How Does Increasing NaOH Mass Affect Heat Evolution in a Lab Experiment?

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The discussion revolves around a lab write-up involving the dissolution of sodium hydroxide (NaOH) in water and calculating the heat evolved. The user is struggling with a specific question regarding the heat change when using 8 grams of NaOH instead of 2 grams. It is clarified that the heat evolved is proportional to the number of moles of NaOH used. With a molecular weight of 40 grams/mol, 2 grams of NaOH corresponds to 1/20 moles, yielding a certain amount of heat (Q1). In contrast, 8 grams equates to 0.2 moles, resulting in a greater heat release (Q2). The user is seeking assistance to understand the relationship between the amount of NaOH used and the resulting heat change, emphasizing the need for clarity on how the increased amount affects the caloric output of the reaction.
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I have been spending nearly 3 hours on this lab write up that is due tomorrow. since i had missed a couple days of school, there are somethings that i was not introduced to therefore this is kinda hard for me. I managed to figure out most part but this is what I'm stuck on right now.

Part of the experiment requires us to mix 2g of NaOH with water and calculate the amount of heat evolved. This question asked: suppose to had used 8g of NaOH instead, what would have been the number of calories evolved in the experiment and what effect would this have on change if Heat? I keep thinking its four times more but that's going to only apply to the temperature. So is there anyone on this late at night taht can help me?
 
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Sodium hydroxide gives x calories per mole upon dissolution; so 2 grams of NaOH (m.w.=40 grams/mol) will give Q1=x/20 calories, since this is 1/20 moles. If you use 8 grams of NaOH, you'll get Q2=x/5 calories. This is because 8 grams of NaOH is 0.2 moles.
 
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