Answer:
% yield of the student's experiment is
[tex]\frac{0.34}{0.60}[/tex] ˣ 100 = 56.67%
Explanation:
given
volume of acetone= 43.8 mL
molar weight of acetone = 58.08 g/mol
density of acetone = 0.791 g/mL
A student mixes 43.8 mL of acetone (58.08 g/mol, 0.791 g/mL)
43.8 mL = 43.8mL × 0.791g/mL
= 34.6458g ≈34.65g
1 mole of acetone = 58.08g
∴34.65g = 34.65g/58.08g
= 0.60mol
molecular weight of the product 1,5-diphenylpenta-1,4-dien-3-one = 234.29 g/mol
mole = mass/ molar weight
mole = 79.4g/ 234.29g/mol
mole(n) = 0.3389mol ≈ 0.34mol
1 mole of acetone will produce 1 mole of the product
∴0.60mol of acetone will produce 0.60mol of the product
but we get 0.34mol of the product
∴ % yield of the student's experiment is
[tex]\frac{0.34}{0.60}[/tex] ˣ 100 = 56.67%
True or False: Adding 4.18 joules to water will increase the temperature more than adding 1 calorie to water.
Answer:
Because one calorie is equal to 4.18 J, it takes 4.18 J to raise the temperature of one gram of water by 1°C. In joules, water's specific heat is 4.18 J per gram per °C. If you look at the specific heat graph shown below, you will see that 4.18 is an unusually large value.
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.
Nitrogen has different oxidation states in the following compounds: nitrite ion, nitrous oxide, nitrate ion, ammonia, and nitrogen gas. Arrange these species in order of increasing nitrogen oxidation state. Select the correct answer below: A. ammonia, nitrogen gas, nitrite, nitrous oxide, nitrate B. nitrogen gas, ammonia, nitrous oxide, nitrite, nitrate C. ammonia, nitrogen gas, nitrous oxide, nitrite, nitrate D. ammonia, nitrogen gas, nitrate, nitrite, nitrous oxide
Answer:
C. ammonia, nitrogen gas, nitrous oxide, nitrite, nitrate
Explanation:
To establish the oxidation number of nitrogen in each compound, we know that the sum of the oxidation numbers of the elements is equal to the charge of the species.
Nitrite ion (NO₂⁻)
1 × N + 2 × O = -1
1 × N + 2 × (-2) = -1
N = +3
Nitrous oxide (NO)
1 × N + 1 × O = 0
1 × N + 1 × (-2) = 0
N = +2
Nitrate ion (NO₃⁻)
1 × N + 3 × O = -1
1 × N + 3 × (-2) = -1
N = +5
Ammonia (NH₃)
1 × N + 3 × H = 0
1 × N + 3 × (+1) = 0
N = -3
Nitrogen gas (N₂)
2 × N = 0
N = 0
The order of increasing nitrogen oxidation state is:
C. ammonia, nitrogen gas, nitrous oxide, nitrite, nitrate
Calculate the heat absorbed by a sample of water that has a mass of 45.00 g when the temperature increases from 21.0oC to 38.5 oC. (s=4.184 J/g.o C)
Answer:
The heat absorbed by the sample of water is 3,294.9 J
Explanation:
Calorimetry is the measurement and calculation of the amounts of heat exchanged by a body or a system.
The sensible heat of a body is the amount of heat received or transferred by a body when it undergoes a temperature variation (Δt) without there being a change of physical state (solid, liquid or gaseous). Its mathematical expression is:
Q = c * m * ΔT
Where Q is the heat exchanged by a body of mass m, made up of a specific heat substance c and where ΔT is the temperature variation.
