Answer:
Atoms of sulfur = 9.60⋅g32.06⋅g⋅mol−1×6.022×1023⋅mol−1
Explanation:
because the units all cancel out, the answer is clearly a number, ≅2×1023 as required.
Answer:
1. n = 0.174mol
2. T= 26.8K
3. P = 1.02atm
4. V = 126.88L
Explanation:
1. P= 2.61atm
V = 1.69L
T = 36.1 °C = 36.1 + 273= 309.1K
R = 0.082atm.L/mol /K
n =?
n = PV / RT = (2.61x1.69)/(0.082x309.1)
n = 0.174mol
2. P = 302 kPa = 302000Pa
101325Pa = 1atm
302000Pa = 302000/101325 = 2.98atm
V = 2382 mL = 2.382L
T =?
n = 3.23 mol
R = 0.082atm.L/mol /K
T= PV /nR = (2.98x2.382)/(3.23x0.082) = 26.8K
3. P =?
V = 0.0250 m³ = 25L
T = 288K
n = 1.08mol
R = 0.082atm.L/mol /K
P = nRT/V = (1.08x0.082x288)/25 = 1.02atm
4. P = 782 torr
760Torr = 1 atm
782 torr = 782/760 = 1.03atm
V =?
T = 303K
n = 5.26 mol
R = 0.082atm.L/mol /K
V = nRT/P
V = (5.26x0.082x303)/1.03 = 126.88L
Answer:
Gallium, Phosphorus, Chlorine, Fluorine
Explanation:
Arrange the elements in order of increasing ionization energy. Use the periodic table to identify their positions on the table.
Drag each tile to the correct box.
Tiles
chlorinefluorinegalliumphosphorus
Sequence
The pH of a 0.260 M solution of ascorbic acid is 0.585. Details about pH can be found below.
<h3>How to calculate pH?</h3>
The pH of a solution can be calculated using the following expression:
pH = - log {H+}
According to this question, ascorbic acid is a diprotic acid and posseses a concentration of 0.260M. The pH can be calculated as follows;
pH = - log {0.260}
pH = 0.585
Therefore, the pH of a 0.260 M solution of ascorbic acid is 0.585.
Learn more about pH at: brainly.com/question/15289741
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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>
