1. 21.67g/ml
2. aluminium
Explanation:
1. density = mass/volume
385.8/17.8= 21.67ml
2. 1g/ml=0.1g/cm^3
21.67g/ml = 2.167g/cm^3
..... substance is probably aluminium
1. the density of the metal is 21.67g/ml
2. This metal is most likely aluminum
The calculation is as follows;
1.
[tex]density = mass \div volume[/tex]
[tex]385.8\div 17.8= 21.67ml[/tex]
2.
1g/ml=0.1g/cm^3
So,
21.67g/ml = 2.167g/cm^3
Therefore, substance is probably aluminum
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Identify each reaction from the citric acid cycle as an oxidation‑reduction reaction, an esterification reaction, an amidation reaction, a hydrolysis reaction, a hydration reaction, or a dehydration reaction.
1. Which type of reaction occurs when succinyl-CoA is converted to succinate in the citric acid cycle?
2. Which type of reaction occurs when malate is converted to oxaloacetate in the citric acid cycle?
3. Which type of reaction occurs when aconitate is converted to isocitrate in the citric acid cycle?
Answer:
1. Oxidation-reduction and hydrolysis
2. Oxidation-reduction
3. Dehydration
Explanation:
Our options for each reaction are:
a) Oxidation‑reduction reaction
b) Esterification reaction
c) Amidation reaction
d) Hydrolysis reaction
c) Hydration reaction
f) Dehydration reaction
In reaction one the have the rupture of the S-CoA bond. This reaction takes place by the addition of a water molecule and the oxidation to a carboxylic acid group. So, for reaction 1 we will have an oxidation-reduction and a hydrolysis reaction.
For reaction 2, the functional group change from alcohol to a carboxylic acid. So, we have an oxidation-reduction reaction.
In the last reaction, we have the production of a double bond by the removal of water. With this in mind, we have a dehydration reaction.
See figure 1
I hope it helps
11. How did the solubility product constant Ksp of KHT in pure water compare to its solubility product constant Ksp of KHT in KCl solution? Are these results what you would expect? Why?
Answer:
Explanation:
KHT is a salt which ionises in water as follows
KHT ⇄ K⁺ + HT⁻
Solubility product Kw= [ K⁺ ] [ HT⁻ ]
product of concentration of K⁺ and HT⁻ in water
In KCl solution , the solubility product of KHT will be decreased .
In KCl solution , there is already presence of K⁺ ion in the solution . So
in the equation
[ K⁺ ] [ HT⁻ ] = constant
when K⁺ increases [ HT⁻ ] decreases . Hence less of KHT dissociates due to which its solubility decreases . It is called common ion effect . It is so because here the presence of common ion that is K⁺ in both salt to be dissolved and in solvent , results in decrease of solubility of the salt .
14. Which group of diamagnetic transition metals exhibits trends in density and melting points that don't match the same trends seen in
other groups?
A. Group 3
B. Group 12
C. Group 7
D. Group 11
Answer:
Group 12
Explanation:
Group 12 transition metals are diamagnetic. They behave properties that distinguish them. They naturally have twelve electrons hence their outermost shell is fully filled.
Transition metals have high densities which increases down the group. However, the increase in density of transition elements of group 12 varies with temperature at a rate that is quite different from other transition elements. Hence the differences in the value of melting points and density changes by only a very small amount as you come down group 12 compared to other groups of transition elements.
Example: One liter of saturated calcium fluoride
solution contains 0.0167 gram of CaFat 25°C.
Calculate the molar solubility of, and Ksp for, CaF2.
