright, it's the mass lost during a reaction that converts to energy

Answer 11

I think it's B

Answer 12

do the same thing as your last problem

Answer 13

yeh binding energy

\[E = \Delta mc^2 \]
we already know the amount of energy released and we also know the speed of light
E = 8.1 x 1016 joules
\[E = \Delta M c^2 = \Delta M (2.9987 \times 10^2 mc^{-1})^2\]

Answer 14

I changed the mass into capital M since I am including units of meter/sec

Answer 15

ok...

Answer 16

\[8.1 \times 10^{16} kgm^2s^{-2}= \Delta M (2.9987 \times 10ms^{-1})^2\]

Answer 17

see how easy it is to solve for the mass defect to be calculated?

Answer 18

hahaha....yeah, it's a freakin cake walk

(puts head between knees)

Answer 19

I keep messing up the exponents, just fix them for me

Answer 20

I still have to get used to latex

Answer 21

make sure you have the right data on energy released

Answer 22

8.1 x 1016 joules is not correct

Answer 23

\[8.1 \times 10^{16}kgm^2s^{-2} = \Delta M (2.9979 \times 10^{8} ms^{-1})^2\]

Answer 24

since \[E = \Delta mc^2\] then \[\Delta m = \frac{E}{c^2}\]and \[\Delta m = \frac{8.1*10^{16}J}{(3.0*10^8\frac{m}{s})^2}\] so \[\Delta m = 0.9kg\]

Answer 25

I don't...I don't....gah...
how do I solve delta?
what do I do there?
It's not an imput on the calculator, and I'm not aware of any value for it..
oh wait, JF just replied..

Answer 26

the units does not match @JFraser you need to fix it first

Answer 27

8.1 x 10^16

Answer 28

the units do match

Answer 29

A

Answer 30

if you fix it in the end
it will match, but he needs to see it

Answer 31

ooohh....intellect battle..

(chews nails excitedly)

Answer 32

i just did it step by step so he can see it

Answer 33

okay

Answer 34

the delta just means "change" it is just a symbol

Answer 35

so...ignore it?

Answer 36

no, don't ignore it, understand it

Answer 37

wel you can't really ignore it because it tells you a great deal

Answer 38

I think you're too smart for the calculator - WolframAlpha can't understand you, lol.

Answer 39

you're not solving for the mass of either the reactants or the products, you're solving for the difference between the two. that's why the delta is important

Answer 40

it tells you that there is a change

Answer 41

A, I think

Answer 42

no wonder you're not understanding the concepts, you're relying on wolframalpha

Answer 43

no, no...maybe

Answer 44

not entirely tho, it helps with longer equations where I'm more liable to make mistakes

Answer 45

read the chapter 14 on that whole chemistry book link I gave you
it is an intro to nuclear chemistry

Answer 46

what do the ms and s stand for?

Answer 47

I just don't understand how are you taking chemistry at this point without knowing what deltas are

Answer 48

m = meters
s = seconds

Answer 49

ok....I think we're over-complicating this - I don't think the equation they want me to use is this complicated...hold on

Answer 50

@JFraser is using a fractional format, I am using a one line
so you will notice negative exponents in my equations

Answer 51

THIS IS NOT COMPLICATED!
this is Algebra 1

Answer 52

BWHAHAHAHAAHA!! I GET Algebra...this is just mixing math with little symbols and numbers that you have to constantly flick back and fro to the textbook to understand.

Answer 53

no...it really wasn't that complicated...I still don't know how to solve this freakin thing, but it most certainly is not the way he was describing it. Overc-omplication - can't judge tho - do it myself.

Answer 54

*over-complication

Answer 55

is it the importance of the delta that's confusing you, or the algebraic rearrangements of the pieces?

Answer 56

I don't understand how to calculate m and s.
It's fine to have variable in an equation, but I need to know how to find them
And they didn't give an equation teaching me how to calculate m and s from the amount of energy released.

Answer 57

I finally sent an email to my teacher about it, hopefully she'll be able to help.

Answer 58

The equation they give me to use is... (1 J = 1 kg m2/s2)

Answer 59

I know J...but not m or s
hence my confusion

Answer 60

the equivalent unit \[1.0 J = 1.0 \frac{ kgm^2 }{ s^2 }\]
(kilograms*meters^2) per seconds^2
this is a unit most students learn from physics

Answer 61

I honesty think that this level of chemistry is not for you
you still need to get your elementary algebra situated

Answer 62

the "m" and "s" inside the equation you gave aren't values to find, they are part of the unit itself. the unit of energy (J) is made up of smaller units (kg, m, and s) in a particular combination that looks like what you posted above. You really need help recognizing units, too