Explanation:
system at equilibrium, will the reaction shift towards reactants ~
--?'
2. (2 Pts) Consider the reaction N2(g) + 3H2(g) =; 2NH3(g). The production of ammonia is an
exothermic reaction. Will heating the equilibrium system increase o~e amount of
ammonia produced? . .co:(
3. (2 Pts) Consider the reaction N2(g) + 3H2(g) =; 2NH3(g). Ifwe use a catalyst, which way will
the reaction shift? ':'\
.1.+- w~t s~,H (o')l r'eo.c. e~ ei~i"liht-,·u.fn\ P~~,
4. (3 Pts) ff 1ven th e o £ 11 owmg d t a a £ or th ere action: A(g) + 2B(s) =; AB2(g)
Temperature (K) Kc
300 1.5x104
600 55 k ' pr, cl l<..J~
e- ~ r fee, ct o. ~ 1<
900 3.4 X 10-3
Is the reaction endothermic or exothermic (explain your answer)?
t d- IS o.,;r-. \4\a..i~1f't~ °the te.Y'il(lf1,:J'u.r-a a•~S. j lrvdu..c,,.) +~H~to{' '\
exothe-rnh't.-- ,.. ..,. (/.., ,~.
5. (4 Pts) Consider the reaction, N2(g) + 3H2(g) =; 2NH3(g). Kc= 4.2 at 600 K.
What is the value of Kc for 4 NH3(g) =; 2N2(g) + 6H2(g)
N ... ~l + 3 H~(ri ~ ~Nli3~) kl,= ~:s;H,J3 # 4. J..
~ ;)N~~) ~ ~ H ~) ~\-_ == [A!;J:t D~~Jb
J. [,v 1+3] ~
I
4,:i.~ = 0,05
5.00 mol of ammonia are introduced into a 5.00 L reactor vessel and when the equilibrium is reached, 1.00 mole remains. The concentration equilibrium constant is 17.3.
Initially, there are 5.00 mol of ammonia in a 5.00 L reactor vessel. The initial concentration of ammonia is:
[tex][NH_3]_i = \frac{5.00mol}{5.00L} = 1.00 M[/tex]
At equilibrium, there is 1.00 mole of ammonia in the 5.00 L vessel. The concentration of ammonia at equilibrium is:
[tex][NH_3]eq = \frac{1.00mol}{5.00L} = 0.200 M[/tex]
We can calculate the concentrations of all the species at equilibrium using an ICE chart.
2 NH₃(g) ⇄ 3 H₂(g) + N₂(g)
I 1.00 0 0
C -2x +3x +x
E 1.00-2x 3x x
Since the concentration of ammonia at equilibrium is 0.200 M,
[tex]1.00-2x = 0.200\\\\x = 0.400 M[/tex]
The concentrations of all the species at equilibrium are:
[tex][NH_3] = 0.200 M\\[H_2] = 3x = 1.20 M\\[N_2] = x = 0.400 M[/tex]
The concentration equilibrium constant (Kc) is:
[tex]Kc = \frac{[H_2]^{2} [N_2]}{[NH_3]^{2} } = \frac{(1.20^{3})(0.400) }{0.200^{2} } = 17.3[/tex]
5.00 mol of ammonia are introduced into a 5.00 L reactor vessel and when the equilibrium is reached, 1.00 mole remains. The concentration equilibrium constant is 17.3.
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What is the balanced form of the chemical equation shown below?
Ca(OH)2(aq) + Na2CO3(aq) → CaCO3(s) + NaOH(aq)
Answer:
D
Explanation:
Double Displacement reaction
Both sides are balanced with option D
The balanced form of the chemical equation shown below is [tex]\rm Ca(OH)_2(aq) + Na_2CO_3(aq) \rightarrow CaCO_3(s) + 2NaOH(aq).[/tex] The correct option is D.
What is a balanced equation?A balanced equation is where the reactant and the product have the number of moles of elements. According to the law, the reaction, and the product have the same number of moles after the reaction, so balancing an equation is important.
