Chemistry Help Really Lost

Chemistry Help! I am lost?

Which of the following is true?

Be is smaller than Mg due mainly to differences in Zeff.

Na is larger than Mg due mainly to differences in the principal quantum number.

C is smaller than F due to an increase in the outermost energy levels of electrons.

P is larger than N due primarily to stronger attractions between electrons in the outermost orbitals.

Ca is larger than Mg due to larger orbitals.

Chemistry HELP..LOST?

1.)8.11 g of sulfur is burned. The sulfur reacts with oxygen gas to produce a sulfur oxide. The weight of the sulfur oxide that is produced is 16.20 g.
What is the empirical formula for this sulfur oxide that is produced?

2.)An unidentified organic compound is determined to be 52.2% C, 13.0% H, and 34.8% O.
What is the empirical formula for this compound?

Chemistry HELP!!!!?

so we got these questions in chemistry today and i am SO lost. if anyone could just help out a bit i would REALLY appreciate it.

1. If you place 10 g (s) of CO (little 2, idk how to type it, but 2 oxygens) on a scale and allow it to sublime, the weight decreases. Does this mean that you have lost CO2 molecules? Why or why not?

2. Suppose that you have 1.4 x 10 (to the 23rd) molecules of solid dry ice. After the solid is allowed to go completely from solid to gas, how many molecules of gaseous CO2 do you have?

3. If 10 g (s) CO2 (dry ice) sublimes, how many grams of CO2 (g) are produced?

Please help with chem, im really lost?

The rate law for any reaction involving the reactants A and B is written as:
Rate = k [A]^x [B]^y
where,
k: rate constant
[A]: concentration of A
x: order of reaction with respect to A
[B]: concentration of B
y: order of reaction with respect to B

Now using the given data for rate, [A] and [B] write the rate law equations:

(1) 1.3 × 10-2 = k* 0.15^x * 0.12^y
(2) 2.6 × 10-2 = k* 0.30^x * 0.12^y
(3) 0.24 × 10-2 = k* 0.15^x * 0.24 ^y

Now, dividing (1) by (2)
1.3/2.6 = (0.15/0.30)^x
=> (1/2) = (1/2)^x
=> x=1

Similarly to find y, divide (1) by (3)
(1.3/0.24) = (0.12/0.24)^y
=> y= -2.44

So, the rate law can be written as:
Rate = k [A] [B]^(-2.44)

Chemistry Question?

You calculate [OH-] and [H3O] from solutions, if you know the concentration becuase when you put those brackets on, that means molarity. so if you havae 1.3M of an acid, the [H3O] is 1.3. [OH-} is or bases - it's the molarity of hydroxide ions. Just watch for di and polypriotic acids, if they can release more than one hydrogen, it changes the [H3O] by the multiple of H in solution.

You find the pH by taking the negative log of Ka. Ka is calaculated by using the formula [C][D]/[A][B] for the reaction A + B = C + D. there are coeffecients in some balanced reactions, and I have NO idea how i could explain it well, but I'll try.

Say your balanced reaction is 1 mol of A plus one mole of B makes 2 moles of C and 1 mole of D. Ka is calculated by products over reactants, but since you get two moles of C in the reactants, you need to SQUARE C. This is confusing to nearly everyone (I teach freshman college chem, and this is a big hang up for a lot of students.,) You should be able to ask your teacher to explain it - it's a consise question.

You find a reaction rate by dividing how much was consumed or created by the time it took to consume or create.

OH! Log = logarythm, negative log is the negative of that number. You (unless you are a supermathgenius) need a scientific calculator to do these calculations (you'll soon get into antilogs, too). If you have that type of calculator, it's a matter of pushing one button. Log is defined as whatever number you need to raise 10 to to achieve the number. Negative log is the opposite of the log.

Hope this helps ~ it's kind of late here and I think I might be getting too sleepy to answer with understandable answers. But that should be a start.

Chemistry Help... 10 Quick question = 10 easy points, I'm Really Lost!?

