Answer:
the acid strength is the order of [tex]\mathsf{HF _{(aq)} }[/tex] > [tex]\mathsf{C_6H_5COOH _{(aq)} }[/tex] > [tex]\mathsf{C_6H_5OH _{(aq)} }[/tex]
Explanation:
Given that :
a . [tex]\mathsf{C_6H_5COO^- _{(aq)} + C_6H_5OH _{(aq)} \to C_6H_5COOH _{(aq)} + C_6H_5O^- _{(aq)}}[/tex]
b. [tex]\mathsf{ F^- _{(aq)} + C_6H_5OH _{(aq)} \to C_6H_5O^- _{(aq)} + HF _{(aq)} }[/tex]
c. [tex]\mathsf{C_6H_5COOH _{(aq)} + F^- _{(aq)} \to HF _{(aq)} + C_6H_5COO^- _{(aq)} }[/tex]
Acid strength is the ability of an acid to dissociate into a proton and an anion. Take for instance.
HA ↔ H⁺ + A⁻
The acid strength of the following compounds above are:
[tex]\mathsf{C_6H_5OH _{(aq)} }[/tex] = 1.00 × 10⁻¹⁰
[tex]\mathsf{HF _{(aq)} }[/tex] = 6.6 × 10⁻⁴
[tex]\mathsf{C_6H_5COOH _{(aq)} }[/tex] = 6.3 × 10⁻⁵
As the acid dissociation constant increases the relative acid strength also increases.
From above, the acid strength is the order of [tex]\mathsf{HF _{(aq)} }[/tex] > [tex]\mathsf{C_6H_5COOH _{(aq)} }[/tex] > [tex]\mathsf{C_6H_5OH _{(aq)} }[/tex]
[tex]\mathsf{C_6H_5COO^- }[/tex], [tex]\mathsf{C_6H_5O^- _{(aq)}}[/tex] and F⁻ are Bronsted- Lowry acid
Bronsted- Lowry acid are molecule or ion that have the ability to donate a proton.
Bomb calorimetry is a poor choice to determine the number of nutritional Calories in food; it consistently overestimates the Caloric content because options: A) dietary fiber isn't used by the body. B) carbohydrates don't burn to completion. C) proteins don't burn. D) water has Calories and isn't burnable.
Answer:
A) dietary fiber isn't used by the body.
Explanation:
The food we eat contains certain nutritional contents that provides energy, measured in calories (CAL) to the body. A procedure called BOMB CALORIMETRY can be used to determine the energy contents of these foods. The energy-supplying macromolecules contained in food substances we eat are carbohydrate, protein, fats etc.
Bomb calorimetry uses the method of burning the food substance in a device called bomb calorimeter, and measure the caloric content of the burnt food. Bomb calorimetry measures all the present calories in a food substance, which can include dietary fibers. Due to this reason, it is considered a poor choice in determining the number of nutritional calories in a food substance.
Dietary fibers are indigestible carbohydrates that cannot be broken down and used by the body. They pass along the alimentary canal until they are egested. Hence, they are no source of nutrients to the body. Since bomb calorimetry measures all calories including dietary fibers, it is said to overestimate the caloric content of food substances.
is the general formula of a certain hydrate. When 256.3 g of the compound is heated to drive off the water, 214.2 g of anhydrous compound is left. Further analysis shows that the percentage composition of the anhydrate is 21.90% Ca, 43.14% Se, and 34.97% O.. (Hint: Treat the anhydrous compound and water just as you have treated elements in calculating in the formula of the hydrate.) (Use an asterisk to enter the dot in the formula. If a subscript is 1, omit it.) Find the empirical formula of the anhydrous compound. Find the empirical formula of the hydrate.
Answer:
The general formula of the hydrate is Caa Seb Oc. nH2O. Based on the given information, the weight of the hydrated compound is 256.3 grams, the weight of the anhydrous compound is 214.2 grams.