To balance an equation, it is significant to see the number of moles of reactant and the same number of moles is in the product side. Here the moles of sodium has to be balanced.
Thus, the correct option is D, [tex]\rm Ca(OH)_2(aq) + Na_2CO_3(aq) \rightarrow CaCO_3(s) + 2NaOH(aq).[/tex]
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For the reaction X + Y → Z, the reaction rate is found to depend only upon the concentration of X. A plot of 1/X verses time gives a straight line. What is the rate law for this reaction?
Answer:
r = k [X]²
Explanation:
X + Y → Z
Generally, the rate of reaction depends on the concentration of reactants. However, the question stated that the rate depends only on reactant X.
The plot of 1/X versus time giving a straight line signifies that this is a second order reaction.
For a second-order reaction, a plot of the inverse of the concentration of a reactant versus time is a straight line with a slope of k.
From this, our rate law is r = k [X]²
Zinc bromide is considered which of the following?
A) molecular compound
B) atomic element
C) molecular element
D) ionic compound
Answer:
D
Explanation:
soluble in water and acidicA multistep reaction can only occur as fast as its slowest step. Therefore, it is the rate law of the slow step that determines the rate law for the overall reaction. Consider the following multistep reaction:
A+B ----- AB(slow)
A+AB-----A2B(fast
....................................................
2A+B ----- A2B(overall)
Based on this mechanism, determine the rate law for the overall reaction.
a) rate = kA2BAB
b) rate = kAB
c) rate = kAAB
d) rate = kA2B
Answer:
b) rate = kAB.
Explanation:
Hello,
In this case, considering the given statement, we can notice that the rate law of the overall reaction will be determined for the slowest step, that is:
[tex]A+B \rightarrow AB\ \ (slow)[/tex]
In such a way, we can infer that the rate law will contain both the concentration of A and B to the first power both, since their stoichiometric coefficients in the chemical equation are both one:
[tex]rate=k[A][B][/tex]
Thereby, answer is b) rate = kAB, that should be better rate = k[A][B] by expressing the concentrations.
Best regards.
Which of these groups of elements show the least electronegativity?
Explanation:
On the periodic table, electronegativity generally increases as you move from left to right across a period and decreases as you move down a group. As a result, the most electronegative elements are found on the top right of the periodic table, while the least electronegative elements are found on the bottom left. The answer is alkali metals.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
Sample gas has a volume of 3.40 L at 10°C what will be its volume at 100°C pressure remaining constant
Answer:
V2 = 4.48L
Explanation:
using charles law
V1/T1=V2/T2
3.4/283=V2/373
0.012=V2/373
V2= 0.012 x 373
V2 = 4.48L
Aspirin (C9H8O4) is produced by the reaction of salicylic acid (C7H6O3, Molar mass = 138.1 g/mol) and acetic anhydride (C4H6O3, Molar mass = 102.1 g/mol) based on the BALANCED equation : C7H6O3(s) + C4H6O3(l ) → C9H8O4(s) + C2H4O2( l) If 63.07 grams of aspirin (Molar mass = 180.2 g/mol) was collected from an experiment when 138.1 grams C7H6O3 reacted with excess C4H6O3, what was the percent yield?
Answer:
35%
Explanation:
Percentage yield = actual yield / theoretical yield × 100.
Given:
Actual yield = 63.07g
Theoretical yield = ?
Mole ratio of C7H6O3 to C4H6O3 = 1 : 1
1 mole of C7H6O3 - 138.1g
Which implies that only 1 mole s[tex]\frac{63.07}{180.2} * 100[/tex]hould be used up in the reaction, yielding 180.2 g of C9H8O4. ⇒ Theoretical yield = 180.2g
∴ % Yield = [tex]\frac{63.07}{180.2} * 100[/tex]
= 35% yield.
Let me know if you found this easy to understand.
Calculate the mass percent of .485g of H, which reacts with O to form 2.32g H2O?