#7 - A (Physical Change in Both)
#10 - A (Bill's)
#11 - D (50,000)
#13 - B (5.0 x 105 calories)
#14 - A (Slope)
#15 - A (dependent variable values)

Those are the only ones I know, hope it helps!

Chemistry help please?????

2MnO4- (aq) + 5H2O2(aq) + 6H+(aq) --> 5O2(g) + 2Mn2+ (aq) + 8H2O(l)

(a)
Each 1 mole of KMnO4 provides 1 mole MnO4-, thus Molarity of MnO4- = Molarity of KMnO4
Molairy mnO4- = 0.105 M

Molarity = moles / litres
Therefore moles = molarity x litres
moles MnO4- = 0.105 M x 0.0432 L
= 0.004536 moles
= 0.00454 moles (3 sig figs)

(b) I am going to assume you means how many moles of H2O2 were present ?????

The balanced equation tells you that
2 moles MnO4- reacts with 5 moles H2O2
So 1 mole MnO4- reacts with 5/2 moles h2O2
Thus 0.004536 moles MnO4- reacts with (5/2 x 0.004536) moles H2O2
= 0.01134 moles H2O2
= 0.0113 moles H2O2 (3 sig figs)

(c)
moles = mass / molar mass
Therefore mass = molar mass x moles
mass H2O2 = 34.016 g/mol x 0.01134 moles
= 0.38574 g
= 0.386 g H2O2 (3 sig figs)

(d)
mass percent = mass H2O2 / total mass x 100/1
= 0.386 g / 13.8 g x 100
= 2.80 % (3 sig figs)

(e)
Oxidation is loss of electrons. The species that undergoes oxidation loses electron. It is called the reducing agent because it causes the other reagent to be reduced

reduction is gain of electrons. The species that undergoes reduction gains electrons. It is called the oxidising agent because it takes the electrons from the other reagent thus causeing it to be oxidised.

Reduction half equation (note MnO4- is gaining electrons to Mn2+. It is reduced and thus is the oxidising agent)
2MnO4- + 16H+ + 10e ----------> 2Mn2+ + 8H20

oxidation half equation (note H2O2 is losing electrons thus being oxidised to O2. It is oxidised and thus is the reducing reagent)
5H2O2 -------> 5O2 + 10H+ + 10e

Can i get some chemistry help please?

The notation mentioned:
The left superscript is the mass number = #protons + #neutrons
It will be 0 for an electron (or beta- particle) since it has neither protons nor neutrons.

The left subscript is the atomic number (mostly). = #protons.
It might be better to think of it as a charge since it can be -1 for the electron (or beta- particle)

The neutron has a mass of 1 and no charge so it's written 1/0n.

In nuclear reactions,
the top numbers on the left added together = the top numbers on the right added together.
Same thing with the bottom numbers.

Alpha decay always produces a helium nucleus (4/2He)
The new element will have a mass number that is 4 less and an atomic number that is 2 less.
Americium is element 95 (that's a lookup)
and is written as 241/95Am.
The reaction will be
241/95Am ---->237/93Np + 4/2He

Beta- decay always produces an electron, 0/-1e,
so the mass number doesn't change, and the atomic number increases by 1.
?/55Cs ---> ?/56Ba + 0/-1e
The particular isotope of Cs and Ba is not specified, so I think that's the most specific answer you can provide.

In electron capture, an electron is taken into the nucleus and so is written as a reactant.
In the nucleus it combines with a proton to make a neutron. The mass number doesn't change, but the atomic number decreases by 1.

?/38Sr + 0/-1 e ----> ?/37Rb.

In the last problem, you're given all the products and have to infer the reactant.
---> 2 (1/0n) + 4/2He + 232/90Th.
2x1 + 4 + 232 = 238
2x0 + 2 + 90 = 92 ==>U
So 238/92U ---> 2 (1/0n) + 4/2He + 232/90Th.

There is also a beta+, or positron, decay.
A positron is the antiparticle for the electron. It has the same mass and opposite charge and is written as 0/1e. (no protons, no neutrons, and a +1 charge.)