Answer:
b) sharpening a pencil
Explanation:
If you melt lead, boil water, or dissolve sugar in water, you can return all of them back to their original state. If you sharpen a pencil, you can't reattach the shavings as they were originally.
The balanced chemical reaction is:
<span>2C4H10(g)+13O2(g)->10H2O(g)+8CO2(g)
</span>
<span>Calculate the mass of water produced when 1.77 grams of butane reacts with excessive oxygen?
</span>1.77 g C4H10 (1 mol C4H10/58.14 g C4H10) (10 mol H2O / 2 mol C4H10) ( 1.01 g H2O / 1 mol H2O ) = <span>0.15 g H2O
</span><span>Calculate the mass of butane needed to produce 71.6 of carbon dioxide.
</span>71.6 g CO2 (1 mol CO2/ 44.01 g CO2) ( 2 mol C4H10 / 8 mol CO2 ) (58.14 g C4H10 / 1 mol C4H10 ) = 23.65 g C4H10
Answer:
25.8
Explanation:
Let's write the reaction between magnesium-phosphide and potassium:
Mg3P2 + K = Mg + K3P
And now let's balance this equation:
Mg3P2+6K=3Mg+2K3P
We see that the ratio of magnesium-phosphide and potassium is 1:6, which means that for every mole of magnesium-phosphide there need to be 6 moles of potassium.
Since we have 4.3 moles of Mg3P2, there need to be 6 • 4.3 = 25.8 moles of potassium.
Answer:
Aluminium is ordinarily classified as a metal. It is lustrous, malleable and ductile, and has high electrical and thermal conductivity. Like most metals, it has a close-packed crystalline structure and forms a cation in an aqueous solution.
Answer:
I can't draw diagrams on this web site but I can do with numbers I think. So an electron is moved from n = 1 to n = 5. I'm assuming I've interpreted the problem correctly; if not you will need to make a correction. I'm assuming that you know the electron in the n = 1 state is the ground state so the 4th exited state moves it to the n = 5 level.
n = 5 4th excited state
n = 4 3rd excited state
n = 3 2nd excited state
n = 2 1st excited state
n = 1 ground state
Here are the possible spectral lines.
n = 5 to 4, n = 5 to 3, n = 5 to 2, n = 5 to 1 or 4 lines.
n = 4 to 3, 4 to 2, 4 to 1 = 3 lines
n = 3 to 2, 3 to 1 = 2 lines
n = 2 to 1 = 1 line. Add 'em up. I get 10.
b. The Lyman series is from whatever to n = 1. Count the above that end in n = 1.
c.The E for any level is -21.8E-19 Joules/n^2
To find the E for any transition (delta E) take E for upper n and subtract from the E for the lower n and that gives you delta E for the transition.
So for n = 5 to n = 1, use -Efor 5 -(-Efor 1) = + something which I'll leave for you. You could convert that to wavelength in meters with delta E = hc/wavelength. You might want to try it for the Balmer series (n ending in n = 2). I think the red line is about 650 nm.
Explanation:
