Initial rate of formation

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#1
Could someone explain the reasoning as to why the initial rate is this pleaseeee?

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#2
(Original post by Yazomi)
Could someone explain the reasoning as to why the initial rate is this pleaseeee?

The answer is meant to be the one in light blue
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7 months ago
#3
Look at the units in the top row of the table - they're all different powers of 10. You need to account for these when you're doing your calculations.

Remember that e.g. 0.5*10^-7 = 5*10^-8 = 0.00000005

You can use the "x10^x" button on your calculator to do these calculations more easily. Try the question again accounting for the different units and let us know how you get on.

If you've never seen numbers like this before, a good term to google is "standard form".
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#4
(Original post by jm4761)
Look at the units in the top row of the table - they're all different powers of 10. You need to account for these when you're doing your calculations.

Remember that e.g. 0.5*10^-7 = 5*10^-8 = 0.00000005

You can use the "x10^x" button on your calculator to do these calculations more easily. Try the question again accounting for the different units and let us know how you get on.

If you've never seen numbers like this before, a good term to google is "standard form".
That makes sense but what’s the reason for why the initial rate for experiment 1 needs to be divided by 4? Thanks!
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7 months ago
#5
(Original post by Yazomi)
That makes sense but what’s the reason for why the initial rate for experiment 1 needs to be divided by 4? Thanks!
Look at the equation for methanal production in the question. You get 2 moles of methanal for every 0.5 moles of oxygen produced. Another way of writing this is you get 4 moles of methanal for every 1 mole of oxygen. In effect, you are getting 4 times as much methanal than oxygen, so the initial rate for methanal/4 will give you the initial rate for oxygen production.
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#6
(Original post by jm4761)
Look at the equation for methanal production in the question. You get 2 moles of methanal for every 0.5 moles of oxygen produced. Another way of writing this is you get 4 moles of methanal for every 1 mole of oxygen. In effect, you are getting 4 times as much methanal than oxygen, so the initial rate for methanal/4 will give you the initial rate for oxygen production.
Ahhhh I see thanks! I get it now
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