Answer:
3 Cu + 8 HNO3 --> 3 Cu(NO3)2 + 2 NO + 4 H2O
Explanation:
Make sure both sides are equal
3 <u>Cu</u> + 8 H<u>N</u>O3 --> 3 <u>Cu</u>(NO3)2 + 2 <u>N</u>O + 4 H2O
// start by those elements that change their oxidation degree
<u>Cu</u> and <u>N</u>
// also you can write reduction-oxidation reactions
+ 2 
--> 
| 2
- 3 --> 
| 3
// write the numbers of electrons that are lost/gained as the coefficients of the opposite elements
// then check if H and O are the same on both sides
// adjust if they aren't.
Answer: The rate of disappearance ( -Δ[B]/Δt) under the same conditions is 0.200 M/s.
Explanation:
Initial Rate of the reaction = 0.100 M/s
Rate of disappearance of B:
![-\frac{1}{2}\frac{\Delta [B]}{\Delta t}=0.100 M/s](https://tex.z-dn.net/?f=-%5Cfrac%7B1%7D%7B2%7D%5Cfrac%7B%5CDelta%20%5BB%5D%7D%7B%5CDelta%20t%7D%3D0.100%20M%2Fs)
![-\frac{\Delta [B]}{\Delta t}=2\times 0.100 M/s=0.200 M/s](https://tex.z-dn.net/?f=-%5Cfrac%7B%5CDelta%20%5BB%5D%7D%7B%5CDelta%20t%7D%3D2%5Ctimes%200.100%20M%2Fs%3D0.200%20M%2Fs)
The rate of disappearance ( -Δ[B]/Δt) under the same conditions is 0.200 M/s.
The answer for Plato will be letter A
I think the answer is D. Periods 6-7, Because t<span>he periods that would contain elements with electrons in s, p, d, and f orbitals are periods 6-7.
(I hope this helps you a lot! :) Have a nice day!)</span>
1) Molarity
M = n / V
n: number of moles of solute
V: volume of the solution in liters
n = mass / molar mass = 0.000333 g / 332.32 g / mol = 1*10 ^ - 6 moles
V = 225 ml * 1 liter / 1000 ml = 0.225 liter
M = 10^-6 mol / 0.225 liter = 0.00000444 M
2) ppm
ppm = parts per million
grams of solute: 0.000333 g
grams of solution = volume * density = 225 ml * 0.785 g / ml = 176.625 g
ppm = [0.00033 g / 176.625 g] * 1,000,000 = 1.868 ppm