In this case:
Q=?m= 45 gc= 4.184 [tex]\frac{J}{g*C}[/tex]ΔT= Tfinal - Tinitial= 38.5 C - 21 C= 17.5 CReplacing:
Q= 4.184 [tex]\frac{J}{g*C}[/tex] * 45 g* 17.5 C
Solving:
Q=3,294.9 J
The heat absorbed by the sample of water is 3,294.9 J
How many moles of aqueous magnesium ions and chloride ions are formed when 0.250 mol of magnesium chloride dissolves in water
Answer:
0.250 mol Mg²⁺
0.500 mol Cl⁻
Explanation:
Magnesium chloride (MgCl₂) dissociates into ions according to the following equilibrium:
MgCl₂ ⇒ Mg²⁺ + 2 Cl⁻
1 mol 1 mol 2 mol
1 mol of Mg²⁺ and 2 moles of Cl⁻ are formed per mole of MgCl₂. If we have 0.250 mol of MgCl₂, the following amounts of ions will be formed:
0.250 mol MgCl₂ x 1 mol Mg²⁺/mol MgCl₂= 0.250 mol Mg²⁺
0.250 mol MgCl₂ x 2 mol Cl⁻/mol MgCl₂= 0.500 mol Cl⁻
Answer:
HEY THE ANSWER ABOVE ME IS RIGHT!! i defientely misclicked my rating :/
5/5 all the way.
Explanation:
If the equilibrium constant of the reaction is 0.85, then which statement is true if the mass of A is 10.5 grams; the density of B is 0.82 g/ml; the concentration of C is 0.64 M; and the concentration of D is 0.38 M.
A(s) + 3 B(l) _____ 2(aq) + D(aq)
Pick the correct statement about this system.
A. Q < K and reaction shifts left
B. Q > K and reaction shifts left
C. Q > K and reaction shifts right
D. Q = K and reaction does not shift
E. Q < K and reaction shifts right
Answer:
E. Q < K and reaction shifts right
Explanation:
Step 1: Write the balanced equation
A(s) + 3 B(l) ⇄ 2(aq) + D(aq)
Step 2: Calculate the reaction quotient (Q)
The reaction quotient, as the equilibrium constant (K), only includes aqueous and gaseous species.
Q = [C]² × [D]
Q = 0.64² × 0.38
Q = 0.15
Step 3: Compare Q with K and determine in which direction will shift the reaction
Since Q < K, the reaction will shift to the right to attain the equilibrium.
Find the [OH−] of a 0.32 M methylamine (CH3NH2) solution. (The value of Kb for methylamine (CH3NH2) is 4.4×10−4.) Express your answer to two significant figures and include the appropriate units.
Answer:
[tex][OH^-]=0.01165M[/tex]
Explanation:
Hello,
In this case, for the dissociation of methylamine:
[tex]CH_3NH_2(aq)+H_2O(l)\rightleftharpoons CH_3NH_3^+(aq)+OH^-(aq)[/tex]
We can write the basic dissociation constant as:
[tex]Kb=\frac{[CH_3NH_3^+][OH^-]}{[CH_3NH_2]}[/tex]
That in terms of the reaction extent [tex]x[/tex], turns out:
[tex]Kb=\frac{x*x}{[CH_3NH_2]_0-x}[/tex]
[tex]4.4x10^{-4}=\frac{x^2}{0.32M-x}[/tex]
That has the following solution for [tex]x[/tex]:
[tex]x_1=-0.01209M\\x_2=0.01165M[/tex]
Yer 0.01165M is valid only as no negative concentrations are eligible. It means that it is the concentration of hydroxyl ions in the solution:
[tex][OH^-]=0.01165M[/tex]
Best regards.
The reaction, 2 SO3(g) <--> 2 SO2(g) + O2(g) is endothermic. Predict what will happen if the temperature is increased.
Explanation:
This reaction is in equilibrium and would hence obey lechatelier's principle. This principle states that whenever a system at equilibrium undergoes a change, it would react in way so as to annul that change.
Since it is an endothermic reaction, increasing the temperature would cause the reaction to shift towards the right.
This means that it favours product formation and more of the product would be formed.
The accepted value of the number of Liters of gas in a mole is 22.4. List two possible reasons on why our experiment yielded a different value for the number of Liters in a mole of a gas.
Hint: Our experiment was conducted in July, in St. Paul, Minnesota.
Answer:
- Pressure in St. Paul, Minnesota
- Temperature in St. Paul, Minnesota
Explanation:
22.4 L or dm³ is the volume for a gas under Standard pressure and temperature conditions.
It is logically to say, that tempereature value at the day of the experiment was not 273.15 K, which is 32°F
We can say, that the pressure was not 1 atm. St Paul Minnesota has a minimum, but a little height, so the pressure differs by few figures from the standard pressure values.
