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3 years ago

"In a test tube was add a piece of Zn and 2 mL of HCl (.1M). This reaction produced ZnCl2 and Hydrogen gas. This reactions is a:

Chemistry

1 answer:

omeli [17]3 years ago

4 0

Answer:

d. A single replacement reaction.

Explanation:

The zinc replaces the hydrogen in the hydrochloric acid.

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Which statement best describes the message of the poem? *

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3 years ago

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Water is being pumped from the bottom of a well 150 feet deep at a rate of 200 gal/hour into a vented storage tank 30 feet above

KengaRu [80]

Explanation:

As the given data is as follows.

Height, H = 150 feet

Heat gain = 30,000 BTU/hr, and Heat loss = 25000 BTU/hr

m = mass of water heated = 700 gallons = 5810 lbs

$C_{p}$ is the heat capacity of water = 1 BTU/lb $^{o}F$ (given)

$\Delta T$ = temperature difference = $120^{o}F - 35^{o}F$

Heat energy required to heat 700 gal can be calculated as follows:

Heat Required = $5810 lbs \times 1 BTU/lb^{o}F \times (120^{o}F - 35^{o}F)$

Thus, water rises till $120^{o}F$ .

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4 years ago

What is this thermodynamic?

VLD [36.1K]

Answer:

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3 years ago

Write the symbols of following elements. a. Phosphorus b. Bromine c. Silver d. Sodium

Vikki [24]

Answer:

a. P

b. Br

c. Ag

d. Na

Explanation:

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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

3 years ago