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Marta_Voda [28]
1 year ago
9

you wish to prepare an hc2h3o2 buffer with a ph of 5.44. if the pka of the acid is 4.74, what ratio of c2h3o2-/hc2h3o2 must you

use?
Chemistry
1 answer:
vazorg [7]1 year ago
5 0

The ratio of buffer C₂H₃O₂ /HC₂H₃O₂  must you use are1:0.199 or 10:2

the ratio of buffer C₂H₃O₂ /HC₂H₃O₂  can be calculate using the Henderson-Hasselbalch Equation which relates the pH to the measure of acidity pKa. The equation is given as:

pH = pKa + log ([base]/[acid]

Where,

[base] = concentration of C₂H₃O₂in molarity or moles

[acid] = concentration of HC₂H₃O₂ in molarity or moles

For the sake of easy calculation, allow us to assume that:

[base] =1

[acid] = x

Therefore using equation 1,

5.44 = 4.74 + log (1 / x)

log [base / acid] = 0.7

1 / x = 5.0118

x = 0.199

The required ratio of buffer C₂H₃O₂ /HC₂H₃O₂ is 1:0.199 or 10:2

learn more about buffer ratio here brainly.com/question/4342532

#SPJ4

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A sample of gas contains 0.1900 mol of CO(g) and 0.1900 mol of NO(g) and occupies a volume of 22.0 L. The following reaction tak
worty [1.4K]

Answer:

V₂ = 16.5 L

Explanation:

To solve this problem we use <em>Avogadro's law, </em>which applies when temperature and pressure remain constant:

V₁/n₁ = V₂/n₂

In this case, V₁ is 22.0 L, n₁ is [mol CO + mol NO], V₂ is our unknown, and n₂ is [mol CO₂ + mol N₂].

  • n₁ = mol CO + mol NO = 0.1900 + 0.1900 = 0.3800 mol

<em>We use the reaction to calculate n₂</em>:

2CO(g) + 2NO(g) → 2CO₂(g) + N₂(g)

  • mol CO₂:

0.1900 mol CO * \frac{2molCO_{2}}{2molCO} = 0.1900 mol CO₂

  • mol N₂:

0.1900 mol NO * \frac{1molN_{2}}{2molNO} = 0.095 mol N₂

  • n₂ = mol CO₂ + mol N₂ = 0.1900 + 0.095 = 0.2850 mol

Calculating V₂:

22.0 L / 0.3800 mol = V₂ / 0.2850 mol

V₂ = 16.5 L

3 0
3 years ago
Calculate the volume of 12.0 g of helium at 100°c and 1.2 atm
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Volume = nRT/P

n = number of particles (moles)

R = universal gas constant (0.0821)

T = temperature (Kelvin)

P = pressure (atm)

(Assuming you have 1 mole of Helium in a chemical reaction) We would need to convert grams to moles: 12.0g He x 1 mol He/4 molar mass of He = 3 mol He

Convert Celsius to Kelvin: 100*C + 273.15 = 373.15 K

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When an unknown amine reacts with an unknown acid chloride, an amide with a molecular mass of 163 g/mol (M = 163 m/z) is formed.
DanielleElmas [232]

Answer:

            As the molecular mass of given amine is 163 g/mol (a odd number) it means that this compound contains a odd number of Nitrogen atoms. We will first apply Rule of Thirteen to get the molecular formula.

Rule of Thirteen:

First divide the parent peak value by 13 as,

                = 163 ÷ 13

                = 12.53

Now, multiply 13 by 12,

                = 13 × 12 (here, 12 specifies number of carbon atoms)

                = 156

Now subtract 156 from 163,

                = 163 - 156

                = 7

Add 7 into 12,

                = 7 + 12

                = 19 (hydrogen atoms)

So, the rough formula we have is,

                                                       C₁₂H₁₉

Now, add one Nitrogen atom to above formula and subtract one Carbon and 2 Hydrogen atoms as these numbers are equal to atomic mass of Nitrogen atom as,

                C₁₂H₁₉   -------N-------->    C₁₁H₁₇N

Also, as shown in ¹³C-NMR there is one peak around 180 ppm and the peak at 1661 cm⁻¹ in IR spectrum is characteristic to carbonyl group hence, we will add one oxygen atom to the chemical formula accordingly. i.e.

                C₁₁H₁₇N   -------O-------->    C₁₀H₁₃NO

Molecular Formula: C₁₀H₁₃NO

Also,

In NMR the the four peaks around 120 ppm are assigned to a mono substituted benzene ring.

The absence of IR peak above 3200 cm⁻³ also confirms that the amine is tertiary in nature and there is no hydrogen attached to the nitrogen atom.

It can be observed that the peaks in upfield are duplicating. This can be due to the presence of rotamers of said compound.

The most plausible structure for given data is shown below, and the resonance structure along with rotamers are also shown.

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