We also have to mention, that 22.4 L is the value for the Ideal gases at standards conditions. Ideal gases does not exisist on practice, we always talk about real gases. Don't forget the Ideal Gases Law equation:
P . V = n . R . T
Pressure . Volume = number of moles . 0.082 L.atm /mol. K . 273.15K
Number of moles must be 1 at STP, to determine a volume of 22.4L
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
Assume that a nickel weighs exactly 5.038650 g for the sets of weights listed below obtained by a single weighing on the balance below
Answer:
afshkkyfugutuiryfyi
For each of the processes, determine whether the entropy of the system is increasing or decreasing. The system is underlined.
1. a snowman melts on a spring day
2. a document goes through a paper shredder
3. a water bottle cools down in a refrigerator
4. silver tarnishes
5. dissolved sigar precipitates out of water to form rock candy
A. Entropy is increasing
B. Entropy is decreasing
Entropy is INCREASING when a snowman melts, a document goes through paper shredder, silver tarnishes, while it is DECREASING when dissolved sugar precipitates, water vapor forms droplets and water cools down.
Entropy can be defined as the degree of randomness or disorder of a particular system.
Entropy is equal to zero (0) for a perfectly ordered system.
Heat increases the entropy of the system because more energy excites the molecules and it increases the amount of random activity.
Moreover, the cooling decreases the entropy of the system because molecules are more ordered and it decreases the amount of random activity.
In conclusion, entropy is INCREASING when a snowman melts, a document goes through paper shredder, silver tarnishes, while it is DECREASING when dissolved sugar precipitates, water vapor forms water droplets and the water cools down.
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Sulfuric acid is commonly used as an electrolyte in car batteries. Suppose you spill some on your garage floor. Before cleaning it up, you wisely decide to neutralize it with sodium bicarbonate (baking soda) from your kitchen. The reaction of sodium bicarbonate and sulfuric acid is
Answer:
The mass of NaHCO3 required is 235.22 g
Explanation:
*******
Continuation of Question:
2NaHCO3(s) + H2SO4(aq) → Na2SO4(aq) + 2CO2(g) + 2H2O(l)
You estimate that your acid spill contains about 1.4 mol H2SO4. What mass of NaHCO3 do you need to neutralize the acid?
********\
The question requires us to calculate the mass of NaHCO3 to neutralize the acid.
From the balanced chemical equation;
1 mol of H2SO4 requires 2 mol of NaHCO3
1.4 would require x?
Upon solving for x we have;
x = 1.4 * 2 = 2.8 mol of NaHCO3
The relationship between mass and number of moles is given as;
Mass = Number of moles * Molar mass
Mass = 2.8 mol * 84.007 g/mol
Mass = 235.22 g
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
Which of the following solutions would have the highest pH? Assume that they are all 0.10 M in acid at 25°C. The acid is followed by its Ka value.
a. HCHO2, 1.8 x 10-4
b. HF, 3.5 x 10-4
c. HClO2, 1.1 x 10-2
d. HCN, 4.9 x 10-10
e. HNO2, 4.6 x 10-4
Answer:
[tex]HCN~~Ka=4.9x10^-^1^0[/tex]
Explanation:
In this case, we have to remember the relationship between the Ka value and the pH. We can use the general reaction for any acid with his Ka value expression:
[tex]HA~->~H^+~+~A^-[/tex] [tex]Ka=\frac{[H^+][A^-]}{[HA]}[/tex]
In the Ka expression, we have a proportional relationship between Ka and the concentration of [tex]H^+[/tex]. Therefore, if we have a higher Ka value we will have a smaller pH (lets keep in mind that with a higher
So, if we have to find the higher pH value we need to search the smaller Ka value in this case [tex]HCN~~Ka=4.9x10^-^1^0[/tex].
I hope helps!
HCN has the highest pH among all the acids listed in the question.
The Ka is called the acid dissociation constant. It shows the extent to which an acid is ionized in water. The pH shows the hydrogen ion concentration of water. The higher the Ka, the higher the hydrogen ion concentration and the lower the pH.
