Answer:
You require 12.8mL of the 0.500M C₂H₃O₂Na and 7.2mL of the 0.500M C₂H₃O₂H
Explanation:
It is possible to obtain pH of a weak acid using H-H equation:
pH = pKa + log₁₀ [A⁻] / [HA]
For the buffer of acetic acid/acetate, the equation is:
pH = pKa + log₁₀ [C₂H₃O₂Na] / [C₂H₃O₂H]
Replacing:
5.00 = 4.75 + log₁₀ [C₂H₃O₂Na] / [C₂H₃O₂H]
1.7783 = [C₂H₃O₂Na] / [C₂H₃O₂H] (1)
Buffer strength is the concentration of the buffer, that means:
0.1M = [C₂H₃O₂Na] + [C₂H₃O₂H] (2)
Replacing (2) in (1):
1.7783 = 0.1M - [C₂H₃O₂H] / [C₂H₃O₂H]
1.7783 [C₂H₃O₂H] = 0.1M - [C₂H₃O₂H]
2.7783 [C₂H₃O₂H] = 0.1M
[C₂H₃O₂H] = 0.036MAlso:
[C₂H₃O₂Na] = 0.1M - 0.036M
[C₂H₃O₂Na] = 0.064MThe moles of both compounds you require is:
[C₂H₃O₂Na] = 0.1L × (0.064mol / L) = 0.0064moles
[C₂H₃O₂H] = 0.1L × (0.036mol / L) = 0.0036moles
Your stock solutions are 0.500M, thus, volume of both solutions you require is:
[C₂H₃O₂Na] = 0.0064moles × (1L / 0.500M) = 0.0128L = 12.8mL
[C₂H₃O₂H] = 0.0036moles × (1L / 0.500M) = 0.0072mL = 7.2mL
You require 12.8mL of the 0.500M C₂H₃O₂Na and 7.2mL of the 0.500M C₂H₃O₂HErbium metal (Er) can be prepared by reacting erbium(III) fluoride with magnesium; the other product is magnesium fluoride. Write and balance the equation.
Answer:
2ErF3 + 3Mg → 2Er + 3MgF2
Explanation:
Erbium metal is a member of the lanthaniod series. It reacts with halogens directly to yield erbium III halides such as erbium III chloride, Erbium III fluoride etc.
Erbium metal (Er) can be prepared by reacting erbium(III) fluoride with magnesium; the products are erbium metal and magnesium fluoride. This is a normal redox process in which the Erbium metal is reduced while the magnesium is oxidized. The balanced reaction equation of this process is; 2ErF3 + 3Mg → 2Er + 3MgF2
If the particles of matter that make up a substance are relatively far apart and can move freely, the substance is in what state?
gaseous
liquid
solid
Answer:
Gaseous
Explanation:
Gasses can move freely and do not form the shape of their containers
Liquids are more free than solids, but they conform to the shape of their container
Solids are not free
How fast are the atoms moving if the temperature of a gas is cold?
A. very, very slowly
B. they are stagnant
C. very, very quickly
Answer:
i think option a is correct answer because when there is low temperature then the kinetic enegry will be very less and the atoms moves very slowly.
Answer:
A. very, very slowly
Explanation:
A is the answer because atoms will move faster in hot gas than in cold gas.
The substance nitrogen has the following properties: normal melting point: 63.2 K normal boiling point: 77.4 K triple point: 0.127 atm, 63.1 K critical point: 33.5 atm, 126.0 K At temperatures above 126 K and pressures above 33.5 atm, N2 is a supercritical fluid . N2 does not exist as a liquid at pressures below atm. N2 is a _________ at 16.7 atm and 56.5 K. N2 is a _________ at 1.00 atm and 73.9 K. N2 is a _________ at 0.127 atm and 84.0 K.
Answer:
- N2 does not exist as a liquid at pressures below 0.127 atm.
- N2 is a solid at 16.7 atm and 56.5 K.
- N2 is a liquid at 1.00 atm and 73.9 K
- N2 is a gas at 0.127 atm and 84.0 K.
Explanation:
Hello,
At first, we organize the information:
- Normal melting point: 63.2 K.
- Normal boiling point: 77.4 K.
- Triple point: 0.127 atm and 63.1 K.
- Critical point: 33.5 atm and 126.0 K.
In such a way:
- N2 does not exist as a liquid at pressures below 0.127 atm: that is because below this point, solid N2 exists only (triple point).
- N2 is a solid at 16.7 atm and 56.5 K: that is because it is above the triple point, below the critical point and below the normal melting point.
- N2 is a liquid at 1.00 atm and 73.9 K: that is because it is above the triple point, below the critical point and below the normal boiling point.
- N2 is a gas at 0.127 atm and 84.0 K: that is because it is above the triple point temperature at the triple point pressure.
Best regards.
The rate at which two methyl radicals couple to form ethane is significantly faster than the rate at which two tert-butyl radicals couple. Offer two explanations for this observation.