Answer:
53.1% of hydrogen reacts
Explanation:
The mixture of 2 atoms of H with 1 atom of O produce 1 molecule of H₂.
The mass of hydrogen in 2.32g of H₂O could be obtained using molar mass of H₂O (18.01g/mol) and molar mass of hydrogen (1.01g/mol) as follows:
Moles H₂O: 2.32g H₂O × (1mole / 18.01g) = 0.1288 moles of water
1 mole of H₂O contains 2 moles of H, moles of hydrogen in 0.1288 moles of water are:
0.1288 moles H₂O × (2 moles H / 1 mole H₂O) = 0.2576 moles of H
In mass:
0.2576 moles H × (1.01g/ mol H) = 0.260g H you have in the formed water
As before reaction you had 0.485g of H and just 0.260g reacted, mass percent is:
(Mass that reacts / Mass added) × 100
(0.260g / 0.485g) × 100 =
53.1% of hydrogen reactsWhat amounts of sodium benzoate would be required to prepare 2.5L of 0.35M benzoic buffer solution with a pH of 6.10? Ka of benzoic acid = 6.5 x 10-5 MW benzoic acid, HC7H5O2, is 122.01 MW sodium benzoate, NaC7H5O2, is 144.01
Answer:
Benzoic acid: 1.288g
Sodium benzoate: 124.48g
Explanation:
Benzoic acid, HC7H5O2 is in equilibrium with its conjugate base, C7H5O2⁻ producing a buffer. The pH of the buffer can be determined following H-H equation:
pH = pKa + log [C7H5O2⁻] / [HC7H5O2] (1)
Where pH is desire pH = 6.10 pKa is -log Ka = 4.187 and [] are molar concentrations of the buffer.
As you want to prepare 2.5L of a 0.35M of buffer, moles of buffer are:
2.5L ₓ (0.35mol / L) = 0.875moles of buffer.
And you can write:
0.875 moles = [C7H5O2⁻] + [HC7H5O2] (2)
Replacing (2) in (1)
pH = pKa + log [C7H5O2⁻] / [HC7H5O2]
6.10 = 4.187 + log [C7H5O2⁻] / [HC7H5O2]
1.913 = log [C7H5O2⁻] / [HC7H5O2]
81.846 = 0.875mol - [HC7H5O2] / [HC7H5O2]
81.846 [HC7H5O2] = 0.875mol - [HC7H5O2]
82.846 [HC7H5O2] = 0.875mol
[HC7H5O2] = 0.01056 molesAnd moles of the benzoate, [C7H5O2⁻]:
[C7H5O2⁻] = 0.875mol - 0.01056mol =
[C7H5O2⁻] = 0.8644molUsing molar mass of benzoic acid and sodium benzoate, amount of each compound you must add to prepare 2.5L of the buffer are:
Benzoic acid: 0.01056mol ₓ (122.01g/mol) = 1.288g
Sodium benzoate: 0.8644mol ₓ (144.01g/mol) = 124.482g
1. The following thermochemical equation is for the reaction of water(l) to form hydrogen(g) and oxygen(g). 2H2O(l)2H2(g) + O2(g) H = 572 kJ How many grams of H2O(l) would be made to react if 110 kJ of energy were provided? _____ grams
2. The following thermochemical equation is for the reaction of carbon monoxide(g) with hydrogen(g) to form methane(g) and water(g). CO(g) + 3H2(g) CH4(g) + H2O(g) H = -206 kJ When 6.27 grams of carbon monoxide(g) react with excess hydrogen(g),_____ kJ of energy are ____ a.evolved b.absorbed
Answer:
1. 6.92 g of H2O
2i. - 46 KJ of energy.
ii. Option A. Evolved.
Explanation:
1. Determination of the mass of H2O that would be made to react if 110 kJ of energy were provided.
This can be obtained as follow:
The equation for the reaction is given below
2H2O(l) —> 2H2(g) + O2(g) H = 572 kJ
Next, we shall determine the mass of H2O required to produce 572 kJ from the balanced equation.
