CHEMISTRY COLLEGE

What happens when this match is struck against the side of the match box?

Answers

Answer 1

Answer:

Answer:

It lights on fire

Explanation:

The friction sparks the match causing it to go on fire.

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You have 0.500 L of an 0.250 M acetate buffer solution (i.e. [HC₂H₃O₂] + [C₂H₃O₂⁻] = 0.250 M) at pH 3.50. How many mL of 1.000 M NaOH must you add in order to change the pH to 5.07? Acetic acid has a pKa of 4.74.

Answers

Answer:

80mL of 1.00M NaOH

Explanation:

Using H-H equation, we can determine oH of a buffer as acetate buffer. First, we need to determine amount of acetate ion and acetic acid at pH 3.50 and 5.07. Then, with the reaction of NaOH with acetic acid we can find the amount of 1.00M NaOH that must be added:

At pH 3.50:

pH = pka + log [C₂H₃O₂⁻] / [HC₂H₃O₂]

3.50 = 4.74 + log [C₂H₃O₂⁻] / [HC₂H₃O₂]

0.057544 = [C₂H₃O₂⁻] / [HC₂H₃O₂] (1)

Using and replacing in (1):

[HC₂H₃O₂] + [C₂H₃O₂⁻] = 0.250 M

[HC₂H₃O₂] + 0.057544[HC₂H₃O₂] = 0.250 M

1.057544 [HC₂H₃O₂] = 0.250M

[HC₂H₃O₂] = 0.2364M * 0.500L = 0.1182 moles of acetic acid at first pH

At pH 5.07:

pH = pka + log [C₂H₃O₂⁻] / [HC₂H₃O₂]

5.07 = 4.74 + log [C₂H₃O₂⁻] / [HC₂H₃O₂]

2.13796= [C₂H₃O₂⁻] / [HC₂H₃O₂] (1)

Using and replacing in (1):

[HC₂H₃O₂] + 2.13796[HC₂H₃O₂] = 0.250 M

3.13796 [HC₂H₃O₂] = 0.250M

[HC₂H₃O₂] = 0.07967M * 0.500L = 0.0398 moles of acetic acid at first pH

Now, NaOH reacts with HC₂H₃O₂ as follows:

NaOH + HC₂H₃O₂ → NaC₂H₃O₂ + H₂O

As moles of acetic acid decreases from 0,1198 moles - 0,0398 moles = 0,08 moles of acetic acid are consumed = 0,08 moles of NaOH

0,08 mol NaOH * (1L / 1mol) = 0,08L of 1.00M NaOH =

80mL of 1.00M NaOH

Consider four small molecules, A–D, which have the following binding affinities for a specific enzyme (these numbers are the equilibrium constants Kd for the dissociation of the enzyme/molecule complex). Which binds most tightly to the enzyme? Which binds least tightly?A) 4.5 μM

B) 13 nM

C) 8.2 pM

D) 6.9 mM

Answers

Answer:

Binding affinity measures the strength of the interaction between a molecule to its ligand; it is expressed in terms of the equilibrium dissociation constant; and the higher value of this constant, the more weaker the binding between the molecule and the ligand is. On the other hand, small constans means that the interaction is tight. So "C" binds most tightly to the enzyme and "D" binds least tightly.

Write the equilibrium expression, calculate KEQ and then tell where the equilibrium lies: Fe (s) + O2 (g) ↔ Fe2O3 (s) In a 2.0 L Container At equilibrium: Fe = 1.0 mol O2 = 1.0 E-3 mol Fe2O3 = 2.0 mol

Answers

Answer:

The value of Keq is 4e-9. See the solution below

Explanation:

We need to balanced rhe equation and use the formula of the Keq

You are asked to prepare 500 mL 0.300 M500 mL 0.300 M acetate buffer at pH 4.904.90 using only pure acetic acid ( MW=60.05 g/mol,MW=60.05 g/mol, pKa=4.76), pKa=4.76), 3.00 M NaOH,3.00 M NaOH, and water. Answer the questions regarding the preparation of the buffer. 1. How many grams of acetic acid will be needed to prepare the 500 mL buffer? Note that the given concentration of acetate refers to the concentration of all acetate species in solution.

Answers

The quantity of acetic acid that is needed to prepare the 500 mL buffer is 9.0075 grams.

Given the following data:

Volume of acetate buffer = 500 mL to L = 0.5 L.

