You have a flask of liquid. You set the flask down near a window. A beam of

) Do you think the pH of 1,0 M tri-methyl ammonium (CH3)3NH+, pKa = 9.80, will be higher or lower than that of 1.0 M phenol, C6H

Elanso [62]

Answer:

1. The pH of 1.0 M trimethyl ammonium (pH = 1.01) is lower than the pH of 0.1 M phenol (5.00).

2. The difference in pH values is 4.95.

Explanation:

1. The pH of a compound can be found using the following equation:

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

First, we need to find [H₃O⁺] for trimethyl ammonium and for phenol.

Trimethyl ammonium:

We can calculate [H₃O⁺] using the Ka as follows:

(CH₃)₃NH⁺ + H₂O → (CH₃)₃N + H₃O⁺

1.0 - x x x

$Ka = \frac{[(CH_{3})_{3}N][H_{3}O^{+}]}{[(CH_{3})_{3}NH^{+}]}$

$10^{-pKa} = \frac{x*x}{1.0 - x}$

$10^{-9.80}(1.0 - x) - x^{2} = 0$

By solving the above equation for x we have:

x = 0.097 = [H₃O⁺]

$pH = -log([H_{3}O^{+}]) = -log(0.097) = 1.01$

Phenol:

C₆H₅OH + H₂O → C₆H₅O⁻ + H₃O⁺

1.0 - x x x

$Ka = \frac{[C_{6}H_{5}O^{-}][H_{3}O^{+}]}{[C_{6}H_{5}OH]}$

$10^{-10} = \frac{x^{2}}{1.0 - x}$

$1.0 \cdot 10^{-10}(1.0 - x) - x^{2} = 0$

Solving the above equation for x we have:

x = 9.96x10⁻⁶ = [H₃O⁺]

$pH = -log([H_{3}O^{+}]) = -log(9.99 \cdot 10^{-6}) = 5.00$

Hence, the pH of 1.0 M trimethyl ammonium is lower than the pH of 0.1 M phenol.

2. The difference in pH values for the two acids is:

$\Delta pH = pH_{C_{6}H_{5}OH} - pH_{(CH_{3})_{3}NH^{+}} = 5.00 - 1.01 = 4.95$

Therefore, the difference in pH values is 4.95.

I hope it helps you!

7 0

4 years ago

A student measures the mass of a 6.0 cm3 block of brown sugar to be 10.0 g. What is the density of the brown sugar?

Alex17521 [72]

Answer:

1.67g/cm3

Explanation:

The formula for density is $d=\frac{m}{v}$ . The m variable stands for mass and the v variable stands for volume.

The mass of the brown sugar is 10.0g and the volume is 6.0cm3, so we can plug those values into the equation.

$d=\frac{10g}{6cm^{3} }$

$d=1.67$ $\frac{g}{cm^{3} }$

Rounded to 3 significant figures, the density of the block of brown sugar is 1.67 g/cm3. If the mass is in grams and the volume is in cm3, the unit for the final answer is $\frac{g}{cm^{3} }$ (grams per centimetres cubed).

8 0

3 years ago

Read 2 more answers

calculate the volume occupied by 10g of propane gas, under normal conditions of temperature and pressure

andriy [413]

Answer:

5.5 L

Explanation:

First we convert 10 g of propane gas (C₃H₈) to moles, using its molar mass:

10 g ÷ 44 g/mol = 0.23 mol

Then we use the PV=nRT formula, where:

P = 1 atm & T = 293 K (This are normal conditions of T and P)

n = 0.23 mol

R = 0.082 atm·L·mol⁻¹·K⁻¹

V = ?

1 atm * V = 0.23 mol * 0.082 atm·L·mol⁻¹·K⁻¹ * 293 K

V = 5.5 L

3 0

3 years ago

The following reaction shows the products when sulfuric acid and aluminum hydroxide react.

Elden [556K]

The correct answer is approximately 11.73 grams of sulfuric acid.

The theoretical yield of water from Al(OH)3 is lower than that of H₂SO₄. As a consequence, Al(OH)3 is the limiting reactant, H₂SO₄ is in excess.

The balanced equation is:

2Al(OH)₃ + 3H₂SO₄ ⇒ Al₂(SO₄)₃ + 6H₂O

Each mole of Al(OH)3 corresponds to 3/2 moles of H₂SO₄. The molecular mass of Al(OH)3 is 78.003 g/mol. There are 15/78.003 = 0.19230 moles of Al(OH)3 in the five grams of Al(OH)3 available. Al(OH)3 is in limiting, which means that all 0.19230 moles will be consumed. Accordingly, 0.19230 × 3/2 = 0.28845 moles of H₂SO₄ will be consumed.

The molar mass of H₂SO₄ is 98.706 g/mol. The mass of 0.28845 moles of H₂SO₄ is 0.28845 × 98.706 = 28.289 g

40 grams of sulfuric acid is available, out of which 28.289 grams is consumed. The remaining 40-28.289 = 11.711 g is in excess, which is closest to the first option, that is, 11.73 grams of H₂SO₄.

6 0

3 years ago

If a combination reaction takes place between potassium and chlorine what is the product

zepelin [54]

It would be potassium chloride or kcl

4 0

3 years ago