Lnettie21 Posted March 29, 2004 Share Posted March 29, 2004 Can you help me with these two questions please If 4.436 g of NaOH was added to 150.0 mL of water and we assume the density of water to be 1.000 g/mL Then what is the total mass? Given that the temperature change observed was 11.5 C when4.436 g of NaOH were added to 150.o mL of water, What is the heat energy in kJ that is produced. (Don't forget the specivic heat of this solution is 4.06j/g C and assume the density of water to be 1.000 g/mL) Thanks Link to comment Share on other sites More sharing options...
blike Posted March 30, 2004 Share Posted March 30, 2004 For the first question, you are given 4.436g as your NaOH mass. You are also given the density and volume of water it has been added to. You need to convert your water's volume to mass: 150.0mL H2O * (1.000 g/mL) = 150.0 g H2O. Now simply add the two masses: 150g + 4.436g = 154.436g You try the second part on your own Link to comment Share on other sites More sharing options...
wolfson Posted March 30, 2004 Share Posted March 30, 2004 Looks good to me blike Link to comment Share on other sites More sharing options...
chemistry Posted March 30, 2004 Share Posted March 30, 2004 You have the total mass of the solution. You also have the specific heat...use q=ms(dT) Link to comment Share on other sites More sharing options...
wolfson Posted March 30, 2004 Share Posted March 30, 2004 q=ms/_\T = m s (tf – ti) tf = final temperature of substance ti = initial temperature of substance where m = mass and s = J/g per degree c Chemistry just thought id tell me/her what the symbols are. Link to comment Share on other sites More sharing options...
Lnettie21 Posted March 30, 2004 Author Share Posted March 30, 2004 Thanks Everyone. For the second part, I got 154.436*4.06*11.5=7,210.617. Is this correct? This number is kind of high Link to comment Share on other sites More sharing options...
wolfson Posted March 31, 2004 Share Posted March 31, 2004 7.2 KJ seems fine Link to comment Share on other sites More sharing options...
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