First find the mass of <span>solute:
Molar mass KNO</span>₃ = <span>101.1032 g/mol
mass = Molarity * molar mass * volume
mass = 0.800 * 101.1032 * 2.5
mass = 202.2064 g of KNO</span>₃
<span>To prepare 2.5 L (0800 M) of KNO3 solution, must weigh 202.2064 g of salt, dissolve in a Beker, transfer with the help of a funnel of transfer to a volumetric flask, complete with water up to the mark, capping the balloon and finally shake the solution to mix.</span>
hope this helps!
Answer:
- Option A): <em>Due to the constraints upton the angular momentum quantum number, the subshell </em><u><em>2d</em></u><em> does not exist.</em>
Explanation:
The <em>angular momentum quantum number</em>, identified with the letter l (lowercase L), number is the second quantum number.
This number identifies the shape of the orbital or <em>kind of subshell</em>.
The possible values of the angular momentum quantum number, l, are constrained by the value of the principal quantum number n: l can take values from 0 to n - 1.
So, you can use this guide:
Principal quantum Angular momentum Shape of the orbital
number, n quantum number, l
1 0 s
2 0, 1 s, p
3 0, 1, 2 s, p, d
Hence,
- <u>the subshell 2d (n = 2, l = 2) is not feasible</u>.
- 2s (option B) is possible: n = 2, l = 0
- 2p (option C) is possible: n = 2, l = 1
A. We can calculate the initial concentrations of each by
the formula:
initial concentration ci = initial volume * initial
concentration / total mixture volume
where,
total mixture volume = 10 mL + 20 mL + 10 mL + 10 mL = 50
mL
ci (acetone) = 10 mL * 4.0 M / 50 mL = 0.8 M
ci (H+) = 20 mL * 1.0 M / 50 mL = 0.4 M (note: there is only 1 H+ per
1 HCl)
ci (I2) = 10 mL * 0.0050 M / 50 mL = 0.001 M
B. The rate of reaction is determined to be complete when
all of I2 is consumed. This is signified by complete disappearance of I2 color
in the solution. The rate therefore is:
rate of reaction = 0.001 M / 120 seconds
rate of reaction = 8.33 x 10^-6 M / s
Answer:
i need a picture or more info to understand
Answer:

one atom of an element = 6.02 \times {10}^{23} atom
The mass of one atom of sulphur = 32g
The mass of one atom of aluminium = 27g
so one atom of aluminium = 6.02 \times {10}^{23}
27g of AL = 6.02 \times {10}^{23} atom
2.70g of AL = X atoms
Then you cross multiply ........
and get the answer
.