Therefore, the weight of water evaporated is 256.3 g - 214.2 g = 42.1 grams
The molecular weight of water is 18 gram per mole. So, the number of moles of water will be,
Moles of water = weight of water/molecular weight
= 42.1 grams / 18 = 2.3
The given composition of calcium is 21.90 %. So, the concentration of calcium in anhydrous compound is,
= 214.2 * 0.2190 = 46.91 grams
The given composition of Se is 43.14 %. So, the concentration of selenium in anhydrous compound is,
= 214.2 * 0.4314 = 92.40 grams
The given composition of oxygen is 34.97%, So, the concentration of oxygen in anhydrous compound is,
= 214.2 * 0.3497 = 74.91 grams
The molecular weight of Ca is 40.078, the obtained concentration is 46.91 grams, stoichiometry will be, 46.91/40.078 = 1.17
The molecular weight of Se is 78.96, the obtained concentration is 92.40, stoichiometry will be,
92.40/78.96 = 1.17
The molecular weight of Oxygen is 15.999, the concentration obtained is 74.91, the stoichiometry will be,
74.91/15.999 = 4.68.
Thus, the formula becomes, Ca1.17. Se1.1e O4.68. 2.3H2O, the closest actual component is CaSeO4.2H2O
243
Am
95
1. The atomic symbol of americium-243 is shown. Which of the following is correct?
• A. The atomic mass is 243 amu, and the atomic number is 95.
B. The atomic mass is 338 amu, and the atomic number is 95.
• C. The atomic mass is 95 amu, and the atomic number is 243.
D. The atomic mass is 243 amu, and the atomic number is 338.
Answer:
A. The atomic mass is 243 amu, and the atomic number is 95.
is a polyprotic acid. Write balanced chemical equations for the sequence of reactions that carbonic acid can undergo when it's dissolved in water.
Answer:
H₂CO₃ H₂O ⇄ HCO₃⁻ + H₃O⁺ Ka1
HCO₃⁻ + H₂O ⇄ CO₃⁻² + H₃O⁺ Ka2
CO₃⁻² + H₂O ⇄ HCO₃⁻ + OH⁻ Kb1
HCO₃⁻ + H₂O ⇄ H₂CO₃ + OH⁻ Kb2
Explanation:
Formula for carbonic acid is: H₂CO₃
It is a dyprotic acid, because it can release two protons. We can also mention that is a weak one. The equilibrums are:
H₂CO₃ H₂O ⇄ HCO₃⁻ + H₃O⁺ Ka1
HCO₃⁻ + H₂O ⇄ CO₃⁻² + H₃O⁺ Ka2
When the conjugate strong bases, carbonate and bicarbonate take a proton from water, the reactions are:
CO₃⁻² + H₂O ⇄ HCO₃⁻ + OH⁻ Kb1
HCO₃⁻ + H₂O ⇄ H₂CO₃ + OH⁻ Kb2
Notice, that bicarbonate anion can release or take a proton to/from water. This is called amphoteric,
What is meant by functional group in an organic compound?
Functional group for
A) keton
B) acid anhydride
C) aldehyde
D) amide
C) alcohol
Answer:
i THINK the answer to this is C) aldehyde
Answer:
The answer is C, the guy above is correct.
Explanation:
Which of the terms heat of vaporization and heat of fusion refers to condensation and which refers to melting?
Answer:
Heat of vaporization refers to condensation and heat of fusion refers to melting.
Explanation:
Heat of vaporization or heat of evaporation, is defined as the energy required to convert liquid substance into a gas which creates condensation. As a gas condenses to a liquid, heat is released.
Heat of fusion refers to melting because heat of fusion is defined as the energy required to change any amount of substance when it melts.
Hence, the correct answer is "Heat of vaporization refers to condensation and heat of fusion refers to melting.".
What chemical bonds hold atoms?
g Which ONE of the following pairs of organic compounds are NOT pairs of isomers? A) butanol ( CH3-CH2-CH2-CH2-OH ) and diethyl ether ( CH3–CH2–O–CH2–CH3 ) B) isopentane ( (CH3)2-CH-CH2-CH3 ) and neopentane ( (CH3)4C ) C) ethanolamine ( H2N-CH2-CH2-OH ) and acetamide ( CH3-CO-NH2 ) D) acrylic acid ( CH2=CH-COOH ) and propanedial ( OHC–CH2–CHO ) E) trimethylamine ( (CH3)3N ) and propylamine ( CH3-CH2-CH2-NH2 )
Answer:
ethanolamine ( H2N-CH2-CH2-OH ) and acetamide ( CH3-CO-NH2 )
Explanation:
Isomers are compounds that have the same molecular formula but different structural formulas. Hence any pair of compounds that can be represented by exactly the same molecular formula are isomers of each other.
If we look at the pair of compounds; ethanolamine ( H2N-CH2-CH2-OH ) and acetamide ( CH3-CO-NH2 ), one compound has molecular formula, C2H7ON while the other has a molecular formula, C2H5ON, hence they are not isomers of each other.