Molarity of acetate buffer = 0.300 M.

pH = 4.90.

MW = 60.05 g/mol.

pKa = 4.76.

How to calculate the mass of acetic acid.

First of all, we would write the equilibrium chemical reaction for acetate-acetic acid as follows:

$CH_3COOH \rightleftharpoons CH_3COO^(-)+H^+$

Next, we would calculate HA by applying Henderson-Hasselbalch equation:

$pH =pka+ log_(10) (A^-)/(HA)$

Where:

HA is acetic acid.

$A^-$ is acetate ion.

Substituting the given parameters into the formula, we have;

$4.90 =4.76+ log_(10) (A^-)/(HA)\n\n4.90 -4.76+ log_(10) (A^-)/(HA)\n\n(A^-)/(HA)=1.38\n\nA^- = 1.38[HA]$

For the concentration of both acids, we have:

$[HA]+[A^-]=0.300M\n\n[HA]+1.38[HA]=0.300M\n\n2.38[HA]=0.300M\n\nHA = 0.126$

For acetate ion:

$A^- = 1.38[HA] = 1.38 * 0.126\n\nA^- =0.174$

At a volume of 0.5 liters, we have:

$HA = 0.5 * 0.126\n\nHA = 0.063 \;moles$

$A^- = 0.5 * 0.174\n\nA^- =0.087 \;moles$

By stoichiometry:

Total moles = $0.063 + 0.087$ = 0.15 moles.

$Mass = number \;of \;moles * molar\;mass\n\nMass =0.15 * 60.05$

Mass = 9.0075 grams.

Read more on moles here: brainly.com/question/3173452

Answer:

You will need 9,0 g of acetic acid

Explanation:

The equilibrium acetate-acetic acid is:

CH₃COOH ⇄ CH₃COO⁻ + H⁺ pka = 4,76

Using Henderson-Hasselbalch you will obtain:

pH = pka + log₁₀ $([A^(-)])/([HA])$

Where HA is acetic acid and A⁻ is acetate ion

4,90 = 4,76 + log₁₀ $([A^(-)])/([HA])$

1,38 = $([A^(-)])/([HA])$ (1)

As acetate concentration is 0,300M:

0,300M = [HA] + [A⁻] (2)

Replacing (2) in (1):

[HA] = 0,126 M

And:

[A⁻] = 0,174 M

As you need to produce 500 mL:

0,5 L × 0,126 M = 0,063 moles of acetic acid

0,5 L × 0,174 M = 0,087 moles of acetate

To produce moles of acetate from acetic acid:

CH₃COOH + NaOH → CH₃COO⁻ + Na⁺ + H₂O

Thus, moles of acetate are equivalents to moles of NaOH and all acetates comes from acetic acid, thus:

0,087 moles of acetate + 0,063 moles of acetic acid ≡ 0,15 moles of acetic acid × $(60,05 g)/(1mol)$ = 9,0 g of acetic acid

I hope it helps!

Why does ammonium nitrate (NH4NO3) dissolve readily in water even though the dissolution process is endothermic by 26.4 kJ/mol? Why does ammonium nitrate (NH4NO3) dissolve readily in water even though the dissolution process is endothermic by 26.4 kJ/mol? The vapor pressure of the water decreases upon addition of the solute. The osmotic properties of the system lead to this behavior. The overall enthalpy of the system decreases upon addition of the solute. The overall entropy of the system increases upon dissolution of this strong electrolyte. The overall enthalpy of the system increases upon dissolution of this strong electrolyte.

Answers

Answer: Option (c) is the correct answer.

Explanation:

Entropy is defined as the degree of randomness that is present within the particles of a substance.

As $NH_(4)NO_(3)$ is ionic in nature. Hence, when it is added to water then it will readily dissociate into ammonium ions ( $NH^}{+}_(4)$ ) and nitrate ions ( $NO^(-)_(3)$ ).

Therefore, it means that ions of ammonium nitrate will be free to move from one place to another. Hence, there will occur an increase in entropy.

Thus, we can conclude that ammonium nitrate ( $NH_(4)NO_(3)$ ) dissolve readily in water even though the dissolution process is endothermic by 26.4 kJ/mol because the overall entropy of the system increases upon dissolution of this strong electrolyte.

The osmotic pressure, π, of a solution of glucose is 132 atm . find the molarity of the solution at 298 k.