<p>I’m self-studying.</p>
<p>If you’re going to answer, can you please answer with an explanation.</p>
<p>“The other four will react with the NaOH.” I already know that, but why?</p>
<p>“The ethene is the limiting reactant, not the oxygen.” Ok, so in the actual reaction, 1 mole will react with 100g/(96g/mol) = 1.04 moles. In the theoretical reaction, 1 mole will react with 3 moles, so what molecule is lacking? Oxygen. It has to be the limiting reactant. Why do you think ethene/ethylene is the limiting reactant?</p>
<p>“Why don’t you try to solve it first?” Because I tried, and I couldn’t?</p>
<p>That picture is a skeletal formula. Therefore, the corners represent C, and balance the num of hydrogen.</p>
<p>@Savvy123</p>
<p>Can you please answer the other questions with a bit of detail…Test date is getting closer, and I still don’t know how to solve those questions. Please man.</p>
<p>For the NaOH one, NaOH, a base, reacts with acidic gases (SO2 and CO2) as well as acids (HBr and HCL). O2 wouldn’t react, so it could be used as a drying agent. The answer is D.
In the ethane and oxygen question, 100 grams of O2 is 100/32=3.125 moles of O2. One mole of ethane would react with only 3 moles of oxygen, so 0.25 moles of unreacted oxygen are left. The answer is B.
For the NaHCO3 question, in every litre of solute there are .001 moles of NaHCO3. So, there are .001 moles of all ions in the solution as well. The concentration of hydrogen ions would be .001 M.
In organic skeletal formula, any corner is said to have CH2 by convention. Only functional groups besides CH2 are labeled on their corners.
Hope it helps :)</p>
<p>You helped a lot; thanks bro. I’m still confused about the ethane and oxygen question.</p>
<p>Why did you do 100g/(32g) when you have 3 moles of O2. I’m looking at other examples in my prep book, and they multiply the molar mass by the number of moles before dividing it by the given mass.
E.g
How many liters of oxygen can you prepare from the decomposition of 42.6 grams of sodium chlorate?</p>
<p>2NaClO3 -> 2NaCl + 3O2</p>
<p>42.6/(2*molar mass of NaClO3) = xL / 67.2L</p>
<p>Note how they multiplied the molar mass by 2. Instead, you just used the molar mass of O2. Could you explain…</p>
<p>C2H4 + 3O2 → 2CO2 + 2H20</p>
<p>You don’t have 3 mol of O2. The balanced chemical equation gives the ratios of reactants so that there will be none left over.</p>
<p>Since you have 100g of O2.</p>
<p>32 g = 1 mol
1 g = 1/32 mol
100 g = 1/32 * 100 mol of O2 which is 3.125 or a little over 3 mol.</p>
<p>The balanced equation states that if you have 1 mol of C2H4, you will need 3 mol of O2 for no excess reactants. Since you have 3.125 mol of O2, you will have 0.125 mol of O2 remaining which can react with more ethene. Hence C2H4 is the limiting reagent.</p>
<p>HypnoTurtle is right, the equation only tells you the ratios in which the reactants will react. The only thing you know is that there are 100 g of O2, from which you must find the number of moles.
Is the answer in your prep book 13.4 litres? Instead of doing it the way your prep book does, I would suggest taking the problem step by step so its less confusing:
1)Number of NaClO3 moles= 42.6/106.44=0.40 moles
2)Number of moles of oxygen produced from 0.40 moles of NaClO3= (3/2)<em>(.4)=0.60 moles
3) Volume of O2 at STP= .60</em>22.4=13.4 litres</p>
<p>In your book they multiplied by 2 because the ratio of NaClO3 to O2 produced is 2:3. On the other side of the equation they multiplied 22.4 litres with 3 to get 67.2 litres.</p>
<p>Ok what about these two:</p>
<p>“If two elements (same group) have similar chemical properties because they have the same number of electrons, why don’t: O2- and Ne have similar chemical properties?”</p>
<p>What would happen if I added helium gas to this reaction:</p>
<p>H2 (g) + I2 (g) + 51.9 kJ → 2HI (g)</p>
<p>“I thought at first that adding concentration to the reactants will shift the equilibrium to the right, so it would increase the rate of the forward reaction. But apparently that’s wrong.”</p>
<p>So the other Image question has also not been answered. </p>
<p>Here’s another image question:
[View</a> image: seriesreaction](<a href=“http://postimg.org/image/4mhutt2mb/]View”>http://postimg.org/image/4mhutt2mb/)</p>
<p>I’m confused about these questions, because they’re not the typical acid-base reactions:
H2O + HA -> H- + H3O</p>
<p>Could someone explain to me 5-9, please. Thanks.</p>
<p>Also, how come the reaction of iron filings with powdered sulfur is an oxidation-reduction reaction?</p>
<p>O2- and Ne do not have similar properties because Ne is a noble gas. All noble gases are highly unreactive, while ions such as O2- are reactive. This is why they have different properties.
I don’t see helium gas in the chemical equation, so maybe the question is wrong. Otherwise, adding helium would have no effect.
Iron filings and sulfur react as so: Fe + S → FeS. The oxidation numbers for iron and sulfur are both 0 on the reactants side. On the products side, the oxidation number of iron is +2 and the ox. no. of sulfur is -2. Since there is a change in the oxidation numbers of the elements, the reaction is considered to be an oxidation-reduction reaction.