A 635 mL NaCl solution is diluted to a volume of 1.13 L and a concentration of 5.00 M . What was the initial concentration C1?
Answer:
8.90 M
Explanation:
Step 1: Given data
Initial concentration (C₁): ?Initial volume (V₁): 635 mL = 0.635 LFinal concentration (C₂): 5.00 MFinal volume (V₂): 1.13 LStep 2: Calculate the initial concentration
We have a concentrated NaCl solution and we want to prepare a diluted one. We will use the dilution rule.
C₁ × V₁ = C₂ × V₂
C₁ = C₂ × V₂ / V₁
C₁ = 5.00 M × 1.13 L / 0.635 L
C₁ = 8.90 M
Answer:
[tex]\large \boxed{\text{8.90 mol/L}}[/tex]
Explanation:
We can use the dilution formula to calculate the concentration of the original solution.
[tex]\begin{array}{rcl}V_{1}c_{1} & = & V_{2}c_{2}\\\text{635 mL }\times c_{1} & = & \text{1130 mL} \times \text{5.00 mol/L}\\635 c_{1}&=& \text{5650 mol/L}\\c_{1}& = & \dfrac{5650}{635}\text{ mol/L}\\\\& = & \textbf{8.90 mol/L}\\\end{array}\\\text{The initial concentration was $\large \boxed{\textbf{8.90 mol/L }}$}[/tex]
Different vinegars can be 5-20% acetic acid solutions and have been used for medicinal purposes for thousands of years. If a person takes 2.0 tablespoons of vinegar a day and the Molarity of the vinegar is .84 M, then how many grams of acetic acid (HC2H3O2) will be consumed? 1 Tablespoon is 15 mL.
.013 g
.026 g
.76 g
1.5 g
Answer:
1.5g
Explanation:
Remember that Molarity = (#moles of solute)/(#liters of solution)
This problem informs us that the Molarity of the vinegar is 0.84 and that the solution is 15mL.
First let's get your SI units to the correct ones.
15mL (1L/1000mL) = 0.015L
Molarity = (#moles of solute)/(#liters of solution) ~
(Molarity)(#liters of solution) = #moles of solute
(0.84M)(.015L) = 0.0126moles of acetic acid per tablespoon
2 tablespoons a day = 0.0126moles*2 = 0.0252 moles of acetic acid.
Now that we have the # of moles of acetic acid we need to get our answer into grams. The molecular weight of HC2H3O2 is 60g/mole.
0.0252mole HC2H3O2 (60g HC2H3O2/1mole HC2H3O2) = 1.512g ~ 1.5g HC2H3O2.
Select True or False: Pi bonds are covalent bonds in which the electron density is concentrated above and below a plane containing the nuclei of the bonding atoms and occurs by sideways overlap of p orbitals.
Answer:
True
Explanation:
In pi bonds, the electron density concentrates itself between the atoms of the compound but are present on either side of the line joining the atoms. Electron density is found above and below the plane of the line joining the internuclear axis of the two atoms involved in the bond.
Pi bonds usually occur by sideways overlap of atomic orbitals and this leads to both double and triple bonds.
28.What is the correct IUPAC name for the following compound?A)12-crown-5B)12-crown-4C)4-crown-12D)12-crown-12E)Cyclododecane tetraether
Answer:
12-crown-4
Explanation:
We must recall that any structural moiety in organic chemistry having the R-O-R unit is an ether. If the oxygen form a ring in which they are sandwiched in between carbon atoms, the compound is known as a crown ether. The name emanates from the close resemblance of the compound to an actual crown.
If we want to name the crown ether, we first count the number of carbon atoms present and the number of oxygen atoms present. The correct name is now, total number of carbon + oxygen atoms -crown- number of oxygen atoms, in this case; 12-crown-4, hence the answer.
Answer:
12-crown-4
Explanation:
10. For the following isotopes that have missing information, fill in the missing informatic
complete the notation: 36P
Answer:
Krypton.
Explanation:
Krypton is an atom which has 36 protons in its nucleus. There are 31 isotopes of Krypton which have same number of protons i. e. 36, same number of electrons i. e. 36 but different number of neutrons. Isotope refers to those atoms having same atomic number i. e. number of proton but different mass number i. e. number of neutron. For example, in Krypton-78, there 36 protons and 42 neutrons.