Answer:
[tex]Molar\ solubility=2.14x10^{-4}M[/tex]
[tex]Ksp=3.91x10^{-11}[/tex]
Explanation:
Hello,
In this case, given that 0.0167 grams of calcium fluoride in 1 L of solution form a saturated one, we can notice it is the solubility, therefore, the molar solubility is computed by using the molar mass of calcium fluoride (78.1 g/mol):
[tex]Molar\ solubility=\frac{0.0167gCaF_2}{1L}*\frac{1molCaF_2}{78.1gCaF_2} \\\\Molar\ solubility=2.14x10^{-4}M[/tex]
Next, since dissociation equation for calcium fluoride is:
[tex]CaF_2(s)\rightarrow Ca^{2+}(aq)+2F^-(aq)[/tex]
The equilibrium expression is:
[tex]Ksp=[Ca^{2+}][F^-]^2[/tex]
We can compute the solubility product by remembering that the concentration of both calcium and fluoride ions equals the molar solubility, thereby:
[tex]Ksp=(2.14x10^{-4})(2*2.14x10^{-4})^2\\\\Ksp=3.91x10^{-11}[/tex]
Regards.
How do protons and neutrons stay together in the nucleus
Explanation:
The nucleus of an atom is held together by the strong nuclear force that binds together protons and neutrons. Although the strong nuclear force is the strongest of the four fundamental forces, it acts only over very short - typically nuclear - distances. It binds together the protons and neutrons in the nucleus.
What is the molecular geometry of the underlined carbon atom in the following compound: HC-C-CH3?
U bent
tetrahedral
linear
trigonal planar
do none of the above
Answer:
linear
Explanation:
If we look at the compound, we will understand that the underlined carbon is in sp hybridization. Recall that sp hybridization leads to a C-C bond angle of about 180°. When two chemical species are at a bond angle if 180°, then the both bonding groups are found to be on a straight line.
Hence, for any carbon in a triply bonded state or indeed any carbon atom that is in an sp will exhibit a linear geometry according to the Valence Shell Electron Pair repulsion Theory since there are two electron pairs present.
Iron(II) is available to bond with chloride ion. How many of each type of ion will bond to form an ionic compound?
A) 3 iron(II), 1 chloride
B) 2 iron(II), 3 chloride
C) 2 iron(II), 1 chloride
D) 1 iron(II), 2 chloride
Answer:
D) 1 iron(II), 2 chloride
Explanation:
Iron II chloride is the compound; FeCl2. It is formed as follows, ionically;
Fe^2+(aq) + 2Cl^-(aq) -----> FeCl2
The formation of one mole of FeCl2 involves the reaction one mole of iron and two moles of chloride ions. This means that in FeCl2, the ratio of iron to chlorine is 1:2 as seen above.
Therefore there is one iron II ion and two chloride ions in each mole of iron II chloride, hence the answer.
8.670 mL + 9.87 mL=
Answer:18.4 ML
Explanation:
easy add
The Handbook of Chemistry and Physics gives solubilities of the following compounds in grams per 100 mL water. Because these compounds are only slightly soluble, assume that the volume does not change on dissolution and calculate the solubility product for each.
(a) BaSeO4, 0.0118 g/100 mL
(b) Ba(BrO3)2 H20, 0.30 g/100 mL
(c) NH4MgAsO4-6H20, 0.038 g/100 mL
(d) La2(MoOs)3, 0.00179 g/100 mL
Answer:
(a) [tex]Ksp=4.50x10^{-7}[/tex]
(b) [tex]Ksp=1.55x10^{-6}[/tex]
(c) [tex]Ksp=2.27x10^{-12}[/tex]
(d) [tex]Ksp=1.05x10^{-22}[/tex]
Explanation:
Hello,
In this case, given the solubility of each salt, we can compute their molar solubilities by using the molar masses. Afterwards, by using the mole ratio between ions, we can compute the concentration of each dissolved and therefore the solubility product:
