Question #0e523

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Answer 1 · 2015-12-02T23:12:28

$9.56 g$ $9.56"g"$

You have quoted molar heat capacities from your data book but you have used $m$ $m$ for mass in your expression for $Q$ $Q$ .

So we must work in moles and then convert back to grams using the idea that no. moles = mass/mass of 1 mole

Or in symbols:

$n = \frac{m}{M_{r}}$ $n=m/M_r$

So:

$Q=nxxC_vxxDeltaT$

Note that $n$ is the no. moles.

Applying this first to helium:

The $M_r$ of helium is 4. (Same as the $A_r$ ).

So from our expression for $n$ we can write:

$Q=(2)/(4)xx12.5xxDeltaT " "color(red)((1))$

Now for oxygen:

The $M_r$ of $O_2$ is 32 so:

$Q=(m)/(32)xx20.9xxDeltaT " "color(red)((2))$

Now we can put $color(red)((1))$ equal to $color(red)((2))rArr$

$2/4xx12.5xxcancel(DeltaT)=m/32xx20.9xxcancel(DeltaT)$

$:.m=(0.5xx12.5xx32)/(20.9)=9.56"g"$

(I have used approximate $M_r$ values. You should use the ones you are given)