Akeem cut his finger during an investigation, and it is bleeding slightly. Before helping him bandage the wound,
which precaution should the teacher take?
O Tell someone to call 911,
O Put on protective gloves.
O Wash Akeem's finger in the shower.
O Apply disinfectant before cleaning.
Answer:
b.) Put on protective gloves
Answer:
2020 Put on protective gloves.
Explanation:
Find the standard enthalpy of formation of iodine atoms. (Round your answer to one decimal place.) Standard enthalpy of formation
Answer:
Enthalpy of formation is the energy change when one mole of a substance is formed from its constituent atoms under standard conditions
21. What are the two main ways of working with clay?
Answer:
Diferentes tipos de arcilla
ARCILLA DE LADRILLOS. Contiene muchas impurezas. ...
ARCILLA DE ALFARERO. Llamada también barro rojo y utilizada en alfarería y para modelar. ...
ARCILLA DE GRES. Es una arcilla con gran contenido de feldespato. ...
ARCILLAS “BALL CLAY” O DE BOLA. ...
CAOLIN. ...
ARCILLA REFRACTARIA. ...
BENTONITA.
Explanation:
Answer:
Coil method and the slab method.
Explanation:
In the following reaction, what element is gaining mass?
Mg(s) + CuSO4(aq) -> MgSO4(aq) + Cu(s)
A. Mg
B. None of these
C. Cu
D. O
Mg gained mass because it went from being a single element (on the reactant side) to being a molecule (on the product side).
Mg element is gaining mass. Hence, option A is correct.
What is a chemical equation?A chemical reaction is a representation of symbols of the elements to indicate the number of substances and moles of reactant and product.
Mg gained mass because it went from being a single element (on the reactant side) to being a molecule (on the product side).
Hence, option A is correct.
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A student sets up the following equation to convert a measurement. The (?) Stands for a number the student is going to calculate. Fill in the missing part of this equation. (0.030 cm^3) x ? =m^3
Answer:
\text{0.30 cm}^{3} \times \left (\dfrac{10^{-2}\text{ m}}{\text{1 cm}}\right )^{3} = 3.0 \times 10^{-7} \text{ m}^{3}
Explanation:
0.030 cm³ × ? = x m³
You want to convert cubic centimetres to cubic metres, so you multiply the cubic centimetres by a conversion factor.
For example, you know that centi means "× 10⁻²", so
1 cm = 10⁻² m
If we divide each side by 1 cm, we get 1 = (10⁻² m/1 cm).
If we divide each side by 10⁻² m, we get (1 cm/10⁻² m) = 1.
So, we can use either (10⁻² m/1 cm) or (1 cm/10⁻² m) as a conversion factor, because each fraction equals one.
We choose the former because it has the desired units on top.
The "cm" is cubed, so we must cube the conversion factor.
The calculation becomes
[tex]\text{0.30 cm}^{3} \times \left (\dfrac{10^{-2}\text{ m}}{\text{1 cm}}\right )^{3} = 0.30 \times 10^{-6}\text{ m}^{3} = \mathbf{3.0 \times 10^{-7}} \textbf{ m}^{\mathbf{3}}\\\\\textbf{0.30 cm}^{\mathbf{3}} \times \left (\dfrac{\mathbf{10^{-2}}\textbf{ m}}{\textbf{1 cm}}\right )^{\mathbf{3}} = \mathbf{3.0 \times 10^{-7}} \textbf{ m}^{\mathbf{3}}[/tex]
what is the valency of element sulphur
sorry not sure wish I can help u
How does the spontaneity of the process below depend on temperature? PCl5(g)+H2O(g)→POCl3(g)+2HCl(g) ΔH=−126 kJ mol−1, ΔS=146 J K−1mol−
The given question is incomplete, the complete question is:
How does the spontaneity of the process below depend on temperature? PCI5(9)+H2O(g)POCI3(g) +2HCI(g) -126 kJ mol1, AS = 146 J K-'mol1 ΔΗ Select the correct answer below: nonspontaneous at all temperatures spontaneous at all temperatures spontaneous at high temperatures and nonspontaneous at low temperatures spontaneous at low temperatures and nonspontaneous at high temperatures
Answer:
The correct answer is spontaneous at all the temperatures.