(a) [tex]BaSeO_4(s)\rightleftharpoons Ba^{2+}(aq)+SeO_4^{2-}(aq)[/tex]
[tex]Molar\ solubility=\frac{0.0188g}{100mL} *\frac{1mol}{280.3g}*\frac{1000mL}{1L}=6.7x10^{-4}\frac{mol}{L}[/tex]
In such a way, as barium and selenate ions are in 1:1 molar ratio, they have the same concentration, for which the solubility product turns out:
[tex]Ksp=[Ba^{2+}][SeO_4^{2-}]=(6.7x10^{-4}\frac{mol}{L} )^2\\\\Ksp=4.50x10^{-7}[/tex]
(B) [tex]Ba(BrO_3)_2(s)\rightleftharpoons Ba^{2+}(aq)+2BrO_3^{-}(aq)[/tex]
[tex]Molar\ solubility=\frac{0.30g}{100mL} *\frac{1mol}{411.15g}*\frac{1000mL}{1L}=7.30x10^{-3}\frac{mol}{L}[/tex]
In such a way, as barium and bromate ions are in 1:2 molar ratio, bromate ions have twice the concentration of barium ions, for which the solubility product turns out:
[tex]Ksp=[Ba^{2+}][BrO_3^-]^2=(7.30x10^{-3}\frac{mol}{L})(3.65x10^{-3}\frac{mol}{L})^2\\\\Ksp=1.55x10^{-6}[/tex]
(C) [tex]NH_4MgAsO_4(s)\rightleftharpoons NH_4^+(aq)+Mg^{2+}(aq)+AsO_4^{3-}(aq)[/tex]
[tex]Molar\ solubility=\frac{0.038g}{100mL} *\frac{1mol}{289.35g}*\frac{1000mL}{1L}=1.31x10^{-4}\frac{mol}{L}[/tex]
In such a way, as ammonium, magnesium and arsenate ions are in 1:1:1 molar ratio, they have the same concentrations, for which the solubility product turns out:
[tex]Ksp=[NH_4^+][Mg^{2+}][AsO_4^{3-}]^2=(1.31x10^{-4}\frac{mol}{L})^3\\\\Ksp=2.27x10^{-12}[/tex]
(D) [tex]La_2(MoOs)_3(s)\rightleftharpoons 2La^{3+}(aq)+3MoOs^{2-}(aq)[/tex]
[tex]Molar\ solubility=\frac{0.00179g}{100mL} *\frac{1mol}{1136.38g}*\frac{1000mL}{1L}=1.58x10^{-5}\frac{mol}{L}[/tex]
In such a way, as the involved ions are in 2:3 molar ratio, La ion is twice the molar solubility and MoOs ion is three times it, for which the solubility product turns out:
[tex]Ksp=[La^{3+}]^2[MoOs^{-2}]^3=(2*1.58x10^{-5}\frac{mol}{L})^2(3*1.58x10^{-5}\frac{mol}{L})^3\\\\Ksp=1.05x10^{-22}[/tex]
Best regards.
how are mass and weight affected in chemical reactions?
Answer:
How the chemical reacts
Explanation:
Explain the Doppler effect using light. Why is the Doppler effect important in astronomy?
oxygen get stable configuration by ____________two electrons
please give the answer as fast as you can
please
Answer:
gaining two electrons
Explanation:
electron configuration
2:6
so add two to 6 to get stable 2:8
Calculate the equilibrium concentrations of N2O4 and NO2 at 25 ∘C in a vessel that contains an initial N2O4 concentration of 0.0655 M . The equilibrium constant Kc for the reaction N2O4(g)⇌2NO2(g) is 4.64×10−3 at 25 ∘C. Express your answers using four decimal places separated by a comma.
Answer:
[N2O4] = 0.0573M
[NO2] = 0.0163M
Explanation:
The equilibrium of N2O4 is:
N2O4(g)⇌2NO2(g)
Where Kc is defined as:
Kc = 4.64x10⁻³ = [NO2]² / [N2O4]
When you add just N2O4, the reaction will occurs until [NO2]² / [N2O4] = 4.64x10⁻³. Here, the system reaches equilibrium.
That means if 0.0655M N2O4 begin reaction, in equilibrium we will have:
[N2O4] = 0.0655M - X
[NO2] = 2X
Where X is defined as reaction coordinate
Replacing in Kc:
4.64x10⁻³ = [NO2]² / [N2O4]
4.64x10⁻³ = [2X]² / [0.0655-X]
3.0392x10⁻⁴ - 4.64x10⁻³X = 4X²
3.0392x10⁻⁴ - 4.64x10⁻³X - 4X² = 0
Solving for X:
X = -0.0093 → False solution. there is no negative concentrations
X = 0.008156M → Right solution.