Hence, HCN has the lowest Ka and the lowest hydrogen ion concentration. Therefore, HCN has the highest pH among all the acids listed in the question.
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Given a fixed amount of gas help at a constant pressure, calculate the temperature to which the gas would have to be changed if a 1.75 L sample at 23.0*C were to have a final volume of 3.50 L.
A. 46.0*C
B. 89.5*C
C. 169*C
D. 319*C
E. 592*C
Answer:
592 K or 319° C
Explanation:
From the statement of Charles law we know that the volume of a given mass of gas is directly proportional to its absolute temperature at constant pressure. Thus;
V1/T1= V2/T2
Initial volume V1 = 1.75 L
Initial temperature T1= 23.0 +273 = 296 K
Final volume V2= 3.50 L
Final temperature T2 = the unknown
T2= V2T1/V1= 3.50 × 296 / 1.75
T2 = 592 K or 319° C
Draw a Lewis structure for one important resonance form of HBrO4 (HOBrO3). Include all lone pair electrons in your structure. Do not include formal charges in your structure.
Answer:
The Lewis structure is attached with the answer -
Explanation:
Lewis structure or Lewis dot diagram are diagrams or representation of showing the bonding between different or same atoms of a molecule in any and also shows lone pairs of electrons that may exist in the molecule as dots.
HBrO₄ is bromine oxoacid which is also known as perbromic acid. It is a unstable inorganic compound.
The Lewis structure is attached in form of image with representation of lone pairs of electrons.
1. Suppose 1.00 g of NaOH is used to prepare 250 mL of an NaOH solution. Compare the expected molarity of this solution to the actual average molarity you measured in the standardization. What do you notice
Answer:
0.1M solution of NaOH
Explanation:
1 mole of NaOH - 40g
? moles - 1 g = 1/40 = 0.025 moles.
Molarity of 1.00g of NaOH in 0.25L (250 mL) = no. of moles/volume
= 0.025/0.25
= 0.1M.
The second-order decomposition of HI has a rate constant of 1.80 · 10-3 M-1s-1. How much HI remains after 27.3 s if the initial concentration of HI is 4.78 M?
Answer: 3.87M of HI remains after 27.3 s
Explanation:
Using the Second order decomposition equation of
1/[H]t =K x t +1/[A]o
Given initial concentration ,[A]o = 4.78M
time, t = 27.3 s
rate of constant , k= 1.80 x 10^-3 M-1s-1
1/[H] t= 1/[A] t= concentration after time, t=?
SOLUTION
1/[A] t =kt +1/[A]o
1/[A] t =(1.80 x 10^-3 (27.3)+1/4.78
0.04914+0.2092=0.2583
1/[A] t =0.2583
[A] t =1/0.2583= 3.87M
If the H+ concentration is 0.00001 M, what is the OH- concentration?
Answer:
1.00x10^-9
Explanation:
Determine the volume occupied by 10 mol of helium at
27 ° C and 82 atm
Answer:
3.00 L
Explanation:
PV = nRT
(82 atm × 101325 Pa/atm) V = (10 mol) (8.314 J/mol/K) (27 + 273) K
V = 0.00300 m³
V = 3.00 L
A 400 mL sample of hydrogen gas is collected over water at 20°C and 760 torr the vapor pressure of water at 20°C is 17.5 torr. what volume will the dry hydrogen gas occupy at 20°C and 760 torr?
Answer:
V2 = 17371.43ml
Explanation:
We use Boyles laws
since temperature is constant
P1V1=P2V2
760 x 400 = 17.5 x V2
304000 = 17.5 x V2
V2 = 304000/17.5
V2 = 17371.43ml
The volume will the dry hydrogen gas occupy at the temperature of 20°C and vapor pressure at 760 torrs will be 18 ml.
What is vapor pressure?
The vapor pressure of a liquid is independent of the volume of liquid in the container, whether one liter or thirty liters; both samples will have the same vapor pressure at the same temperature.
