Hmb28
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I don't really understand this question
What volume of the following solution will be oxidised by 25.0cm3 of 0.0200mol dm-3 acidified potassium manganate (vii) solution ? A) 0.0200moldm-3 tin(ii) nitrate
So far I did moles of MnO4-2= 0.02*25/1000=5x10-4
So 5x10-4 / 0.02= 0.025dm3 is this right ?
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metaljoe
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Yes the moles are correct. Now I believe you need to work out the equation regarding tin and acidified potassium manganate (vii). Use this to calculate the mole ratio and use this amount of moles for tin along with its concentration - 0.02moldm^-3 in orer to work out the volume. Now you have the answer!!
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Hmb28
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(Original post by metaljoe)
Yes the moles are correct. Now I believe you need to work out the equation regarding tin and acidified potassium manganate (vii). Use this to calculate the mole ratio and use this amount of moles for tin along with its concentration - 0.02moldm^-3 in orer to work out the volume. Now you have the answer!!
Thanks !
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metaljoe
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Not a problem my friend, if you need any help with other questions just give me a bell.
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