Explanation:
Gibbs Free energy is an essential relation that determines the spontaneity of any reaction, that is, ΔG = ΔH - TΔS
When ΔG is less than zero, that is, negative, the reaction is considered to be in spontaneous state. Based on the given information, ΔH = -126 kJ/mol
= -126000 J/mol, it is negative
ΔS = 146 J/K/mol, it is positive
Now, ΔG = ΔH-TΔS
= (-ve) - T (+ve), Thus, when ΔH, is -ve, ΔS is +ve, -TΔS is -ve, the ΔG will be -ve. Therefore, reaction will be spontaneous at all the temperatures.
Evaluate the exponential expression (−2)6.
A general exponential expression is something like:
A^n
This means that we need to multiply the number A by itself n times.
Using that we will get (-2)^6 = 64
With that definition, we can rewrite:
(-2)^6 = (-2)*(-2)*(-2)*(-2)*(-2)*(-2)
So we just need to solve the above expression.
Also, remember the rule of signs:
(-)*(-) = (+)
We will get:
(-2)*(-2)*(-2)*(-2)*(-2)*(-2) = [(-2)*(-2)]*[(-2)*(-2)]*[(-2)*(-2)]
= 4*4*4 = 16*4 = 64
Then we got:
(-2)^6 = 64
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A balloon is filled to a volume of 1.50 L with 3.00 moles of gas at 25 °C. With pressure and temperature held constant, what will be the volume (in L) of the balloon if 0.50 moles of gas are released?
Answer:
Volume : 1.25 L
Explanation:
We are given here that the volume ( V[tex]_1[/tex] ) = 1.50 Liters, the initial moles ( held at 25 °C ) = 3.00 mol, and the final moles ( n[tex]_2[/tex] ) = 3.00 - 0.5 = 2.5 mol. The final mol is calculated given that 0.50 mol of gas are released from the prior 3.00 moles of gas.
Volume ( V[tex]_1[/tex] ) = 1.50 L,
Initial moles ( n[tex]_1[/tex] ) = 3.00 mol,
Final Volume ( n[tex]_2[/tex] ) = 3.00 - 0.5 = 2.5 mol
Applying the combined gas law, we can calculate the final volume ( V[tex]_2[/tex] ).
P[tex]_1[/tex]V[tex]_1[/tex] / n[tex]_1[/tex]T[tex]_1[/tex] = P[tex]_2[/tex]V[tex]_2[/tex] / n[tex]_2[/tex]T[tex]_2[/tex] - we know that the pressure and temperature are constant, and therefore we can apply the following formula,
V[tex]_1[/tex] / n[tex]_1[/tex] = V[tex]_2[/tex] / n[tex]_2[/tex] - isolate V[tex]_2[/tex],
V[tex]_2[/tex] = V[tex]_1[/tex] n[tex]_2[/tex] / n[tex]_1[/tex] = 1.50 L [tex]*[/tex] 2.5 mol / 3.00 mol = ( 1.5 [tex]*[/tex] 2.5 / 3 ) L = 1.25 L
The volume of the balloon will be 1.25 L.
Consider these metal ion/metal standard reduction potentials Cd2+(aq)|Cd(s) Zn2+(aq)|Zn(s) Ni2+(aq)|Ni(s) Cu2+(aq)|Cu(s) Ag+(aq)|Ag(s) -0.40 V -0.76 V ‑0.25 V +0.34 V +0.80 V Based on the data above, which species is the best reducing agent?
Answer:
The best reducing agent is Zn(s)
Explanation:
A reducing agent must to be able to reduce another compound, by oxidizing itself. Consequently, the oxidation potential must be high. The oxidation potential of a compound is the reduction potential of the same compound with the opposite charge. Given the reduction potentials, the best reducing agent will be the compound with the most negative reduction potential. Among the following reduction potentials:
Cd₂⁺(aq)|Cd(s) ⇒ -0.40 V
Zn²⁺(aq)|Zn(s) ⇒ -0.76 V
Ni²⁺(aq)|Ni(s) ⇒‑0.25 V
Cu²⁺(aq)|Cu(s) ⇒ +0.34 V
Ag⁺(aq)|Ag(s) ⇒ +0.80 V
The most negative is Zn²⁺(aq)|Zn(s) ⇒ -0.76 V
From this, the most reducing agent is Zn. Zn(s) is oxidized to Zn²⁺ ions with the highest oxidation potential (0.76 V).
A solution of HCOOH has 0.16M HCOOH at equilibrium. The Ka for HCOOH is 1.8×10−4. What is the pH of this solution at equilibrium? Express the pH numerically.
