If you wish to prepare a 1.13 M solution of NaNo3, to what volume (in liters) would you have to dilute 25.4 mL of 3.21 M NaNO3?
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The empirical formula is obtained by calculating the mole ratios of the atoms in the elements.
The number of moles =mass/ R.A.M
For hydrogen, no. of moles=8/1=8
For oxygen, no. Of moles=64/16=4
The tabular solution is attached.
The number of moles =mass/ R.A.M
For hydrogen, no. of moles=8/1=8
For oxygen, no. Of moles=64/16=4
The tabular solution is attached.