The temperature has an exponential connection with vapor pressure, which means that as the temperature rises, the vapor pressure rises as well the equation is -
P1 V1 / T1 = P2 V2 / T1
here, P = pressure
T = temperature
V = volume
substituting the value in the equation,
400 ×760 / 20 = 17.5× V / 20
V = 400× 760 / 20 × 17.5 / 20
V = 18 ml
Therefore the volume of the hydrogen gas remaining at this temperature will be 18 ml.
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Identify the elements that have the following abbreviated electron configurations.
A) [Ne] 3s23p5.
B) [Ar] 4s23d7.
C) [Xe] 6s1.
Answer:
A) Chlorine (Cl)
B) Cobalt (Co)
C) Caesium (Cs)
Hope this helps.
The abbreviated electron configurations that was given in the question belongs to
Chlorine (Cl)
Cobalt (Co)
Caesium (Cs) respectively.
Electronic configurations can be regarded as the electronic structure, which is the way an electrons is arranged in energy levels towards an atomic nucleus.The electron configurations is very useful when describing the orbitals of an atom in its ground state.To calculate an electron configuration, we can put the periodic table into sections, and this section will represent the atomic orbitals which is the regions that house the electrons. Groups one of the period table and two belongs to s-block, group 3 through 12 belongs to the d-block, while 13 to 18 can be attributed to p-block ,The rows that is found at bottom are the f-blockTherefore, electron configurations explain orbitals of an atom when it is in it's ground state.
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Aluminum and oxygen react according to the following equation: 4Al + 3O2 -> 2Al2O3 In a certain experiment, 4.6g Al was reacted with excess oxygen and 6.8g of product was obtained. What was the percent yield of the reaction?
Answer:
Percent yield: 78.2%
Explanation:
Based on the reaction:
4Al + 3O₂ → 2Al₂O₃
4 moles of Al produce 2 moles of Al₂O₃
To find percent yield we need to find theoretical yield (Assuming a yield of 100%) and using:
(Actual yield (6.8g) / Theoretical yield) × 100
Moles of 4.6g of Al (Molar mass: 26.98g/mol) are:
4.6g Al × (1mol / 26.98g) = 0.1705 moles of Al.
As 4 moles of Al produce 2 moles of Al₂O₃, theoretical moles of Al₂O₃ obtained from 0.1705 moles of Al are:
0.17505 moles Al × (2 moles Al₂O₃ / 4 moles Al) = 0.0852 moles of Al₂O₃,
In grams (Molar mass Al₂O₃ = 101.96g/mol):
0.0852 moles of Al₂O₃ × (101.96g / mol) =
8.7g of Al₂O₃ can be produced (Theoretical yield)Thus, Percent yield is:
(6.8g / 8.7g) × 100 =
78.2%If D+2 would react with E-1, what do you predict to be the formula?
Answer:
DE2
Explanation: for every one D+2 you need two E-1 because +2=-2
Given a fixed amount of gas in a rigid container (no change in volume), what pressure will the gas exert if the pressure is initially 1.50 atm at 22.0oC, and the temperature is changed to 11.0oC?
A. 301 atm
B. 1.56 atm
C. 0.750 atm
D. 1.44 atm
E. 3.00 atm
Answer:
The pressure the gas will have if the pressure is initially 1.50 atm at 22.0 ° C and the temperature changes at 11.0 ° C is 1.44 atm (option D)
Explanation:
Gay Lussac's law indicates that, as long as the volume of the container containing the gas is constant, as the temperature increases, the gas molecules move more rapidly. Then the number of collisions against the walls increases, that is, the pressure increases. That is, the gas pressure is directly proportional to its temperature.