Answer:
[tex]pH=2.28[/tex]
Explanation:
Hello,
In this case, for the acid dissociation of formic acid (HCOOH) we have:
[tex]HCOOH(aq)\rightarrow H^+(aq)+HCOO^-(aq)[/tex]
Whose equilibrium expression is:
[tex]Ka=\frac{[H^+][HCOO^-]}{[HCOOH]}[/tex]
That in terms of the reaction extent is:
[tex]1.8x10^{-4}=\frac{x*x}{0.16-x}[/tex]
Thus, solving for [tex]x[/tex] which is also equal to the concentration of hydrogen ions we obtain:
[tex]x=0.00528M[/tex]
[tex][H^+]=0.00528M[/tex]
Then, as the pH is computed as:
[tex]pH=-log([H^+])[/tex]
The pH turns out:
[tex]pH=-log(0.00528M)\\\\pH=2.28[/tex]
Regards.
Calculate the pH of a solution formed by mixing 250.0 mL of 0.15 M NH4Cl with 200.0 mL of 0.12 M NH3. The Kb for NH3 is 1.8 × 10-5.
Answer:
The pH of the solution is 9.06.
Explanation:
The reaction of the dissociation of NH₃ in water is:
NH₃(aq) + H₂O(l) ⇄ NH₄⁺(aq) + OH⁻(aq) (1)
[NH₃] - x [NH₄⁺] + x x
The concentration of NH₃ and NH₄⁺ is:
[tex] [NH_{3}] = \frac{n_{NH_{3}}}{V_{T}} = \frac{C_{i}_{(NH_{3})}*Vi_{NH_{3}}}{V_{NH_{3}} + V_{NH_{4}^{+}}} = \frac{0.12 M*0.2 L}{0.2 L + 0.25 L} = 0.053 M [/tex]
[tex] [NH_{4}^{+}] = \frac{C_{i}_{(NH_{4}^{+})*V_{NH_{4}^{+}}}}{V_{NH_{3}} + V_{NH_{4}^{+}}} = \frac{0.15 M*0.25 L}{0.2 L + 0.25 L} = 0.083 M [/tex]
From equation (1) we have:
[tex]Kb = \frac{[NH_{4}^{+}][OH^{-}]}{[NH_{3}]}[/tex]
[tex] 1.8 \cdot 10^{-5} = \frac{(0.083 + x)*x}{0.053 - x} [/tex]
[tex] 1.8 \cdot 10^{-5}(0.053 - x) - (0.083 + x)*x = 0 [/tex]
By solving the above equation for x we have:
x = 1.15x10⁻⁵ = [OH⁻]
The pH of the solution is:
[tex] pOH = -log([OH^{-}]) = -log(1.15 \cdot 10^{-5}) = 4.94 [/tex]
[tex] pH = 14 - pOH = 14 - 4.94 = 9.06 [/tex]
Therefore, the pH of the solution is 9.06.
I hope it helps you!
What is the mass of 7.2 moles of titanium to the nearest tenth?
Answer:
7.2 moles of titanium weigh 344.7 grams.
Explanation:
Molar mass of titanium = 47.88 g
[tex]1 \: mole \: of \: titanium = 47.88 \: grams \\ 7.2 \: moles \: of \: titanium = g[/tex]
solve for g
[tex]g = ( \frac{7.2 \times 47.88}{1} ) \\ g = 344.7 \: grams[/tex]
Answer:
[tex]\boxed {\boxed {\sf 344.7 \ g \ Ti}}[/tex]
Explanation:
We are asked to find the mass of 7.2 moles of titanium.
To convert from moles to grams, we use the molar mass. This value is the mass of 1 mole of a substance. They are equivalent to the atomic masses found on the Periodic Table, but the units are grams per moles instead of atomic mass units.
Look up titanium's molar mass on the Periodic Table.
Ti: 47.87 g/molConvert moles to grams using dimensional analysis. First, set up a ratio using the molar mass.
[tex]\frac {47.87 \ g \ Ti}{1 \ mol \ Ti}[/tex]
We are converting 7.2 moles to grams, so we multiply the ratio by this value.
[tex]7.2 \ mol \ Ti *\frac {47.87 \ g \ Ti}{1 \ mol \ Ti}[/tex]
The units of moles of titanium cancel.