Answer:
1. stability factor
2. steric hindrance factor
Explanation:
stability of ethane is lesser to that of two tert-butyl, so ethane will be more reactive and faster.
ethane is less hindered and more reactive, while two tert-butyl is more hindered and less reactive
When 75.5 grams of phosphorus pentachloride react with an excess of water, as shown in the unbalanced chemical equation below, how many moles of hydrochloric acid will be produced? Please show all your work for the calculations for full credit. PCl5 + H2O --> H3PO4 + HC
Answer:
Explanation: M(PCL5)= 31 + 5(35.5)
=208.5g/mol
M(H20)= 18g/mol
n(PCL5) = 75.5÷208.5
= 0.362mol
n(HCl)/n(PCL5)= 5/1
n(HCl)= 5×0.362
=1.81mol of HCl
If the vinegar were measured volumetrically (e.g., a pipet), what additional piece of data would be needed to complete the calculations for the experiment?
Answer:
the density if vinegar will also be needed
Explanation:
Because this is an experiment of volumetric analysis
A 5.00 gram sample of an oxide of lead PbxOy contains 4.33 g of lead. Determine simplest formula for the compund
Answer: The empirical formula is [tex]PbO_2[/tex]
Explanation:
Mass of Pb = 4.33 g
Mass of O = (5.00-4.33) g = 0.67 g
Step 1 : convert given masses into moles
Moles of Pb =[tex]\frac{\text{ given mass of Pb}}{\text{ molar mass of Pb}}= \frac{4.33g}{207g/mole}=0.021moles[/tex]
Moles of O =[tex]\frac{\text{ given mass of O}}{\text{ molar mass of O}}= \frac{0.67g}{16g/mole}=0.042moles[/tex]
Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.
For Pb = [tex]\frac{0.021}{0.021}=1[/tex]
For O = [tex]\frac{0.042}{0.021}=2[/tex]
The ratio of Pb O= 1: 2
Hence the empirical formula is [tex]PbO_2[/tex]
A 1.00 liter solution contains 0.31 M sodium acetate and 0.40 M acetic acid. If 0.100 moles of barium hydroxide are added to this system, indicate whether the following statements are TRUE or FALSE . (Assume that the volume does not change upon the addition of barium hydroxide.)
a. The number of moles of CH3COOH will remain the same.
b. The number of moles of CH3COO- will increase.
c. The equilibrium concentration of H3O+ will decrease.
d. The pH will decrease.
e. The ratio of [CH3COOH] / [CH3COO-] will remain the same.
Answer and Explanation:
The buffer solution is composed by sodium acetate (CH₃COONa) and acetic acid (CH₃COOH). Thus, CH₃COOH is the weak acid and CH₃COO⁻ is the conjugate base, derived from the salt CH₃COONa.
If we add a strong base, such as barium hydroxide, Ba(OH)₂, the base will dissociate completely to give OH⁻ ions, as follows:
Ba(OH)₂ ⇒ Ba²⁺ + 2 OH⁻
The OH⁻ ions will react with the acid (CH₃COOH) to form the conjugate base CH₃COO⁻.
Initial number of moles of CH₃COOH = 0.40 mol/L x 1 L = 0.40 mol
Initial number of moles of CHCOO⁻= 0.31 mol/L x 1 L = 0.31 mol
moles of OH- added: 2 OH-/mol x 0.100 mol/L x 1 L = 0.200 OH-
According to this, the following are the answers to the sentences:
a. The number of moles of CH₃COOH will remain the same ⇒ FALSE
The number of moles of CH₃COOH will decrease, because they will react with OH⁻ ions
b. The number of moles of CH₃COO⁻ will increase ⇒ TRUE
Moles of CH₃COO⁻ will be formed from the reaction of the acid (CH₃COOH) with the base (OH⁻ ions)
c. The equilibrium concentration of H₃O⁺ will decrease ⇒ FALSE
The equilibrium concentration of OH⁻ is increased
d. The pH will decrease⇒ FALSE
pKa for acetic acid is 4.75. We add the moles of base to the acid concentration and we remove the same number of moles from the conjugate base in the Henderson-Hasselbach equation to calculate pH:
[tex]pH= pKa + log \frac{[conjugate base + base]}{[acid - base]}[/tex]
pH = 4.75 + log (0.31 mol + 0.20 mol)/(0.40 mol - 0.20 mol) = 5.15
Thus, the pH will increase.
Given the information below, which is more favorable energetically, the oxidation of succinate to fumarate by NAD+ or by FAD? Fumarate + 2H+ + 2e- → Succinate E°´ = 0.031 V NAD+ + 2H+ + 2e- → NADH + H+ E°´ = -0.320 FAD + 2H+ + 2e- → FADH2 E°´ = -0.219
Answer:
Oxidation by FAD
Explanation:
1. Oxidation by NAD⁺
Succinate ⇌ Fumarate + 2H⁺ + 2e⁻; E°´ = -0.031 V
NAD⁺ + 2H⁺ + 2e⁻ ⇌ NADH + H⁺; E°´ = -0.320 V
Succinate + NAD⁺ ⇌ Fumarate + NADH + H⁺; E°' = -0.351 V
2. Oxidation by FAD
Succinate ⇌ Fumarate + 2H⁺ + 2e⁻; E°´ = -0.031 V
FAD + 2H⁺ + 2e⁻ ⇌ FADH₂; E°´ = -0.219 V
Succinate + FADH₂ ⇌ Fumarate + FAD; E°' = -0.250 V
Neither reaction is energetically favourable, but FAD has a more positive half-cell potential.
FAD is the stronger oxidizing agent.
The oxidation by FAD has a more positive cell potential, so it is more favourable energetically.