Gay-Lussac's law can be expressed mathematically as follows:
[tex]\frac{P}{T}=k[/tex]
Where P = pressure, T = temperature, K = Constant
You have a gas that is at a pressure P1 and at a temperature T1. When the temperature varies to a new T2 value, then the pressure will change to P2, and then:
[tex]\frac{P1}{T1}=\frac{P2}{T2}[/tex]
In this case:
P1= 1.50 atmT1= 22 °C= 295 °K (being 0°C= 273 °K)P2= ?T2= 11 °C= 284 KReplacing:
[tex]\frac{1.5 atm}{295 K}=\frac{P2}{284 K}[/tex]
Solving:
[tex]P2= 284 K*\frac{1.5 atm}{295 K}[/tex]
P2=1.44 atm
The pressure the gas will have if the pressure is initially 1.50 atm at 22.0 ° C and the temperature changes at 11.0 ° C is 1.44 atm (option D)
The pH of a solution prepared by mixing 40.00 mL of 0.10 M NH3 with 50.00 mL of 0.10 M NH4Cl and 30mL of 0.05 M H2SO4 is 5.17. Assume that the volume of the solutions are additive . What would be the Ka for NH4
Answer:
Following are the answer to this question:
Explanation:
The value of pH solution is =5.17 So, the p^{OH}:
[tex]p^{OH}[/tex]=14-56.17
=8.823
The volume of the [tex]NH_{3}[/tex] = 40.00 ml
convert into the liter= 0.040L
The value of the concentrated [tex]NH_{3}[/tex] =0.10 M
The volume of the [tex]NH_{4}Cl[/tex]= 50.00 ml
convert into the liter= 0.050L
The value of concentrated [tex]NH_{4}Cl[/tex]= 0.10 M
The volume of the [tex]H_{2}So_{4}[/tex]= 30 ml
convert into the liter= 0.030L
The value of concentrated [tex]H_2So_4[/tex]=0.05 M
Calculating total volume=(0.40+0.050+0.030)
=0.120 L
calculating the new concentrated value of [tex]NH_3[/tex] = [tex]\frac{0.10\times 0.040}{0.120}= 0.33 \ M[/tex]
calculating the new concentrated value of [tex]NH_4Cl[/tex]= [tex]\frac{0.050\times 0.10}{0.120}= 0.04166 \ M[/tex]calculating the new concentrated value of [tex]H_2So_4= \frac{0.030\times 0.05}{0.120}= 0.0125 \ M[/tex] when 1 mol [tex]H_2So_4[/tex] produced 2 mols [tex]H^{+}[/tex] so, 0.0125 in [tex]H_2So_4[/tex]produced:
[tex]=4 \times (2 \times 0.0125) \ mol H^{+}\\\\= 0.025 mol H^{+}[/tex]
create the ICE table:
[tex]NH_3 \ \ \ \ \ \ \ \ + H^{+} \ \ \ \ \ \ \longrightarrow NH_4^{+}[/tex]
I (m) 0.033(m) 0.025 0.04166
C -0.025 -0.025 + 0.025
E 8.3\times 10^{-3} 0 0.0667
now calculating pH:
when ph= 8.83:
[tex]P^{H}= p^{kb}|+ \log\frac{[NH_4^{+}]}{[NH_3]}\\\\8.83=p^{kb}+\log\frac{0.0667}{8.3 \times 10^{-3}}\\\\p^{kb}=8.83-0.9069\\\\ \ \ \ =7.7231 \\\\\ The P^{kb} \ for \ NH_3 \ is =7.7231\\\\\ The P^{kb} \ for N^{+}H_4=14-7.7231\\\\\ \ \ \ \ \ =6.2769[/tex]
Which of these are elimination reactions? Check all that apply.
CH3OH + CH3COOH → CH3CO2CH3 + H20
C3H7OH → C3H6 + H20
H9C2Br + NaOH → C2H4 + NaBr + H20
Answer:
C3H7OH → C3H6 + H20
Explanation:
If we look at the reactant and the product we will realize that the reactant is an alcohol while the product is an alkene. The reaction involves acid catalysed elimination of water from an alcohol.
Water is a good leaving group, hence an important synthetic route to alkenes is the acid catalysed elimination of water from alcohols. Hence the conversion represented by C3H7OH → C3H6 + H20 is an elimination reaction in which water is the leaving group.
Answer:
B and C. Just finished my lesson on Edge.
Review the reversible reactions given, along with the associated equilibrium constant Kat room temperature. In each case, determine whether the forward or reverse reaction is favored.