[tex]7.2 *\frac {47.87 \ g \ Ti}{1 }[/tex]
The denominator of 1 can be ignored (a number over 1 is just the number).
[tex]7.2 *47.87 \ g \ Ti}[/tex]
[tex]344.664 \ g \ Ti[/tex]
We are asked to round to the nearest tenth. The 6 in the hundredth place tells us to round the 6 in the tenth place up to a 7.
[tex]344.7 \ g \ Ti[/tex]
There are approximately 344.7 grams of titanium in 7.2 moles of titanium.
Draw the major condensation product obtained by treatment of ethyl 3-methylbutanoate with sodium ethoxide in ethanol.
Answer:
ethyl 3-ethoxy-3-hydroxy-2-isopropyl-5-methyl hexanoate
Explanation:
In this case, we have a very strong base (sodium ethoxide). Therefore, this compound will remove a hydrogen from ethyl 3-methyl butanoate generating a carbanion.
This carbanion, can attack another ethyl 3-methyl butanoate molecule on the carbonyl group generating a new C-C bond and producing a negative charge in the oxygen.
Then the ethanol can protonate the molecule generating an "OH" group and the ethoxide.
See figure 1
I hope it helps!
Write an equation to show how the base NaOH(s) behaves in water. Include states of matter in your answer. Click in the answer box to open the symbol palette.
Answer:
The reaction is given as:
[tex]NaOH(s)\rightarrow Na^+(aq)+OH^-(aq)[/tex]
Explanation:
Bases are defined as those chemical substances which give hydroxide ions in their aqueous solutions.
[tex]BOH(s)\rightarrow B^+(aq)+OH^-(aq)[/tex]
When sodium hydroxide is added to water it gets dissociated into two ions that are sodium ions and hydroxide ions. Along with this heat energy also releases during this reaction.
The reaction is given as:
[tex]NaOH(s)\rightarrow Na^+(aq)+OH^-(aq)[/tex]
The equation to show how NaOH behaves in water is NaOH → Na⁺ + (OH)⁻
The compound that produce negative hydroxide (OH−) ions when dissolved
in water are called bases .
This compounds NaOH (sodium hydroxide) is an example of a base.
When it dissolves in water it dissociate to form negative hydroxide (OH−)
ions and positive sodium (Na+) ions.
It can be represented by the following equation:
NaOH → Na⁺ + (OH)⁻
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A chemist fills a reaction vessel with 0.978 g aluminum hydroxide AlOH3 solid, 0.607 M aluminum Al+3 aqueous solution, and 0.396 M hydroxide OH− aqueous solution at a temperature of 25.0°C.
Under these conditions, calculate the reaction free energy ΔG for the following chemical reaction:
Al(OH)3(s) = A1+ (aq) +30H (aq)
Use the thermodynamic information in the ALEKS Data tab. Round your answer to the nearest kilojoule.
KJ
Answer: [tex]\Delta G^{0}[/tex] = 168.12 kJ
Explanation: Gibbs Free Energy, at any time, is defined as the enthalpy of the system minus product of temperature and entropy of the reaction, i.e.:
[tex]\Delta G^{0} = \Delta H^{0} - T.\Delta S^{0}[/tex]
Enthalpy is defined as internal heat existent in the system. It is calculated as:
[tex]\Delta H^{0} = \Sigma H^{0}_{product} - \Sigma H^{0}_{reagent}[/tex]
Using Enthalpy Formation Table:
[tex]\Delta H^{0} = [3*(-299.9)+(-524.7)] - (-1277)[/tex]
[tex]\Delta H^{0} = 62,6 kJ[/tex]
Entropy is the degree of disorder in the system. It is found by:
[tex]\Delta S^{0} = \Sigma S^{0}_{products} - \Sigma S^{0}_{reagents}[/tex]
Calculating:
[tex]\Delta S^{0} = (-321.7) + 3(-10.8) - 0[/tex]
[tex]\Delta S^{0} = -354.1J[/tex]
And so, Gibbs Free energy will be:
[tex]\Delta G^{0} = \Delta H^{0} - T.\Delta S^{0}[/tex]
[tex]\Delta G^{0} = 62600 - [298.(-354.1)][/tex]
[tex]\Delta G^{0} = 168121.8 J[/tex]
Rounding to the nearest kJ:
[tex]\Delta G^{0}[/tex] = 168.12 kJ
The entropy of a substance above absolute zero will always be:
a. Negative
b. Positive
c. Neither Negative nor positive