Molar mass of H2O = (2x1) + 16 = 18 g/mol
Mass of H2O from the balanced equation = 2 x 18 = 36 g
From the balanced equation above, 36 g of H2O reacted to produce 572 kJ of energy.
Finally, we shall determine the mass of water (H2O) needed to produce 110 kJ of energy.
This is illustrated below:
From the balanced equation above, 36 g of H2O reacted to produce 572 kJ of energy.
Therefore, Xg of H2O will react to 110 kJ of energy i.e
Xg of H2O = (36 x 110)/572
Xg of H2O = 6.92 g
Therefore, 6.92 g of H2O is needed to react in order to produce 110 KJ of energy.
2i. Determination of the energy.
The balanced equation for the reaction is given below:
CO(g) + 3H2(g) —> CH4(g) + H2O(g) H = -206 kJ
Next, we shall determine the mass of CO that reacted to produce -206 kJ of energy from the balanced equation.
This is illustrated below:
Molar mass of CO = 12 + 16 = 28 g/mol
Mass of CO from the balanced equation = 1 x 28 = 28 g
From the balanced equation above,
28 g of CO reacted to produce -206 kJ of energy.
Finally, we shall determine the amount of energy produced by reacting 6.27 g of CO. This is illustrated below:
From the balanced equation above,
28 g of CO reacted to produce -206 kJ of energy.
Therefore, 6.27 g of CO will react to produce = (6.27 x -206)/28 = - 46 KJ of energy.
Therefore, - 46 KJ of energy were produced from the reaction.
2ii. Since the energy obtained is negative, it means heat has been given off to the surroundings.
Therefore, the heat is evolved.
What do chemists use percent yield calculations for in the real world?
A. To balance the reaction equation.
B. To determine how much product they will need.
C. To determine how efficient reactions are.
D. To determine how much reactant they need.
Answer:
C. To determine how efficient reactions are.
D. To determine how much reactant they need.
Explanation:
When you are doing a reaction, you are hoping for a percent yield to close of 100%. You make the reaction and determine how many product you obtain. If you know the percent yield of a reaction you can calculate the amount of reactant you need to obtain a determined amount of product.
Having this in mind:
A. To balance the reaction equation. false. To calculate percent yield you need to balance the reaction before. You don't use percent yield to balance the reaction
B. To determine how much product they will need. false. You determine how much product you obtain after the reaction. How much product you need is independent of percent yield
C. To determine how efficient reactions are. true. A way to determine efficience of a reaction is with percent yield. An efficient reaction has a high percent yield.
D. To determine how much reactant they need. true. If you know percent yield of a reaction you can know how many reactant you must add to obtain the amount of product you want.
A sample of an unknown gas effuses in 11.1 min. An equal volume of H2 in the same apparatus at the same temperature and pressure effuses in 2.42 min. What is the molar mass of the unknown gas
Answer:
Molar mass of the gas is 0.0961 g/mol
Explanation:
The effusion rate of an unknown gas = 11.1 min
rate of [tex]H_{2}[/tex] effusion = 2.42 min
molar mass of hydrogen = 1 x 2 = 2 g/m
molar mas of unknown gas = ?
From Graham's law of diffusion and effusion, the rate of effusion and diffusion is inversely proportional to the square root of its molar mass.
from
[tex]\frac{R_{g} }{R_{h} }[/tex] = [tex]\sqrt{\frac{M_{h} }{M_{g} } }[/tex]
where
[tex]R_{h}[/tex] = rate of effusion of hydrogen gas
[tex]R_{g}[/tex] = rate of effusion of unknown gas
[tex]M_{h}[/tex] = molar mass of H2 gas
[tex]M_{g}[/tex] = molar mass of unknown gas
substituting values, we have
[tex]\frac{11.1 }{2.42 }[/tex] = [tex]\sqrt{\frac{2 }{M_{g} } }[/tex]
4.587 = [tex]\sqrt{\frac{2 }{M_{g} } }[/tex]
[tex]\sqrt{M_{g} }[/tex] = [tex]\sqrt{2}[/tex]/4.587
[tex]\sqrt{M_{g} }[/tex] = 0.31
[tex]M_{g}[/tex] = [tex]0.31^{2}[/tex] = 0.0961 g/mol
The molar mass of the unknown gas will be "0.0961 g/mol".