Replacing X, equilibrium concentrations are:
[N2O4] = 0.0655M - X
[NO2] = 2X
[N2O4] = 0.0573M[NO2] = 0.0163MHow many mL of calcium hydroxide are required to neutralize 25.0 mL of 0.50 M
nitric acid?
Answer:
6.5 mL
Explanation:
Step 1: Write the balanced reaction
Ca(OH)₂ + 2 HNO₃ ⇒ Ca(NO₃)₂ + 2 H₂O
Step 2: Calculate the reacting moles of nitric acid
25.0 mL of 0.50 M nitric acid react.
[tex]0.0250L \times \frac{0.50mol}{L} = 0.013 mol[/tex]
Step 3: Calculate the reacting moles of calcium hydroxide
The molar ratio of Ca(OH)₂ to HNO₃ is 1:2. The reacting moles of Ca(OH)₂ are 1/2 × 0.013 mol = 6.5 × 10⁻³ mol
Step 4: Calculate the volume of calcium hydroxide
To answer this, we need the concentration of calcium hydroxide. Since the data is missing, let's suppose it is 1.0 M.
[tex]6.5 \times 10^{-3} mol \times \frac{1,000mL}{1.0mol} = 6.5 mL[/tex]
Calculate the maximum volume in mL of 0.18 M HCl that a tablet containing 340 mg Al(OH)3 and 516 mg Mg(OH)2 would be expected to neutralize. Assume complete neutralization.
Answer:
171 mL of HCl
Explanation:
The first thing we want to do is consider the reaction between Al(OH)3 and water - as that is the expected reaction that is taking place,
Al(OH)3 + 3HCl → AlCl3 + 3H2O
Knowing this, let's identify the mass of Al(OH)3. Aluminum = 27 g / mol, Oxygen( 3 ) = 16 [tex]*[/tex] 3 = 48, Hydrogen ( 3 ) = 1 [tex]*[/tex] 3 = 3 - 27 + 48 + 3 = 78 g / mol. This value is approximated however ( 78 g / mol ), as the molar mass of each substance is rounded as well. Another key thing we need to do here is to convert 340 mg → grams, considering that that unit is a necessity with respect to moles, as you might know - 340 mg = 0.340 g.
Now we can calculate how much moles of HCl will be present in solution, provided we have sufficient information for that,
(0.340 g Al(OH)3) / (78.0036 g / mol Al(OH)3) [tex]*[/tex] (3 mol HCl / 1 mol Al(OH)3)
⇒ (.004358773185 g^2 / mol Al(OH)3) [tex]*[/tex] (3 HCl / Al(OH)3 )
⇒ .01307632 mol HCl
We can apply this same concept on the reaction of Mg(OH)2 and water, receiving the number of moles of HCl when that takes place. Then we can add the two ( moles of HCl ) and divide by the value " 0.18 mol / L " given to us.
" Mg(OH)2 + 2HCl → MgCl2 + 2H2O "
Molar mass of Mg(OH)2 = 58.3197 g / mol,
516 mg = 0.516 g
(0.516 g Mg(OH)2) / (58.3197 g / mol Mg(OH)2) [tex]*[/tex] (2 mol HCl / 1 mol Mg(OH)2)
= .017695564 mol HCL
___________
( .01307632 + .017695564 ) / ( 0.18 M HCl )
= 0.170954911 L
= 171 mL of HCl
Which of the following is not part of the proper protocol for using acids and bases?
A. Storing acids and bases below 10°C
B. Add acid to water, not water to acid
C. Storing acids and bases in separate areas
D. Wearing protective clothing while handling
Answer:
A. Storing acids and bases below 10C
Explanation:
I took the exam and got it correct :)
Storing acids and bases below 10°C is not part of the proper protocol for using acids and bases.
What is an acid?
