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I am kind of having trouble with these problems...I get to a certain point then I'm stuck.

1. When 4.21 grams of hydroxide are added to 250 mL of water, the temperature rises by 4.14 °C. Assume that the density and specific heat of the dilute aqueous solutions are the same as those of H2O and calculate the molar heat of solution of potassium hydroxide.

**Would this eq be right?
q=(254.21g)(4.184J/gK)(4.14K)

2.Calculate the standard heat of formation of copper(I) oxide using the following data:
CuO → Cu + ½ O2 ΔH° = 157.3 kJ/mol
4 CuO → 2 Cu2 O + O2 ΔHo = 292.0 kJ/mol

2007-03-14 21:52:54 · 2 answers · asked by Jay 2 in Science & Mathematics Chemistry

2 answers

1.
4.21 g are added to 250 ml water. To get the specific heat, you need to calculate it for a kg of water, so multiply 4.21 by 4 and replace the first factor. Other than that, the equation looks ok for me.

2.
ΔHo in kJ/mol seems to apply to one mole of the compound left of the arrow. Since you need 2 CuO to form 1 Cu2O, I'd multiply the both equations by 2 and then calculate the difference.

2007-03-14 22:31:51 · answer #1 · answered by jorganos 6 · 0 0

a million)undecided yet based on the formula of q=mct, it sounds almost surprising. 2)i) opposite first eq and multiply via 2 ii) Divide 2nd eq via a million/2 ( carry out a similar for the warmth values and upload them as much as get the fee of the time-honored warmth of formation for Cu(a million)oxide) usual eq: a million/2O2 +2Cu -> Cu2O be conscious that the customary eq complies to its definition

2016-10-18 10:34:25 · answer #2 · answered by Anonymous · 0 0

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