help with chemistry worksheet :x

[Tifa's Real Lover(really]

no idea how to do these 3 types of problems, my notes seem practically useless

1. determine the empirical formula for a compund made from 26.6% potassium, 35.4% chromium, and 38% oxygen. show all steps and stuff

2. determine the net ionic equation for AgNO3(subscript)(aq) + ZnCl(aq) -------> AgCl + Zn(NO3(subscript))2(subscript)
you must show the ionic equation with spectator ions and all that

3. how many liters of chlorine gas are found in 7.4 x10^24 particles of diatomic chlorine?

please help me!!

[Tavrobel]

Question 1: assume that what you are given is in the same proportion as what you have to use. Short story cut short: pretend you have a 100 gram sample, and that the percentages are the same number of grams that you have. So 26.6% of K becomes 26.6 grams of K. Now treat the problem as though you were doing any other empirical formula problem. You have grams of each element; turn that into moles by dividing by the full mass of the element, and then multiply the moles by numbers until you get integers for everything.

Question 2: determine which item is your precipitate (any term that is a solid or water). I'm guessing they want both the full ionic and net ionic equation. For the full ionic, split up every aqueous molecule into its component ions on both sides of the equation. To determine your net ionic equation, take what you just got, and cross out every ion that appears on both sides.

Question 3: divide by Avogadro's Number to determine how many moles of Chlorine you have, then use the formula you were given to get from moles of a gas to liters of a gas. Don't forget that diatomic Chlorine is Cl₂.

If you have no idea about what I just said, get someone to translate for you, and if they don't know, then post saying so.

[trancekuja]

Damn...The amount of knowledge one has to possess to be granted the title Site Staff:Eek:

[qwertysaur]

Too add to what tav said about number 2, you need to remember to write the charges for each ion on both sides of the equation. Remember any ion or atom that keeps the same charge is a spectator ion. For example the chlorine in ZnCl would be written as [Cl]^-1

Also to get superscript type out <sup>Text here</sup>
For subscript type out <sub>Text here</sub>

[Tifa's Real Lover(really]

so far answer i got for number 1. K₂Cr₂O₇

ok help me if im wrong but number 2 is, (NO₃) + Zn --> Zn(NO₃)

what is a spectator ion?


attempted number 3, no idea

[rubah]

You might be able to take a shortcut with #3.

One mole of an ideal gas will occupy a volume of 22.4 liters at STP

where STP = 1 atmosphere of pressure and 0C. (I think chlorine qualifies as an ideal gas then)

[qwertysaur]

so far answer i got for number 1. K₂Cr₂O₇

ok help me if im wrong but number 2 is, (NO₃) + Zn --> Zn(NO₃)

what is a spectator ion?


attempted number 3, no idea

Here is number 2. Work in spoiler tags for each step.

Ag(NO₃)_(aq) + ZnCl₂_(aq) -------> AgCl + Zn(NO3)₂

(SPOILER)Balance the equation first. You then get

2Ag(NO₃)_(aq) + ZnCl₂_(aq) -------> 2AgCl + Zn(NO3)₂

(SPOILER)Find the precipitate. Split everything else into it's ions. Precipitate is AgCl.

2Ag⁺¹ + 2(NO₃)^-1 + Zn⁺² + 2Cl^-1 -------> 2AgCl + Zn⁺² + 2(NO₃)^-1
(SPOILER)Color coded here. Red means spectator ion.

2Ag⁺¹ + 2(NO₃)^-1+ Zn⁺² + 2Cl^-1 -------> 2AgCl + Zn⁺² + 2(NO₃)^-1

Number 3, try PV = nRT

[Tavrobel]

so far answer i got for number 1. K₂Cr₂O₇

ok help me if im wrong but number 2 is, (NO₃) + Zn --> Zn(NO₃)

what is a spectator ion?


attempted number 3, no idea

That looks right, as it is an actual compound. If your mole ratios were ~.680/.680/2.375, then yes, it should be 2/2/7.

A spectator ion is, in short, an ion that appears on both sides of the equation. They play no physical part in the reaction. qwerty's assessment of the problem is correct. The last step should be your final solution for a Full Ionic Equation. Cross out the ions in red for your Net Ionic Equation.

Given: quantity of Chlorine, 7.4*10^23/2 (since it is diatomic)
Divide by Avogadro's number, 6.022*10^23
This should give you the number of moles of Chlorine that you have: 6.144
Then use PV = nRT at Standard Temperature and Pressure, 1 atm, 273 Kelvin (as rubah said)
Solving for "V" (volume), and "n" has been calculated, "r" is a constant, and the units should drop out of the equation
V = whatever rubah said because my calculator is gone

[rubah]

1 atm = 101.325 kilopascals = 14.7 psia (0 psig )

[Tavrobel]

1 rubah = 101.325 kPa

[rubah]

I think he just implied I have no gauge pressure >:x

[Tifa's Real Lover(really]

thanks you guys

[Suvious]

This all looks like jibberish to me x_x

[Fonzie]

I loveee chemistry.