Acids are defined as substances which on dissociation yield H+ ions , and these substances are sour in taste.Compounds such as HCl, H₂SO₄ and HNO₃ are acids as they yield H+ ions on dissociation.
According to the number of H+ ions which are generated on dissociation acids are classified as mono-protic , di-protic ,tri-protic and polyprotic acids depending on the number of protons which are liberated on dissociation.
Acids are widely used in industries for production of fertilizers, detergents batteries and dyes.They are used in chemical industries for production of chemical compounds like salts which are produced by neutralization reactions.
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1. What is the frequency of light with a wavelength of 1064 nm?
2. How many photons of 1064 nm light are contained in a 535 Joules of this light?
Answer:
1) 2.8×10^14 Hz
2) 2.88×10^21 photons
Explanation:
Recall that the formula for the speed of a wave is v=λf, since we are talking about light, we can replace v with c hence; c=λf. Where;
c= speed of light = 3 ×10^8 m/s
λ= wavelength of light= 1064nm= 1064×10^-9 m
f= frequency of light= the unknown
Hence;
f= c/λ= 3×10^8/1064×10^-9
f= 2.8×10^14 Hz
Energy of a single photon=hf= 6.626×10^-34 Js × 2.8×10^14 s^-1 = 18.55×10^-20J
If 1 photon contains 18.55×10^-20J of energy
x photons contains 535 J of energy
x= 535/18.55×10^-20J
x= 2.88×10^21 photons
Which of the following statements is not true for an exothermic reaction? Question options: The products have a higher heat content than the reactants. The temperature of the reaction system increases. The temperature of the surroundings increases. Heat passes from the reaction system to the surroundings. The enthalpy change for the reaction is negativ
Answer:
The products have a higher heat content than the reactants.
Explanation:
The statement above is not true for an exothermic reaction because in an exothermic reaction heat is released to the surroundings. This simply means that the total energy of the products is less than that of the reactants.
Which best describes the total mass of a sample of water when it condenses
from a liquid to a gas?
A. The mass is less because the water molecules get closer together
and take up more space.
B. The mass is the same because the decrease in energy equals the
increase in the number of molecules.
C. The mass is the same because water molecules are not created or
destroyed during a phase change.
D. The mass is greater after water condenses because the mass of
the molecules increases.
Answer:
Its C I hopefully help you
Question 14 of 25
What type of reaction is BaCl2 + Na,504 → 2NaCl + Baso,?
A. Single-replacement
B. Synthesis
C. Double-replacement
D. Decomposition
double displacement
bcoz each of the reactants combines with other reactants to obtain the product
The diagram shows two waves.
How do the frequencies of the waves compare?
Wave A has a lower frequency because it has a
smaller amplitude.
Wave A has a higher frequency because it has a
shorter wavelength.
The waves have the same frequency because they
have the same wavelength.
The waves have the same frequency because they
have the same amplitude.
Answer:
Wave A has a higher frequency because it has a shorter wavelength.
Explanation:
The frequency of a wave and the wave length are related by the following equation:
Velocity (v) = wave length (λ) x frequency (f)
v = λf
If we make frequency (f) the subject of the above equation, we will have:
f = v/λ
Let the velocity (v) be constant.
f = v/λ
f & 1/λ
From the equation above,
We can see that the frequency (f) is inversely proportional to the wavelength (λ).
This implies that a wave with a high frequency, will have a short wavelength and a wave with a short frequency will have a longer wavelength.
Now considering wave A and B in the diagram above,
Wave A will have a higher frequency because it has a shorter wavelength as explained above.
Answer:
it is the second option
Explanation:
Consider the following system, which is at equilibrium, 3C(s) + 3H2(g) <--> CH4(g) + C2H2(g) The result of removing some C(s) from the system will be:
Answer:
The equilibrium position will shift to the left, thus, favouring the backward reaction.
Explanation:
The equation for the reaction is given below:
3C(s) + 3H2(g) <==> CH4(g) + C2H2(g)
According to Le Chatelier's principle, if an external constrain such as change in concentration, temperature or pressure is imposed on a chemical system in equilibrium, the equilibrium will shift in order to neutralize the effect.
