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Aleks [24]
4 years ago
5

For the hypothetical reaction: A + B ---> 2C, write an expression that relates the disappearance of A and B to the appearance

of C. You can use the word delta instead of the symbol .
Chemistry
1 answer:
forsale [732]4 years ago
8 0

Answer:

Rate expression has been given below

Explanation:

According to the given equation, 1 molecule of A reacts with 1 molecule of B and produces 2 molecules of B at a time.

So, rate of disappearance of both A and B are one half of rate of appearance of B

Hence rate expression can be represented as:

Rate=\frac{-\Delta [A]}{\Delta t}=\frac{-\Delta [B]}{\Delta t}=\frac{1}{2}\frac{\Delta [C]}{\Delta t}

where \frac{-\Delta [A]}{\Delta t} is rate of disappearance of A, \frac{-\Delta [B]}{\Delta t} is rate of disappearance of B and \frac{\Delta [C]}{\Delta t} rate of appearance of C

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Calculate the pH of each of the following aqueous solutions. (Enter your answers to two decimal places.) (a) 10.0 mL deionized w
Anna007 [38]

Answer:

a. pH = 7.0

b. pH = 12.52

c. pH = 12.70

d. pH = 12.78

Explanation:

a. Deionized water has the [H⁺] of pure water = 1x10⁻⁷ (Kw = 1x10⁻¹⁴ = [H⁺][OH⁻] - [H⁺] = [OH⁻ -)

pH = -log[H⁺] = 7

b. Moles NaOH = 5x10⁻³L * (0.10mol / L) = 5x10⁻⁴moles OH⁻ / 0.015L = 0.0333M = [OH⁻]

<em>-Total volume = 10mL+5mL = 15mL = 0.015L</em>

pOH = -log[OH⁻] = 1.48

pH = 14-pOH

pH = 12.52

c. Moles NaOH = 0.010L * (0.10mol / L) = 1x10⁻³moles OH⁻ / 0.020L = 0.0500M = [OH⁻]

<em>-Total volume = 10mL+10mL = 20mL = 0.020L</em>

pOH = -log[OH⁻] = 1.30

pH = 14-pOH

pH = 12.70

d. Moles NaOH = 0.015L * (0.10mol / L) = 1.5x10⁻³moles OH⁻ / 0.025L = 0.060M = [OH⁻]

<em>-Total volume = 10mL+15mL = 25mL = 0.025L</em>

pOH = -log[OH⁻] = 1.22

pH = 14-pOH

pH = 12.78

4 0
3 years ago
A buffer contains 0.19 mol of propionic acid (C2H5COOH) and 0.26 mol of sodium propionate (C2H5COONa) in 1.20 L. You may want to
Snezhnost [94]

Explanation:

It is known that pK_{a} of propionic acid = 4.87

And, initial concentration of  propionic acid = \frac{0.19}{1.20}

                                                                       = 0.158 M

Concentration of sodium propionate = \frac{0.26}{1.20}[/tex]

                                                             = 0.216 M

Now, in the given situation only propionic acid and sodium propionate are present .

Hence,      pH = pK_{a} + log(\frac{[salt]}{[acid]})

                       = 4.87 + log \frac{0.216}{0.158}

                        = 4.87 + log (1.36)

                        = 5.00

  • Therefore, when 0.02 mol NaOH is added  then,

     Moles of propionic acid = 0.19 - 0.02

                                              = 0.17 mol

Hence, concentration of propionic acid = \frac{0.17}{1.20 L}

                                                                 = 0.14 M

and,      moles of sodium propionic acid = (0.26 + 0.02) mol

                                                                  = 0.28 mol

Hence, concentration of sodium propionic acid will be calculated as follows.

                        \frac{0.28 mol}{1.20 L}

                           = 0.23 M

Therefore, calculate the pH upon addition of 0.02 mol of NaOH as follows.

             pH = pK_{a} + log(\frac{[salt]}{[acid]})

                       = 4.87 + log \frac{0.23}{0.14}

                        = 4.87 + log (1.64)

                        = 5.08

Hence, the pH of the buffer after the addition of 0.02 mol of NaOH is 5.08.

  • Therefore, when 0.02 mol HI is added  then,

     Moles of propionic acid = 0.19 + 0.02

                                              = 0.21 mol

Hence, concentration of propionic acid = \frac{0.21}{1.20 L}

                                                                 = 0.175 M

and,      moles of sodium propionic acid = (0.26 - 0.02) mol

                                                                  = 0.24 mol

Hence, concentration of sodium propionic acid will be calculated as follows.

                        \frac{0.24 mol}{1.20 L}

                           = 0.2 M

Therefore, calculate pH upon addition of 0.02 mol of HI as follows.

             pH = pK_{a} + log(\frac{[salt]}{[acid]})

                       = 4.87 + log \frac{0.2}{0.175}

                        = 4.87 + log (0.114)

                        = 4.98

Hence, the pH of the buffer after the addition of 0.02 mol of HI is 4.98.

7 0
3 years ago
Equal numbers of moles of He(g), Ar(g), and Ne(g) are placed in a glass vessel at room temperature. If the vessel has a pinhole-
mrs_skeptik [129]

Answer:

VP as function of time => VP(Ar) > VP(Ne) > VP(He).

Explanation:

Effusion rate of the lighter particles will be higher than the heavier particles. That is, the lighter particles will leave the container faster than the heavier particles. Over time, the vapor pressure of the greater number of heavier particles will be higher than the vapor pressure of the lighter particles.

=> VP as function of time => VP(Ar) > VP(Ne) > VP(He).

Review Graham's Law => Effusion Rate ∝ 1/√formula mass.

4 0
3 years ago
Magnesium nitrate contains chemical bonds that are
SVETLANKA909090 [29]

Answer:

Magnesium Nitrate is an ionic compound formed by:

Magnesium;Mg+2 (<em>the cation)</em> and Nitrate NO−3 (<em>the poly-atomic anion)</em>.

In order for these two ions to bond, the charges must be equal and opposite. Therefore, it will take:

     <u>two -1 nitrate ions</u> <u>to balance the</u> <u>one +2 magnesium ion</u>.

3 0
2 years ago
A sample of sugar contains 1.505 × 1023 molecules of sugar. How many moles of sugar are present in the sample?
alisha [4.7K]
1 mole ------------ 6.02 x10²³ molecules
? mole ----------- 1.505 x10²³ molecules

1.505x10²³ / 6.02x10²³ => 0.25 moles

hope this helps!


4 0
3 years ago
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