<p>
</p>
<p>There are about 0.1 mols NaOH, 0.1 mols 5.61 KOH, and 0.01 mols RbOH. Since these are all strong bases, we assume they completely dissociate, giving us 0.21 mols of OH-. Since molality is (mols solute/kg of solvent) and we know there are 90.0g of water, it’s just 0.21/0.090, which gives you 2.33 m.</p>
<p>Let me walk you through</p>
<p>Convert all of these hydroxide compounds to moles. For example, for NaOH, do 4.00g X (1 mol/40 g)= 0.1 moles NaOH</p>
<p>Do the same for the other two hydroxide compounds and get 0.1 moles KOH and 0.01 moles RbOH.</p>
<p>Since there is only one hydroxide molecule per each compound and they are all strong bases (i.e. dissociate completely), you don’t have to worry about doing more stoichiometry. For example, 2 moles of Ba(OH)2 would result in 4 moles of OH-.</p>
<p>Therefore, there is a 1:1 stoichoimetric relationship between moles of compound and moles of OH-.</p>
<p>So now add them up to get 0.21 moles OH-. Now, molality is moles of solute/Kg of solvent. We have 90.0 grams of water and therefore 0.09 Kg of water.</p>
<p>Now do 0.21 moles/0.09 Kg to get molality. The answer is A. 2.33 m. You should be able to do this without a calculator on the AP chem exam FYI.</p>
<p>Thanks guys. Somehow I got 0.10 moles OH- for all three.</p>
<p>Rank the following species in order of increasing H-N-H bond angles:</p>
<p>I. NH2-
II. NH3
III. NH4+</p>
<p>A. I<II<III
B. I<III<II
C. II<III<I
D. III<II<I
E. III<I<II</p>
<p>Why is the answer A and not B? I thought I should have the largest bond angle since it starts out with 180 degrees goes down because of the 2 lone pairs. And I thought III starts out with 109.5 degrees and goes down because of 1 lone pair, so I thought it has the lowest bond angle.</p>
<p>It is asking for bond angles between the NH bonds. NH2- has a bent shape with 2 lone pairs, causing the H’s to bunch up and have a low angle between each other.</p>
<p>NH3 has a single lone pair, so it is trigonal pyramidal which bunches up the H’s. </p>
<p>NH4 has no lone pairs and is tetrahedral so the H’s are evenly spaced.</p>
<p>Remember that the high concentration of negative charge around lone pairs causes them to repel other bonded pairs. I’m not sure if this makes much sense as I am having doubts myself as to WHY lone pairs repel bonded pairs more than lone pairs.</p>
<p>NH2(-) doesn’t start out with 180, it’s got tetrahedral electron geometry(two atoms, two lone pairs) so it “starts out” with 109.5, which is decreased by the two lone pairs. Its VSEPR geometry is bent (or angled, whatever you want to call it).</p>
<p>NH3 has a similar effect, but it’s only got one lone pair, so its angle is slightly larger than NH2(-). Its VSEPR geometry is trigonal pyramidal.</p>
<p>NH4+ is tetrahedral with NO lone pairs (hence the positive charge), so it will obviously have the largest angle of 109.5.</p>
<p>edit: afruff got to it first.</p>
<ol>
<li>Two solid objects, X and Y, of equal masses are placed in a molten lead bath and allowed to reach the bath temperature. (The objects themselves do not melt). Each object is placed in a separate fish tank containing 1000 kg of water. Object X raises the water temperature by 20degsC while object Y raises the water temperature 80degsC. Which of the following is true?</li>
</ol>
<p>My Question: Why does Y have a larger specific heat than X? I thought it was the other way around.</p>
<ol>
<li>Calculate the molecular weight of a small protein (a non-electrolyte) if a 200 mg sample dissolved in 100 mL of water has an osmotic pressure of 9.8 mmHg at 25degsC?</li>
</ol>
<p>Note: I have my second chemistry exam tomorrow morning. I would like clear explanations for these (especially for the first question since I spent a lot of time thinking about it but can’t quite get it), so I feel confident tomorrow. Thanks.</p>
<p>I can’t answer your first question because I cannot, for the life of me, remember chemical thermodynamics very well.</p>
<p>The second one though I can help with.