From Le Chatelier's principle, removing some C(s) implies removing some of the concentration of the reactants.
This will shift the equilibrium position to the left, thus, favouring the backward reaction (i.e forming more reactants) because removing C(s) implies that the products are now more than the reactants and as such, they will react to form more reactants.
The equilibrium position will shift to the left, thus, favouring the backward reaction.
The following information should be considered:
The equation for the reaction is given below:[tex]3C(s) + 3H2(g) <==> CH4(g) + C2H2(g)[/tex]
In the case when an external constrain like as change in concentration is applied on a chemical system in equilibrium, the equilibrium will shift in order to neutralize the effect.This will shift the equilibrium position to the left, thus, favouring the backward reaction.Learn more: https://brainly.com/question/10283285?referrer=searchResults
A plot of 1/[BrO-] vs. time is linear and the slope is equal to 0.056 M-1s-1. If the initial concentration of BrO- is 0.65 M, how long will it take one-half of the BrO- ion to react
Answer:
time taken for one-half of the BrO⁻ ion to react is t= 27.45 secs
Explanation:
equation of reaction
3BrO⁻(aq) → BrO₃⁻(aq) + 2Br⁻(aq) (second order reaction)
given
the slope of the graph is 0.056M⁻¹s⁻¹ = k(constant)
initial concentration [A]₀ = 0.65M
for second order reaction,we can calculate the time taken for one-half of the BrO- ion to react using:
[tex]\frac{1}{[A]}[/tex] =[tex]\frac{1}{[A]}[/tex]₀ ⁺ k × t
where initial concentration [A]₀ = 0.65M
[A] = [A]₀÷2 = 0.325M
[tex]\frac{1}{0.325M}[/tex] = [tex]\frac{1}{0.65M}[/tex] + 0.056M⁻¹s⁻¹ × t
3.077= 1.54 + 0.056t
3.077-1.54=0.056t
1.537=0.056t
t= 27.45 secs
Reduction occurs at which electrode?
Answer:
negative charge electrode
Explanation:
In cathode positive ions are picked up to perform reduction.At the same time negative ions are picked up at anode to get oxidized from electrolyte.
Answer:
The electrode that removes ions from the solution :) a p e x
1. If a question states "The patching material costs NASA $306/in2 …", the conversion factor in this statement is: Recognizing — Equation statement: $306 = _____ ___________
Answer:
$306 = Cost of 1 square inch of the patching material in question.
$306 = 1 in²
Explanation:
The conversion factor is am expression that is used to prove the equivalence of some quantities with different units.
The conversion factor basically converts from one quantity to another.
For this question, the conversion factor given for the patching material is $306/in².
This means that the patching material costs $306 for every square inch, the equation for the conversion is thus
$306 = 1 in² of the patching material.
Hope this Helps!!!
Liquid octane reacts with gaseous oxygen gas to produce gaseous carbon dioxide and gaseous water . What is the theoretical yield of carbon dioxide formed from the reaction of of octane and of oxygen gas
Answer:
24.6g of CO₂ is theoretical yield
Explanation:
The reaction of 8.00g of octane with 38.9g of oxygen.
The reaction of octane with oxygen is:
C₈H₁₈(l) + 25/2O₂ → 9H₂O + 8CO₂
1 mole of octane reacts with 25/2 moles of oxygen to produce 8 moles of CO₂
Theoretical yield is the amount of carbon dioxide formed assuming a yield of 100%. To calculate theoretical yield, first, we need to find limiting reactant and, with the chemical reaction, we can obtain the theoretical moles of CO₂ produced and its mass to obtain theoretical yield.
Limiting reactant:
Moles octane (Molar mass: 114.23g/mol) in 8.00g:
8.00g × (1mol / 114.23g) = 0.0700 moles octane.
Moles oxygen (Molar mass: 32g/mol) in 38.9g:
38.9g × (1mol / 32g) = 1.2156 moles oxygen.
