Balanced equation for solid magnesium oxide

A compound has a pka of 7.4. to 100 ml of a 1.0 m solution of this compound at ph 8.0 is added 30 ml of 1.0 m hydrochloric acid.

Monica [59]

Using the Henderson-Hasselbalch equation on the solution before HCl addition: pH = pKa + log([A-]/[HA]) 8.0 = 7.4 + log([A-]/[HA]); [A-]/[HA] = 4.0. (equation 1) Also, 0.1 L * 1.0 mol/L = 0.1 moles total of the compound. Therefore, [A-] + [HA] = 0.1 (equation 2) Solving the simultaneous equations 1 and 2 gives: A- = 0.08 moles AH = 0.02 moles Adding strong acid reduces A- and increases AH by the same amount. 0.03 L * 1 mol/L = 0.03 moles HCl will be added, soA- = 0.08 - 0.03 = 0.05 moles AH = 0.02 + 0.03 = 0.05 moles Therefore, after HCl addition, [A-]/[HA] = 0.05 / 0.05 = 1.0 Resubstituting into the Henderson-Hasselbalch equation: pH = 7.4 + log(1.0) = 7.4, the final pH.

6 0

3 years ago

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From what you've learned so far, how is molecular structure related to smell?

Sidana [21]

Answer:

Changing the shape of the molecules that create fragrances in a flower or fruit may influence our perception of their smell. The reaction pattern produced, olfactory code, is sent as a signal to the brain, which which is how you smell things.

Explanation:

7 0

3 years ago

Use bond energies to determine δhrxn for the reaction between ethane and chlorine. ch3ch3(g)+cl2(g)→ch3ch2cl(g)+hcl(g)

STALIN [3.7K]

For this reaction to proceed, the following bond breaking should occur:

*one C-H bond
* one Cl-Cl bond

After, the following bond formations should occur:
*one C-Cl bond
*one H-Cl bond

Now, add the bond energies for the respective bond energies which can be found in the attached picture. For bond formations, energy is negative. For bond breaking, energy is positive.

ΔHrxn = (1)(413) + (1)(242) + 1(-328) + 1(-431) = <em>-104 kJ</em>

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3 years ago

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A mixture of XO2 (P = 3.00 atm) and O2 (P = 1.00 atm) is placed in a container. This elementary reaction takes place at 27 °C: 2

sukhopar [10]

Answer:

a) $\triangle G^{0} = 7.31 kJ/mol$

b) $K_{-1} = 0.0594 m^{-1} s^{-1}$

Explanation:

Equation of reaction:

$2 XO_{2} (g) + O_{2} (g) \rightleftharpoons 2XO_{3} (g)$

Initial pressure 3 1 0

Pressure change 2P 1P 2P

Total pressure = (3-2P) + (1-P) + (2P)

Total Pressure = 3.75 atm

(3-2P) + (1-P) + (2P) = 3.75

4 - P = 3.75

P = 4 - 3.75

P = 0.25 atm

Let us calculate the pressure of each of the components of the reaction:

Pressure of XO2 = 3 - 2P = 3 - 2(0.25)

Pressure of XO2 =2.5 atm

Pressure of O2 = 1 - P = 1 -0.25

Pressure of O2 = 0.75 atm

Pressure of XO3 = 2P = 2 * 0.25

Pressure of XO3 = 0.5 atm

From the reaction, equilibrium constant can be calculated using the formula:

$K_{p} = \frac{[PXO_{3}] ^{2} }{[PXO_{2}] ^{2}[PO_{2}] }$

$K_{p} = \frac{0.5^2}{2.5^2 *0.75} \\K_{p} = 0.0533 = K_{eq}$

Standard free energy:

$\triangle G^{0} = - RT ln k_{eq} \\\triangle G^{0} = -(0.008314*300* ln0.0533)\\\triangle G^{0} = 7.31 kJ/mol$

b) value of k−1 at 27 °C, i.e. 300K

$K_{1} = 7.8 * 10^{-2} m^{-2} s^{-1}$

$K_{c} = K_{p}RT\\K_{c} = 0.0533* 0.0821 * 300\\K_{c} = 1.313 m^{-1}$

$K_{-1} = \frac{K_{1} }{K_{c} } \\K_{-1} = \frac{7.8 * 10^{-2} }{1.313 }\\K_{-1} = 0.0594 m^{-1} s^{-1}$

6 0

3 years ago

If 75.0 g of a liquid has a volume of 62.4 mL ,calculate the liquid’s density

Galina-37 [17]

<span>To calculate the density of a liquid, you have to first know that density is the amount of substance per unit of volume. In this specific question, density will be found with units of g/mL. Now, the density can be found by dividing the amount of liquid, 75.0g, by the volume, 62.4mL. Doing this we get: 75.0g/62.4mL= 1.2 g/mL as the density of the liquid.</span>

4 0

3 years ago