A size dependent property is a physical property that changes when the size of an object changes.
The molarity of NaOH needed is calculated as follows
calculate the moles of KhC8h4O4
that is moles = mass/molar mass of KhC8h4O4(204.22 g/mol)
=0.5632g /204.22g/mol= 2.76 x10^-3 moles
write the equation for reaction
khc8h4O4 + NaOH ---> KNaC8h4O4 + H2O
from the equation above the reacting ratio of KhC8h4O4 to NaOh is 1:1 therefore the moles of Naoh is also 2.76 x10^-3 moles
molarity of NaOh = (moles of NaOh / volume ) x 1000
that is { (2.76 x10^-3) / 23.64} x100 =0.117 M
Answer:
1.62 L
Explanation:
T= 97+273.15= 370.15
R= 0.08206 atm/mol⋅K
V= 45 L
n= 2.4 mol
P= (n⋅R⋅T)/V
= (2.4 x 0.08206 x 370.15)/(45) = 1.61997 = 1.62
Answer: D
Explanation:
A reducing agent is a species that reduces other compounds, and is thereby oxidized. The whole compound becomes the reducing agent. In other words, of a compound is oxidized, then they are the reducing agent. On the other hand, if the compound is reduced, it is an ozidizing agent.
Since we have established that a reducing agent is the compound being oxidized, we know that A is not our answer. An oxidized compound is losing electrons. Choice A states exactly this.
For B, this is true as we have established this already.
C is also correct. Since a reducing agent loses electrons, it becomes more positive. This makes the oxidation number increase.
D would be our correct answer. It is actually a good oxidizing agent is a metal in a high oxidation state, such as Mn⁷⁺.
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:
