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3 years ago

Which weak acid would be best to use when preparing a buffer solution with a ph of 9.20?

Chemistry

2 answers:

Murrr4er [49]3 years ago

8 0

Answer:

The acid with about $Ka=6.31x10^{-10}$ .

Explanation:

Hello,

By assuming that the dissociation of a weak acid is:

$H^+X^-H^++X^-$

From the following equation one computes pKa as long as the concentration of the conjugate base equals the concentration of the acid for this buffered solution:

$pH=pKa+Log(\frac{[Conj.Base]}{[Acid]})\\pKa=pH-Log(1)\\pKa=pH=9.2\\Ka=10^{-9.2}\\Ka=6.31x10^{-10}$

So the acid will be that with a $Ka=6.31x10^{-10}$ or close since the possible answers are not shown.

Best regards.

wariber [46]3 years ago

6 0

Answer: -

6.31 x 10⁻¹⁰

Explanation: -

The equation for buffer pH

pH = pKa + log $\frac{Conjugate Base}{Acid}$

If [Conjugate base] = [Acid], then preparing a buffer is best.

Then pH = pKa

pKa = - log Ka

Ka = $10^{-9.2}$

= 6.31 x 10⁻¹⁰

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Sulfur undergoes combustion to yield sulfur trioxide by the following reaction equation:

12345 [234]

Answer:

Therefore, the amount of heat produced by the reaction of 42.8 g S = (-5.2965 × 10²) kJ = (-5.2965 × 10⁵) J

Explanation:

Given reaction: 2S + 3O₂ → 2 SO₃

Given: The enthalpy of reaction: ΔH = - 792 kJ

Given mass of S: w₂ = 42.8 g, Molar mass of S: m = 32 g/mol

In the given reaction, the number of moles of S reacting: n = 2

As, Number of moles: $n = \frac{mass\: (w_{1})}{molar\: mass\: (m)}$

∴ mass of S in 2 moles of S: $w_{1} = n \times m = 2\: mol \times 32\: g/mol = 64\: g$

Given reaction: 2S + 3O₂ → 2 SO₃

In this reaction, the limiting reagent is S

⇒ 2 moles S produces (- 792 kJ) heat.

or, 64 g of S produces (- 792 kJ) heat.

∴ 42.8 g of S produces (x) amount of heat

⇒ The amount of heat produced by 42.8 g S:

$x = \frac{(- 792\: kJ) \times 42.8\: g}{64\: g} = (-529.65)\: kJ$

$\Rightarrow x = (-5.2965 \times 10^{2})\: kJ = (-5.2965 \times 10^{5})\: J$

$(\because 1 kJ = 10^{3} J)$

Therefore, the amount of heat produced by the reaction of 42.8 g S = (-5.2965 × 10²) kJ = (-5.2965 × 10⁵) J

8 0

3 years ago

Can somebody help me, please!

zavuch27 [327]

Answer:

Rubidium-85=61.2

Rubidium-87=24.36

Atomic Mass=85.56 amu

Explanation:

To find the atomic mass, we must multiply the masses of the isotope by the percent abundance, then add.

Rubidium-85

This isotope has an abundance of 72%.

Convert 72% to a decimal. Divide by 100 or move the decimal two places to the left.

72/100= 0.72 or 72.0 --> 7.2 ---> 0.72

Multiply the mass of the isotope, which is 85, by the abundance as a decimal.

mass * decimal abundance= 85* 0.72= 61.2

Rubidium-85=61.2

Rubidium-87

This isotope has an abundance of 28%.

Convert 28% to a decimal. Divide by 100 or move the decimal two places to the left.

28/100= 0.28 or 28.0 --> 2.8 ---> 0.28

Multiply the mass of the isotope, which is 87, by the abundance as a decimal.

mass * decimal abundance= 87* 0.28= 24.36

Rubidium-87=24.36

Atomic Mass of Rubidium:

Add the two numbers together.

Rb-85 (61.2) and Rb-87 (24.36)

61.2+24.36=85.56 amu

4 0

3 years ago

How many moles is 22.4 liters of oxygen gas at standard temperature and pressure represent

makkiz [27]

Answer:

So 1 mole

Explanation:

PV = nRT

P = Pressure atm

V = Volume L

n = Moles

R = 0.08206 L·atm·mol−1·K−1.

T = Temperature K

standard temperature = 273K

standard pressure = 1 atm

22.4 liters of oxygen

Ok so we have

V = 22.4

P = 1 atm

PV = nRT

n = PV/RT

n = 22.4/(0.08206 x 273)

n = 22.4/22.40

n = 1 mole

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3 years ago

2. What effect does adding a neutron have on the atom’s mass?

anyanavicka [17]

It does, however, change the mass of the nucleus. Adding or removing neutronsfrom the nucleus are how isotopes are created. Protons carry a positive electrical charge and they alone determine the charge of the nucleus.

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3 years ago

A price support is a type of _____. price fixing price floor rent control price ceiling

Oksi-84 [34.3K]

Subsidy

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3 years ago

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