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Calculations - HSN forum

# Calculations

### #1x-amz-x

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Posted 30 June 2009 - 07:00 PM

when 100cm3 of dilute nitric acid, concentration 0.1 mol l-1, is neutralised by sodium hydroxide solution, 570 joules of heat is liberated
calculate the enthalpy of neutralisation of nitric acid

i know it's a simple question but i struggle with calculations !

xx

### #2ginneswatson

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Posted 24 May 2010 - 09:34 PM

x-amz-x, on 30 June 2009 - 07:00 PM, said:

when 100cm<sup>3</sup> of dilute nitric acid, concentration 0.1 mol l<sup>-1</sup>, is neutralised by sodium hydroxide solution, 570 joules of heat is liberated
calculate the enthalpy of neutralisation of nitric acid

i know it's a simple question but i struggle with calculations ! <img src="http://www.hsn.uk.net/forum/public/style_emoticons/<#EMO_DIR#>/sad.gif" style="vertical-align:middle" emoid="" border="0" alt="sad.gif" />

xx

"Enthalpy of Neutralisation" means energy released from one mole of water produced. 100cm3 of dilute nitric acid, concentration 0.1 mol l-1 is 0.1 x 0.1 = 0.01 mol of nitric acid. 1 mole of nitric acid would react to produce 1 mole of water.

In other words the 570 joules represents the formation of 0.01 mol of water so multiply by 100 to solve for one mole. 57000 J or 57 kJ.

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