In questions 5-9, the answers are 5.A 6.E 7.B 8.C 9.D. 7 and 8 are obvious,as the reaction shows the dissociation of BaCrO4 in water. So, OH- and H+ are needed. In 6, Ba2+ will be needed to form BaCrO4 from CrO42-. That leaves 5 and 9. If I had to guess, I would put 9 as Cr3+, because it is unlikely that Cr2O7 would be in the products and reactants.</p>
<p>What type of reaction is N.5 ? </p>
<p>OH- + (Cr2O7)2- –> (CrO4)2- + H+</p>
<p>How did you know it would be (Cr2O7)2-?</p>
<p>For N.8, does putting the reactants in OH- cause them to form a solid? Is that a general rule? I can see how in N.7, lowering the pH will cause the solid to ionize, but what about N.8?</p>
<p>Here’s the other image question that has not been answered:
[View</a> image: concentrationgraph](<a href=“http://postimg.org/image/dca2jic73/]View”>http://postimg.org/image/dca2jic73/)</p>
<p>Why is this true:</p>
<p>When the pressure drops sufficiently, the water starts to boil.
BECAUSE
At the boiling point, the vapor pressure of the water equals the pressure inside the bell jar.</p>
<p>I thought, according to P/T = k, if P decreases, T will also decrease, and the H2O(g) would condense, not boil.
What about the vapor pressure? When is there equilibrium vapor pressure?</p>
<p>A solution that is prepared by dissolving 3.150g of substance in 25g of benzene has a freezing point of 1.12 degrees Celsius. (Normal freezing point of benzene is 5.50 degrees Celsius and molal freezing point depression constant is 5.12 degrees Celsius/molal). Calculate the molar mass of the unknown substance.</p>
<p>I suppose you use the equation T = Kf*m (m=molality). But how? Could you solve it?</p>
<p>In the first image question, numbers 7 and 8 are used to show the ionization of barium chromate in water. The H+ and OH- are meant to represent the ions in water, not separate hydroxide and H+ ions.
In the second image question, the answer is A. Adding X and Y to the mixture would cause a sharp increase in their concentrations after T2. Then, the equilibrium would shift to the right so that more product is formed- this explains the slow increase in the concentration of Z as well.</p>
<p>When pressure drops, water molecules in the gas phase and the liquid phase are allowed to move freely in the bell jar. So, as molecules spread out, water vapour would be formed. This is essentiall boiling of water- the large separation between the molecules would make them form a gas from a liquid.
In the second question:
- The molality of the substance in benzene is (3.15/25)*1000=126 molal.
- The change in freezing point is (-)4.38 degrees celsius.
- Number of moles is 4.38/5.12=0.855 moles.
- Molar mass= 3.15/0.855= 3.7 grams per mole.
(I’m not sure if I am right; I haven’t studied the topic yet)</p>
<p>Thanks.</p>
<p>I understand why A is correct for the concentration image question, but why is E wrong.
This is my confusion, when I first posted that question:
“To increase pressure, the reaction has to shift to the side with more molecules, which is the reactant side. Increasing the concentration of the reactant side and simultaneously decreasing the product side is what’s happening in the graph. I was stuck between A and E. What’s the correct explanation?”</p>
<p>What about this:
“At the boiling point, the vapor pressure of the water equals the pressure inside the bell jar.” Does equilibrium vapor pressure always occur at the boiling point?</p>
<p>You were close with the freezing question. I just figured it out.
“1) The molality of the substance in benzene is (3.15/25)<em>1000=126 molal” -That should have been 126 grams.
“4) Molar mass= 3.15/0.855= 3.7 grams per mole.” -Molar mass should be: 126g/0.855 moles = 147 g/mol (I only had the answer to the question, and this is the correct one).
Oh, and btw, how did you get 0.855 moles. Shouldn’t it be 0.855 m (m=molality) because of this: T = Kf</em>m (m=molality) ?</p>
<p>Here’s another image question:
[View</a> image: equilibrium](<a href=“http://postimg.org/image/5qrnn5sg3/]View”>http://postimg.org/image/5qrnn5sg3/)
Answer says volume of beaker A and concentration of beaker B will both decrease.
Maybe the volume will decrease because the water will evaporate as the system approaches equilibrium, but why would Beaker B’s concentration decrease?</p>
<p>Haha, I get it 2200andbeyondXD. Just a mild sarcasm. And ccuser001: when are you giving your satII? Seems like there’s quite a few who are giving chem as subject test, considering there are no replies for your queries. I, myself will go for physics.</p>
<p>In the equilibrium question:
Water will evaporate from A, and that water will go into B. As the volume of B increases but the amount of sugar stays the same, the concentration would decrease.
Equilibrium vapour pressure does not always occur at the boiling point: if you keep a beaker of water in a bell jar, equilibrium vapour pressure will be reached- the rate at which liquid water changes to gas would be the same as the rate at which water vapour changes to liquid water.
Increasing the pressure in question 58 would cause a shift in equilibrium to the side with less molecules-more product would be formed, and the conc. of the reactants would not change. In the graph, the concentrations of the reactants changes more than that of the products, which would only happen if A was correct.</p>
<p>I think I dod the molality question wrong and have no idea how I got 0.855 moles. Must have confused degrees celsius/molal for degrees celsius/mole :P</p>