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If your hot coffee loses 50 kJ of energy as it cools and the volume shrinks because of thermal contraction, how large is the change in the internal energy of the coffee? How large is the change in the enthalpy? Explain. a) +50 kJ b) -50 kJ c) >50 kJ d)>-50kJ e)<50 kJ f)<-50 kJ (I really don't know where to get started on this problem. If someone could give me a thorough explanation, I'd greatly appreciate it!) Thanks in advance!

What happens when we react [MATH]\text{H}_2[/MATH] with [MATH]\text{O}_2[/MATH]? What is the product, and why? Thanks in advance! (Sorry if this isn't in the right forum. I'm not sure if this is classified as HW Help or General Chemistry.)

Could someone please comment on the following solution: [MATH]126 \ \text{g B}_5\text{H}_9 \cdot \frac{1 \ \text{mol} \ \text{B}_5\text{H}_9}{63.14 \ \text{g B}_5\text{H}_9}=2.00 \ \text{mol B}_5\text{H}_9[/MATH] [MATH]192\ \text{g O}_2 \cdot \frac{1 \ \text{mol} \ \text{O}_2}{31.99 \ \text{g O}_2}=6.00 \ \text{mol O}_2 \Rightarrow \times \frac{2 \ \text{mol B}_5\text{H}_9}{12 \ \text{mol O}_2}=0.5 \ \text{mol B}_5\text{H}_9[/MATH] Thus, [MATH]\text{O}_2[/MATH] is the limiting reactant. [MATH]192 \ \text{g O}_2 \cdot \frac{1 \ \text{mol O}_2}{31.99 \ \text{g O}_2} \cdot \frac{9 \ \text{mol H}_2\text{O}}{1 \ \text{mol O}_2} \cdot \frac{18.01 \ \text{g H}_2\text{O}}{1 \ \text{mol H}_2\text{O}}=973 \ \text{g H}_2\text{O}[/MATH]

Could someone please show me how to do this problem? Thanks in advance!

How would I do the following question? Thanks!

Could someone please help me with the following problem? Thanks in advance!