Question

It takes 3.2×103 J of heat to melt an amount of gold (ΔHf=6.30×104 Jkg/) . How much gold do you have?

5.1 g

0.051 g

51 g

19 g

Answer 1

i think its B or c ? im not sure

Answer 2

This is basically a problem that can be answered using dimensional analysis, but can you clarify something real quick? Is the thing in the parenthesis saying \( H°_f = 6.30×10^4 J/kg\)?

Answer 3

I'll just assume it says that it is

It says that 3.2x10^3 J is needed, and we have an H°f of that value given, so we can use dimensional analysis. There isn't any exact formula, but if you look at the units, you can notice that we can do something to cancel some of the terms.

\(\left(3.23\times10^{3}\ J\right)\ \times\frac{1\ kg}{6.30\times10^{4}\ J}\times\frac{1000\ g}{1\ kg} = \text{mass of gold}\) Calculating that, you would get 51.269 grams, and to the nearest significant figure, that's 51.3 g. The closest answer is C.

Answer 4

yes thats what it is saying in parenthesis

Answer 5

oh okay so its more so having a basic definition of whats given ?

Answer 6

to solve for it

Answer 7

There isn't really a basic definition for it, if you notice the units for the enthalpy of formation, \(H°_f\), it is J/kg, correct? And you are given the J to melt the gold. Since they are like units, you can take the reciprocal and find kg/J, and then multiple J * kg/J, to have kg in the end.

Now after you do that, you can convert to g. Basically:

\(\left(\frac{J}{kg}\right)^{-1}=\frac{kg}{J}\)
\(\frac{kg}{J}\times J=kg\)
\(kg\cdot1000=g\)

that's my logic

Answer 8

Basically, you can use the units to understand where you need to go, and then manipulate them around.

Answer 9

ohhh okay, yes that makes more sense. i always had a hard time manipulating equations. However, you explained it to where it makes sense. thank you again!

Answer 10

NP