Given:
Effusion rate of unknown gas,
[tex]R_g = 11.1 \ min[/tex]Effusion rate of [tex]H_2[/tex],
[tex]R_h = 2.42 \ min[/tex]Molar mass of hydrogen,
[tex]M_h = 1\times 2[/tex][tex]= 2 \ g/m[/tex]
According to the Graham's law, we get
→ [tex]\frac{R_g}{R_h} = \sqrt{\frac{M_h}{M_g} }[/tex]
By substituting the values, we get
→ [tex]\frac{11.1}{2.42} = \sqrt{\frac{2}{M_g} }[/tex]
→ [tex]4.587=\sqrt{\frac{2}{M_g} }[/tex]
→ [tex]\sqrt{M_g} = \sqrt{\frac{2}{4.587} }[/tex]
[tex]\sqrt{M_g} = 0.31[/tex]
[tex]M_g = 0.0961 \ g/mol[/tex]
Thus the above solution is right.
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What was Ernest Rutherford experiment
Medical implants and high-quality jewelry items for body piercings are frequently made of a material known as G23Ti or surgical-grade titanium. The percent composition of the material is 64.39% titanium, 24.19% aluminum, and 11.42% vanadium. What is the empirical formula for surgical-grade titanium
Answer:
The Empirical Formular is given as; Ti₆Al₄V
Explanation:
The percent composition of the material is 64.39% titanium, 24.19% aluminum, and 11.42% vanadium.
Elements Titanium Aluminium Vanadium
Percentage 64.39 24.19 11.42
Divide all through by their molar mass
64.39 / 47.87 24.19 / 27 11.42 / 50.94
= 1.345 = 0.896 = 0.224
Divide all though by the smallest number (0.224)
1.345 / 0.224 0.896 / 0.224 0.224 / 0.224
= 6 = 4 = 1
The Empirical Formular is given as; Ti₆Al₄V
Using the stepwise procedure for obtaining the empirical formula of a compound, the empirical formula is [tex] T_{6}Al_{4}V[/tex]
Titanium :
Percentage composition = 64.39%Molar mass = 47.87Divide by Molar mass : = 64.39/47.87 = 1.345
Aluminum :
Percentage composition = 24.19%Molar mass = 27Divide by Molar mass : = 24.19/27 = 0.896
Vanadium :
Percentage composition = 11.42%Molar mass = 50.94%Divide by Molar mass : = 11.42/50.94 = 0.224
Divide by the smallest :
Titanium = 1.345 / 0.224 = 6.00
Aluminum = 0.896 / 0.224 = 4
Vanadium = 0.224 / 0.224 = 1
Hence, the empirical formula is [tex] T_{6}Al_{4}V[/tex]
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What is the purpose of reacting 2.0mL of HNO3 with 2.0 mL of H2SO4 in a separate test tube, prior to adding it to the solution containing the substrate
The question is incomplete, the complete question is;
What is the purpose of reacting 2.0mL of HNO3 with 2.0 mL of H2SO4 in a separate test tube, prior to adding it to the solution containing the substrate? more than one answer is possible
A) The release of a water molecule that acts as an electrophile in the reaction with methyl benzoate.
B) The formation of nitronium ion, which acts an electrophile in the reaction with methylbenzoate.
C)The formation of bisulfate (hydrogen sulfate), which acts as an electrophile in the reaction with methylbenzoate.
D)The release of a water molecule that acts as a nucleophile in the reaction with methyl benzoate.