For a complete reaction of 1.2156 moles of O₂ there are necessaries:
1.2156 moles O₂ ₓ (1mol C₈H₁₈ / 25/2 moles O₂) = 0.0973 moles octane
As we have just 0.0700 moles,
octane is limiting reactant.Moles and mass of carbon dioxide:
As limiting reactant is octane, 0.0700 moles of C₈H₁₈ will produce:
0.0700mol C₈H₁₈ × (8 moles CO₂ / 1 mol C₈H₁₈) = 0.56 moles of CO₂ are theoretically produced. In mass (Molar mass CO₂ = 44.01g/mol):
0.56moles CO₂ × (44.01g / mol) =
24.6g of CO₂ is theoretical yield-Theoretical yield because we are assuming all octane is reacting. In real life, never happens like that-
Please help me out ASAP!
What is meant by concentration?
Answer:
concentration is the abundance of a constituent divided by the total volume of a mixture.
The half‑equivalence point of a titration occurs half way to the equivalence point, where half of the analyte has reacted to form its conjugate, and the other half still remains unreacted. If 0.480 moles of a monoprotic weak acid (Ka=3.0×10−5) is titrated with NaOH, what is the pH of the solution at the half‑equivalence point?
Answer:
pH = 4.52
Explanation:
A monoprotic acid, HA, reacts with NaOH as follows:
HA+ NaOH → A⁻ + H₂O
When the weak acid HA, is in solution with its conjugate base, A⁻, a buffer is produced. That means in the titration of the weak acid with NaOH you are producing a buffer.
The pH of a buffer can be found using H-H equation:
pH = pKa + log [A⁻] / [HA]
Where pKa is -logKa = 4.52 and [] can be understood as the moles of A⁻ and HA.
pH = 4.52 + log [A⁻] / [HA]
In the half-equivalence point, the half of HA was converted in A⁻ and the other half still remains as HA.
That means moles of A⁻ and HA are: 0.480/2 = 0.24 moles of both A⁻ and HA
Replacing in H-H equation:
pH = 4.52 + log [A⁻] / [HA]
pH = 4.52 + log [0.24] / [0.24]
pH = 4.52-In the half-equivalence point of a titration of a weak acid, pH = pKa-
Provide the structures of the fragments that result when the molecular ion of 2-heptanone undergoes fragmentation via McLafferty rearrangement. Include charges and single electrons.
Answer:
See explanation
Explanation:
We have to start, remembering the mechanism behind the McLafferty rearrangement. The hydrogen in the gamma carbon (in this case, carbon 5) would be removed by a heterolytic rupture due to the cation-radical placed in the oxygen of the carbonyl group. Then we will have several heterolytic ruptures. Between carbons alpha and beta (in this case, 4 and 3) and a rupture in the carbonyl group. Due to these ruptures, two double bonds would be formed. One double bond in the alcohol cation-radical and the other one in the alkene.
See figure 1
I hope it helps!
Potassium iodide reacts with lead(II) nitrate in the following precipitation reaction: 2 KI(aq) + Pb(NO3)2(aq) → 2 KNO3(aq) + PbI2(s) What minimum volume of 0.400 M potassium iodide solution is required to completely precipitate all of the lead in 310.0 mL of a 0.112 M lead(II) nitrate solution?
Answer:
0.1736 L or 173.6 ml
Explanation:
Number of moles of lead II nitrate is obtained by;
Number of moles = concentration × volume of solution
Concentration= 0.112 M
Volume of solution= 310 ml
n= 0.112 × 310/1000
n= 0.03472 moles
From the reaction equation;
2 moles of potassium iodide reacted with 1 mole of lead II nitrate
x moles of potassium iodide will react with 0.03472 moles of lead II nitrate
x= 2 × 0.03472 moles= 0.06944 moles of potassium iodide
Volume of potassium iodide solution = number of moles/ concentration = 0.06944/ 0.4
Volume of potassium iodide solution= 0.1736 L or 173.6 ml