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kirill115 [55]
3 years ago
10

What statement is an accurate comparison of the properties of compounds and the properties of the elements that the compounds co

ntain?
A.
They are always identical.







B.
They are different.







C.
They are similar.







D.
They represent opposite characteristics.
Chemistry
2 answers:
Darya [45]3 years ago
6 0
When elements are combined to form compounds, they undergo chemical change in which the properties of the composing elements are always different from the compounds formed after reaction. The change is not only applicable to the chemical properties of the elements but also the physical states of the substances. Answer then is B.
Sedbober [7]3 years ago
4 0

Answer is: B. They are different.

For examle, balanced chemical reaction of forming water from hydrogen and oxygen:

2H₂ + O₂ → 2H₂O.

During chemical reaction no particles are created or destroyed, the atoms are simply rearranged from the reactants to the products.

Oxygen (element) has boiling point of -183°C and hydrogen has boiling point of -253°C. In this chemical change water (compound) is produced and it has new boiling point, boiling point of water is 100°C.

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An unknown compound contains only carbon, hydrogen, and oxygen (CxHyOz). Combustion of 2.50g of this compound produced 3.67g of
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4 0
3 years ago
Benzoic acid is a natural fungicide that naturally occurs in many fruits and berries. The sodium salt of benzoic acid, sodium be
AlexFokin [52]

Answer:

a. pH = 2.52

b. pH = 8.67

c. pH = 12.83

Explanation:

The equation of the titration between the benzoic acid and NaOH is:

C₆H₅CO₂H + OH⁻ ⇄ C₆H₅CO₂⁻ + H₂O    (1)

a. To find the pH after the addition of 20.0 mL of NaOH we need to find the number of moles of C₆H₅CO₂H and NaOH:

\eta_{NaOH} = C*V = 0.250 M*0.020 L = 5.00 \cdot 10^{-3} moles

\eta_{C_{6}H_{5}CO_{2}H}i = C*V = 0.300 M*0.050 L = 0.015 moles  

From the reaction between the benzoic acid and NaOH we have the following number of moles of benzoic acid remaining:

\eta_{C_{6}H_{5}CO_{2}H} = \eta_{C_{6}H_{5}CO_{2}H}i - \eta_{NaOH} = 0.015 moles - 5.00 \cdot 10^{-3} moles = 0.01 moles

The concentration of benzoic acid is:

C = \frac{\eta}{V} = \frac{0.01 moles}{(0.020 + 0.050) L} = 0.14 M

Now, from the dissociation equilibrium of benzoic acid we have:

C₆H₅CO₂H + H₂O ⇄ C₆H₅CO₂⁻ + H₃O⁺  

0.14 - x                            x                x

Ka = \frac{[C_{6}H_{5}CO_{2}^{-}][H_{3}O^{+}]}{[C_{6}H_{5}CO_{2}H]}

Ka = \frac{x*x}{0.14 - x}

6.5 \cdot 10^{-5}*(0.14 - x) - x^{2} = 0   (2)  

By solving equation (2) for x we have:          

x = 0.0030 = [C₆H₅CO₂⁻] = [H₃O⁺]

Finally, the pH is:

pH = -log([H_{3}O^{+}]) = -log (0.0030) = 2.52

b. At the equivalence point, the benzoic acid has been converted to its conjugate base for the reaction with NaOH so, the equilibrium equation is:

C₆H₅CO₂⁻ + H₂O  ⇄  C₆H₅CO₂H + OH⁻     (3)

The number of moles of C₆H₅CO₂⁻ is:

\eta_{C_{6}H_{5}CO_{2}^{-}} = \eta_{C_{6}H_{5}CO_{2}H}i = 0.015 moles

The volume of NaOH added is:

V = \frac{\eta}{C} = \frac{0.015 moles}{0.250 M} = 0.060 L

The concentration of C₆H₅CO₂⁻ is:

C = \frac{\eta}{V} = \frac{0.015 moles}{(0.060 L + 0.050 L)} = 0.14 M

From the equilibrium of equation (3) we have:

C₆H₅CO₂⁻ + H₂O  ⇄  C₆H₅CO₂H + OH⁻  

0.14 - x                              x               x

Kb = \frac{[C_{6}H_{5}CO_{2}H][OH^{-}]}{[C_{6}H_{5}CO_{2}^{-}]}

(\frac{Kw}{Ka})*(0.14 - x) - x^{2} = 0

(\frac{1.00 \cdot 10^{-14}}{6.5 \cdot 10^{-5}})*(0.14 - x) - x^{2} = 0

By solving the equation above for x, we have:

x = 4.64x10⁻⁶ = [C₆H₅CO₂H] = [OH⁻]

The pH is:

pOH = -log[OH^{-}] = -log(4.64 \cdot 10^{-6}) = 5.33

pH = 14 - pOH = 14 - 5.33 = 8.67

     

c. To find the pH after the addition of 100 mL of NaOH we need to find the number of moles of NaOH:

\eta_{NaOH}i = C*V = 0.250 M*0.100 L = 0.025 moles

From the reaction between the benzoic acid and NaOH we have the following number of moles remaining:                              

\eta_{NaOH} = \eta_{NaOH}i - \eta_{C_{6}H_{5}CO_{2}H} = 0.025 moles - 0.015 moles = 0.010 moles                          

The concentration of NaOH is:

C = \frac{\eta}{V} = \frac{0.010 moles}{0.100 L + 0.050 L} = 0.067 M

Therefore, the pH is given by this excess of NaOH:         

pOH = -log([OH^{-}]) = -log(0.067) = 1.17

pH = 14 - pOH = 12.83

I hope it helps you!    

4 0
2 years ago
A chemical change involves the formation or the breaking of bonds. true or false
dusya [7]
The answer would be true.
5 0
2 years ago
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