1st semester rapid review Flashcards
(13 cards)
Give the chemical formula for sulfurous acid.
a. H2SO4
b. H2SO3
c. H2SO2
d. H2S
e. H2Se
b.
H2SO2 isn’t a polyatomic ion so it’s really just between H2SO4 and H2SO3.
the ‘ous’ ending is for the less charged and ‘ic’ would be for the strongest.
What mass of carbon is present in 1.5 x 10^22 molecules of C4H10O2?
a. 0.12 grams
b. 1.2 grams
c. 0.3 grams
d. 0.6 grams
e. 3 grams
b. 1.2
There’s a lot of conversion here so it would be easy to mess up a decimal and accidentally come out with a.
But basically, you convert the whole thing to moles using Avagadro’s number. Then you break it down further by saying that there are 4 moles of C in 1 mole of C4H10O2. Then you convert C to grams.
Which term best describes the relationship between 12C, 13C, and14C?
a. Isomers
b. Isotopes
c. Allotropes
d. Diatomics
e. Triatomics
b. Isotopes
Which of the following is the balanced chemical reaction for the combustion of C2H6O?
a. 2C2H6O + 5O2 = 4CO2 + 6H2O
b. C2H6O = 2C + 3H2O + 2H2
c. 2C2H6O = 4C + 1/2O2 +6H2
d. C2H6O + 3O2 = 2CO2 + 3H2O
e. 4C + 1/2O2 + 6H2 = 2C2H6O
c.
Three answers can easily be eliminated. b., c., and e.
You know that your reactant is going to combine with oxygen to form CO2 and H2O
Answer a. is close but I think there are too many O’s on one side for it to be balanced.
What is the mass percent of nitrogen in N2O4?
a. 15%
b. 30%
c. 40%
d. 55%
e. 70%
b.
I’m good with percentages for some reason so this is one of the few things in chemistry that I can just look at and work out. The total mass is 92. The total mass of N in the molecule is 28. Just divide 28 by 92. Simple.
If 11 moles of P and 20 moles of Cl2 react completely, what is the reagent in excess and how much remains after completion?
2P + 5Cl2 = something but it doesn’t matter
a. P, 3 moles
b. P, 6 moles
c. P, 9 moles
d. Cl2, 6 moles
e. Cl2, 9 moles
a.
This is the other thing in chemistry that I can do just by thinking about it. Considering just P, you can run the reaction 5 times. Considering just Cl2, you can run the reaction only 4 times. So… P is your excess. But how much P do you have left if you can only run the reaction 4 times?
What volume of 4M NaOH is required to produce 800 ml of 0.25M NaOH?
a. 12.5 ml
b. 25 ml
c. 50 ml
d. 100 ml
e. 200 ml
c.
Got it right but it’s really hard to picture a real-life example of this question.
But basically, it’s a simple equation.
(4)(x) = (800)(.25)
This is one of those where I don’t think the units matter as long as they’re the same.
A 20.0 gram sample of a metal was heated to 100 degrees C and then added to 20.0 grams of water at 10 degrees C in a calorimeter. If the temperature reached at thermal equilibrium was 20 degrees C, then what is the specific heat of the metal?
(cWater = 4.184 J/gK)
a. 0.26 J/gK
b. 0.52 J/gK
c. .78 J/gK
d. 1.2 J/gK
e. 33.5 J/gK
b. 0.52 J/gK
Phew! Got it. Couldn’t quite remember how to set up the equation. Really had to think about it. So it’s like metal on one side and water on the other. But instead of the given temperature of each, it’s the change in temp of each compared to the equilibrium. So…
(20 grams)(-80 change in temp)(x) = (20 grams)(10 change in temp)(4.184 specific heat of water)
How much heat would be required to heat 108 grams of solid aluminum at 2419 degrees C to liquid aluminum at 2519 degrees C?
aluminum m.p. = 2519 degrees C
aluminum cAl(s) = 0.90 J/gK (specfic heat)
aluminum changeHeatdegrees fusion = 660 kJ/mole
a. 39,000 kJ
b. 39 kJ
c. 2.6 kJ
d. 2650 kJ
e. 12,640 kJ
d. 2650 kJ
I got 2656.2 but who’s counting
There is a lot going on in this problem. First, you have to convert the temps to K right off the bat. Then, using specific heat, you have to figure out how much energy it takes to raise the Al 100 degrees K. I converted this answer to kJ from J here. Then you’re going to add this answer to how much energy it’s going to take to make it go through a phase change. For this, you need to convert 108 grams to moles.
What is the ground state electron configuration of Ni^2+
(the graphic in the study guide definitely helps with this)
a. 1s2 2s2 2p6 3s2 3p6 4s2 3d8
b. 1s2 2s2 2p6 3s2 3p6 3d8
c. 1s2 2s2 2p6 3s2 3p6 4s1 3d8
d. 1s2 2s2 2p6 3s2 3p6 4s2 3d6
e. 1s2 2s2 2p6 3s2 3p6 4s2 3d10
b.
This took a lot of thinking because I knew there was something ‘extra’ that needed to be applied. There’s the normal configuration. But things change when you remove electrons. The 2 electrons removed are going to be the ones from 4s
What is the maximum number of electrons that can occupy a ‘d’ orbital?
(Jesus. This is a trick question so really look at the wording.)
a. 1
b. 2
c. 5
d. 10
e. none of these
b. 2
Any orbital can hold 2 electrons max. The d ‘subshell’ could hold 10
Which of the following groups has a valence shell electron configuration of ns2 np5
a. Alkali metals
b. Alkaline earth metals
c. Chalcogens
d. Halogens
e. Noble gases
d. Halogens
It’s really just asking which group has 7 valence electrons