CH3COOH → CH3C00^- + H^+
Ka=1.8 x 10^-5
AgCl → Ag^+ + Cl^-
Ksp=1.6 x 10^-10
Al(OH)3 → Al^3+ + 3OH^-
Ksp=3.7 x 10^-15
A+B → C
K=4.9 x 10^3
Answer:
The answers to your questions are given below
Explanation:
The following data were obtained from the question:
CH3COOH → CH3C00^- + H^+
Equilibrium constant, Ka = 1.8 x 10^-5
AgCl → Ag^+ + Cl^-
Equilibrium constant, Ksp = 1.6 x 10^-10
Al(OH)3 → Al^3+ + 3OH^-
Equilibrium constant, Ksp = 3.7 x 10^-15
A+B → C
Equilibrium constant, K = 4.9 x 10^3
When the value of the equilibrium constant is grater than 1, it shows that the concentration of product is higher than that of the reactant and it implies that the forward reaction is favored.
When the value of the equilibrium constant is 1, it shows that the the concentration of the product and reactant are the same. Therefore neither the forward nor the reverse reaction is favored.
When the value of the equilibrium constant is lesser than 1, it shows that the concentration of the reactant is higher than the concentration of the product. Therefore, the reversed reaction is favored.
Now, we shall the question given above as follow:
A. CH3COOH → CH3C00^- + H^+
Equilibrium constant, Ka = 1.8 x 10^-5
Since the value of the equilibrium constant is lesser than 1, it means that the reverse reaction is favored.
B. AgCl → Ag^+ + Cl^-
Equilibrium constant, Ksp = 1.6 x 10^-10
Since the value of the equilibrium constant is lesser than 1, it means that the reverse reaction is favored.
C. Al(OH)3 → Al^3+ + 3OH^-
Equilibrium constant, Ksp = 3.7 x 10^-15
Since the value of the equilibrium constant is lesser than 1, it means that the reverse reaction is favored.
D. A+B → C
Equilibrium constant, K = 4.9 x 10^3
Since the value of the equilibrium constant is greater than 1, it means that the forward reaction is favored.
The reaction conditions are:
A. The reverse reaction is favored.
B. The reverse reaction is favored.
C. The reverse reaction is favored.
D. The forward reaction is favored.
Chemical reaction:
A. [tex]CH_3COOH[/tex] → [tex]CH_3COO^- + H^+[/tex]
Equilibrium constant, Ka = [tex]1.8 * 10^{-5}[/tex]
B. [tex]AgCl[/tex] → [tex]Ag^+ + Cl^-[/tex]
Equilibrium constant, Ksp = [tex]1.6 * 10^{-10}[/tex]
C. [tex]Al(OH)_3[/tex] → [tex]Al^{3+} + 3OH^-[/tex]
Equilibrium constant, Ksp = [tex]3.7 * 10^{-15}[/tex]
D. A+B → C
Equilibrium constant, K = [tex]4.9 * 10^3[/tex]
Conditions for Equilibrium constant:When the value of the equilibrium constant is greater than 1, it shows that the concentration of product is higher than that of the reactant and it implies that the forward reaction is favored.
When the value of the equilibrium constant is 1, it shows that the the concentration of the product and reactant are the same. Therefore neither the forward nor the reverse reaction is favored.
When the value of the equilibrium constant is lesser than 1, it shows that the concentration of the reactant is higher than the concentration of the product. Therefore, the reversed reaction is favored.
Thus, the reactions will be:
A. The reverse reaction is favored.
B. The reverse reaction is favored.
C. The reverse reaction is favored.
D. The forward reaction is favored.
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brainly.com/question/12858312
Solid iron(II) oxide reacts with oxygen gas to produce solid iron(III) oxide. Balance the equation for this reaction (in lowest multiple integers). Write the unbalanced equation for this reaction.
Answer
Hello
I think the reaction is like this FeO+OFe²O³
And the balance reaction is 2Fe+OFe²O³
Explanation:
At first we should find sth that has more atoms than the other then for example we realized that we have two atoms of Fe in Fe²O³ then put 2 before FeO and now we have 2 atoms of Fe in right side and 2 atoms of Fe in left then Oxygen in FeO change to 2 atoms of Oxygen and we have an other one in right side that they become 3 atoms of Oxygen and now we have 3 atoms of Oxygen in both right and left side.
Finally our reaction balanced.
Good luck