Answer:
[OH⁻] = 3.34x10⁻³M; Percent ionization = 0.54%; pH = 11.52
Explanation:
Kb of the reaction:
NH3 + H2O(l) ⇄ NH4+ + OH-
Is:
Kb = 1.8x10⁻⁵ = [NH₄⁺] [OH⁻] / [NH₃]
<em>As all NH₄⁺ and OH⁻ comes from the same source we can write: </em>
<em>[NH₄⁺] = [OH⁻] = X</em>
<em>And as </em>[NH₃] = 0.619M
1.8x10⁻⁵ = [X] [X] / [0.619M]
1.11x10⁻⁵ = X²
3.34x10⁻³ = X = [NH₄⁺] = [OH⁻]
<h3>[OH⁻] = 3.34x10⁻³M</h3><h3 />
% ionization:
[NH₄⁺] / [NH₃] * 100 = 3.34x10⁻³M / 0.619M * 100 = 0.54%
pH:
As pOH = -log [OH-]
pOH = 2.48
pH = 14 - pOH
<h3>pH = 11.52</h3>
The much of the sample that would remain unchanged after 140 seconds is 2.813 g
Explanation
Half life is time taken for the quantity to reduce to half its original value.
if the half life for Scandium is 35 sec, then the number of half life in 140 seconds
=140 sec/ 35 s = 4 half life
Therefore 45 g after first half life = 45 x1/2 =22.5 g
22.5 g after second half life = 22.5 x 1/2 =11.25 g
11.25 g after third half life = 11.25 x 1/2 = 5.625 g
5.625 after fourth half life = 5.625 x 1/2 = 2.813
therefore 2.813 g of Scandium 47 remains unchanged.
Answer:
First, find out how many moles of N2I6 you have. Then convert that to grams.
molar mass N2I6 = 789 g
moles N2I6 = 8.2x1022 molecules N2I6 x 1 mole/6.02x1023 molecules = 1.36x10-1 moles = 0.136 moles
grams N2I6 = 0.136 moles x 789 g/mole = 107 g = 110 g (to 2 significant figures)
The answer is 18 you multiply wavelength (6) times frequency (3)
Use the universal gas formula
PV=nRT
where
P=pressure ( 0.980 atm)
V=volume (L)
T=temperature ( 23 ° C = 23+273.15 = 296.15 ° K)
n=number of moles of ideal gas (0.485 mol)
R=universal gas constant = 0.08205 L atm / (mol·K)
Substitute values,
Volume, V (in litres)
=nRT/P
=0.485*0.08205*296.15/0.980
= 12.0256 L
= 12.0 L (to three significant figures)
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