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Studentka2010 [4]
2 years ago
14

Two substances, A and Z, are to be identified. Substance A can not be broken down by a chemical change. Substance Z can be broke

n down by a chemical change. What can be concluded about these substances?
(1) Both substances are elements.
(2) Both substances are compounds.
(3) Substance A is an element and substance Z is a compound.
(4) Substance A is a compound and substance Z is an element.
Chemistry
1 answer:
I am Lyosha [343]2 years ago
4 0
The answer is 3). This is because elements are the simplest form of a substance, and cannot be broken down any further. Compounds on the other hand are much more complex than elements and can be broken down INTO elements. For example, Na, sodium, is an element and cannot be broken down further. H2O, water, is a compound and can be broken down into Hydrogen and Oxygen.
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What is the ph of a 0.25 m solution of c6h5nh2 given that its kb is 1.8 x 10-6?
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The pH a 0.25 m solution of C₆H₅NH₂ is equal to 3.13.

<h3>How do we calculate pH of weak base?</h3>

pH of the weak base will be calculate by using the Henderson Hasselbalch equation as:

pH = pKb + log([HB⁺]/[B])

pKb = -log(1.8×10⁻⁶) = 5.7

Chemical reaction for C₆H₅NH₂ is:

                          C₆H₅NH₂ + H₂O → C₆H₅NH₃⁺ + OH⁻

Initial:                     0.25                           0            0

Change:                    -x                             x             x

Equilibrium:        0.25-x                           x             x

Base dissociation constant will be calculated as:
Kb = [C₆H₅NH₃⁺][OH⁻] / [C₆H₅NH₂]

Kb = x² / 0.25 - x

x is very small as compared to 0.25, so we neglect x from that term and by putting value of Kb, then the equation becomes:

1.8×10⁻⁶ = x² / 0.25

x² = (1.8×10⁻⁶)(0.25)

x = 0.67×10⁻³ M = [C₆H₅NH₃⁺]

On putting all these values on the above equation of pH, we get

pH = 5.7 + log(0.67×10⁻³/0.25)

pH = 3.13

Hence pH of the solution is 3.13.

To know more about Henderson Hasselbalch equation, visit the below link:
brainly.com/question/13651361

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5 0
2 years ago
The osmotic pressure of a solution formed by dissolving 75.0 mg of aspirin (c9h8o4) in 0.250 l of water at 25 °c is ________ atm
Igoryamba

Answer is: the osmotic pressure of aqueous solution of aspirin is 0.0407 atm.

m(C₉H₈O₄) = 75 mg ÷ 1000 mg/g = 0.075 g.

n(C₉H₈O₄) = 0.075 g ÷ 180.16 g/mol.

n(C₉H₈O₄) = 0.000416 mol.

c(C₉H₈O₄) = 0.000416 mol ÷ 0.250 L.

c(C₉H₈O₄) = 0.00167 M; concentration of solution.

T(C₉H₈O₄) = 25°C = 298.15 K; temperature in Kelvins.

R = 0.08206 L•atm/mol•K; universal gas constant.

π = c(C₉H₈O₄) • T(C₉H₈O₄) • R.

π = 0.00167 mol/L • 298.15 K • 0.08206 L•atm/mol•K.

π = 0.0407 atm.


3 0
3 years ago
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