<p>This is what I did.</p>
<p>2.
a. (1.18)(.843)= 9.95 x 10^-1 g</p>
<p>b. 157.70-.995= 156.7 g</p>
<p>c. 158.08-156.7= 1.38g</p>
<p>d. Use PV=nRT or (PV)/T=(PV)/T, either are fine
n= 1.38/x (x is the molar mass)
(750/760)(.843)= (1.38/x)(.0821)(273+23)
x= 40.3 g</p>
<p>e. (44-40.3)/44 x 100= 8.41%</p>
<p>f.
OC 1:Yes, since dry air is less dense than the gas, the gas may appear lighter
OC 2: No, since gasses are more dense at cooler temperatures, that would make the gas heavier, which is not what happened</p>
<p>g. Measure 843 mL of water out and see if it overflows, fits perfectly, or doesn't fill the container. The container would be less than 843mL, exactly 843 mL, or more than 843 mL, respectively.</p>
<p>3.
a. i. 25/16=1.56 mole; since we need 3.125 mol of Cl2 to react with 1.56 mol of methane, Cl2 is the LR.</p>
<p>ii. 2.58 mol of Cl2 reacts with 1.29 mol of CH2Cl2, we need 1.29 mol of CH2Cl2.</p>
<p>b. (242000)/(6.022x10^23)=4.02x10^-19</p>
<p>c. I’m not really sure about this one.
E=hv (v=c/lambda) so E=h(c/lambda)
4.02x10^-19=(Plank’s)((3x10^8)/lambda)
4.95x10^-7 meters</p>
<p>d. Intermediate, since it is consumed.</p>
<p>e. Not sure about these.
i. 2
ii.1</p>
<p>4.
a.
i. 2(Fe2O3) + 3C = 4Fe +3(CO2)
ii. 0 to +4
b.
i. NH3+HCL=NH4Cl
ii.acidic, since NH4Cl can donate protons (not sure about that)
c.
i. 2(HgO)=2Hg+O2
ii. less than since gas escapes</p>
<p>5.
a. [H2][CO]/[H2O]
b. decrease, since there are less gas molecules on the left side, which is where equilibrium will shift
c. less than one, since gibbs is positive, it is not spontaneous, and will therefore prefer the reactants
d. the products, since it is endothermic, and the products have gained energy compared to the reactants
e. yes, because the gibbs for y is closer to 0 than for x, and thus will occur more easily
f.
i. it’s negative, since the products have less entropy than the reactants
ii. i forgot what i put, but it was negative (I think -172kJ/mol)
iii. increase, since it will react to turn into two moles of carbon monoxide, which has more volume than solid carbon and one mole of carbon dioxide</p>
<p>6.
a. i. basically 1s2 up to 3p4 (it’s hard to type out) for S, and up to 3p6 for S2-
ii. since the same pull of the nucleus is now spread out over more electrons
iii. I put S2-, since it will be more strongly attracted to the positive end of the magnet</p>
<p>b. it is easier to remove from S2-, since the Ar atom has more protons and therefore a stronger pull</p>
<p>c. I put the p atoms. Not sure.</p>
<p>d.<br>
i. I put the H2S, but not sure.
ii. I put H2S, but I think H2O is stronger. I think I got d wrong.</p>