Osmotic pressure=iMRT. Since the protein is a non-electrolyte and doesn’t dissociate, i=1. So we have 9.8mmHg=(1)M(0.0821 L<em>atm/mol</em>K)(298K).</p>
<p>You need to convert the osmotic pressure to atm. 1 atm=760 mmHg… you can do the conversion. So after converting to atm, we have 0.0129 atm=M(0.0821)(298).</p>
<p>Solving for M (the molarity), we get M=5.273<em>10^-4. Remembering that molarity is mols of solute/liters of solution, we can write 5.273</em>10^-4=x/0.100L, where “x” is the number of mols.</p>
<p>Solving for “x”, we get 5.273<em>10^-5 mols. We are also given the mass, so we can write this as 5.273</em>10^-5 mols=0.200g. Divide to get the molecular weight in grams/mole: 3792.9 g/mol.</p>
<p>I’m not sure, but this is how I look at 7.</p>
<p>If Y has a higher specific heat than X, then more energy is needed to be transfered to raise Y to the lead bath temp. Which means it should have more energy to release into the water tank, so it should raise the water temp higher, which it does.</p>
<p>I don’t remember thermodynamics very well either! So, this is the only crude explanation I can think of. Sorry!</p>
<p>Yeah, the thermo question baffles me.</p>
<p>Oh no. I just came back from my chemistry exam, and I know I had at least 2/25 wrong. Now I need to get 91 or better on the final exam to get an A in the class.</p>
<ol>
<li>Which one of the following gas samples would be expected to behave MOST like an ideal gas at room temperature?
A. 1L of He at 10 atm pressure
B. 1L of N2 at 50 atm pressure
C. 1L of CO2 at 50 atm pressure
D. 1L of H2O at 1 atm pressure
E. 1L of Xe at 1 atm pressure</li>
</ol>
<p>I know it’s between D and E (because ideal gases are under low pressure/high temperature). I kind of chose E, because I didn’t think H2O at room temperature and 1atm (that’s like regular water that we drink) would behave ideally. However, some other students say they chose D. What do you guys think is the answer?</p>
<p>It’s definitely not H2O. There’s too much H-bonding for it to behave like an ideal gas. A noble gas will behave most like an ideal gas… I would go with E.</p>
<p>Isn’t H2O at 1atm and room temp liquid?</p>
<p>Oh yeah, it is. I didn’t even notice the “room temperature” part of the question.</p>
<ol>
<li>When a student mixes 50 mL of 1.0 M HCl and 50 mL of 1.0 M NaOH in a coffee-cup calorimeter, the temperature of the resultant solution increases from 21.0 °C to 27.5 °C. Calculate the enthalpy change for the reaction in kJ per mol of HCl, assuming that the calorimeter loses only a negligible quantity of heat. The total volume of the solution is 100 mL, its density is 1.0 g/mL, and its specific heat is 4.18 J/g-K.
A. 2.7 kJ/mol
B. −2.7 kJ/mol
C. 54.4 kJ/mol
D. −54.4. kJ/mol
E. −108 kJ/mol</li>
</ol>
<p>It would appear that the reaction is exothermic since the temperature increased.
Use Q = mc(delta T)
m = density(volume), so m = 100 g
delta T = 27.5 - 21 = 6.5
Q = 100(6.5)(4.18) = 2717 J,
but we’re not done here, because this is for 50 mL of a 1.0 M solution
We need to find out how many moles of HCl were used.
So, 1 mol/1000mL x 50mL=.05mol HCl.</p>
<p>Now we have 2.717 kJ for every .05 mol HCl used, so divide #kJ by #mol and we get 54.34 kJ/mol, and since the reaction is exothermic, the answer is D.</p>
<p>If the temperature of the solution increased, doesn’t that mean the reaction is endothermic? It gained heat; that’s why the temperature went up.</p>
<p>If a reaction is exothermic, it releases heat into the water. So the water temp goes up.</p>