Answer:
B) The formation of nitronium ion, which acts an electrophile in the reaction with methylbenzoate.
Explanation:
The benzene ring is known to be stable hence it can only undergo a substitution reaction with the aromatic ring still intact. When the substitution reaction involves an electrophile we refer to the process as electrophillic aromatic substitution. Electrophilic aromatic substitution is a useful synthetic route for many organic compounds.
In the electrophilic substitution of methyl benzoate using the 1:1 volume ratio mixture of H2SO4/HNO3, the nitronium ion (NO2+) is the electrophile generated in the test tube. It is this NO2+ that now reacts with the methyl benzoate to yield the reaction product.
If a 275 mL gas container had pressure of 732.6 mm Hg at -28°C and the gas was condensed into a liquid with a mass of 1.95 g, what is the molar mass of the gas?
Answer:
THE MOLAR MASS OF THE GAS IS 147.78 G/MOLE
Explanation:
Using PV = nRT
n = Mass / molar mass
P = 732.6 mmHg = 1 atm = 760 mmHg
So therefore 732.6 mmHg will be equal to 732.6 / 760 = 0.964 atm
P = 0.964 atm
V = 275 mL = 275 *10 ^-3 L
R = 0.082 Latm/ mol K
T = -28 C = 273 - 28 K = 245 K
mass = 1.95 g
molar mass = unknown
Having known the other variables in the formula, the molar mass of the gas can be obtained.
PV = m R T/ molar mass
Molar mass = m RT / PV
Molar mass = 1.95 * 0.082 * 245 / 0.964 * 275 *10^-3
Molar mass = 39.1755 / 265.1 *10^-3
Molar mass = 39.1755 / 0.2651
Molar mass = 147.78 g/mol
The molar mass of the gas is 147.78 g/mol
Determine the pH during the titration of 25.5 mL of 0.276 M hydroiodic acid by 0.105 M barium hydroxide at the following points: (1) Before the addition of any barium hydroxide .55 (2) After the addition of 16.7 mL of barium hydroxide (3) At the equivalence point (4) After adding 40.7 mL of barium hydroxide
Answer:
1) before the addition of barium hydroxide
pH = -log[H⁺] = -log (0.276) = 0.559≈0.56
2)after the addition of barium hydroxide
pH = -log [H⁺] = -log(0.0857) = 1.067
3)at equivalent point, the solution will be neutral
pH = 7.0
4) after adding 40.7mL barium hydroxide
Explanation:
equation of reaction
2HCl(aq) + Ba(OH)₂(aq) ------->BaCl₂(aq) + 2H₂O(l)
1) Before the addition of barium hydroxide
concentration of HBr = 0.276M
[H⁺] = 0.276M
pH = -log[H⁺] = -log (0.276) = 0.559≈0.56
2) After adding 16.7mL barium hydroxide
moles of [OH⁻] = 16.7mL × 0.105 × 2
=3.507m mol = 3.507 × 10³mol
moles of [H⁺] = 25.5mL × 0.276M
=7.038m mol = 7.038 × 10³mol
moles of [H⁺] remaining = (7.038 - 3.421)m mol
= 3.617m mol = 3.617 × 10³mol
[H⁺]= [tex]\frac{3.617}{25.5 + 16.7}[/tex] = 0.0857
pH = -log [H⁺] = -log(0.0857) = 1.067
3) At equivalent point, the solution will be neutral
pH = 7.0
4) After adding 40.7mL barium hydroxide
moles of [OH⁻] = 40.7mL × 0.105M × 2
=8.547
moles of [OH⁻] remaining = 8.547 - 7.038
= 1.509m mol = 1.509 × 10³mol
pOH= -log[OH⁻]= 2.82
pH = 14 - 2.82 = 11.18
An endothermic reaction proceeds in the forward direction. Which of the following statements will be true if the temperature at which the reaction occurs is decreased?
a. The equilibrium constant increases.
b. The reaction shifts toward the products.
c. The concentrations of the products decrease.
d. The equilibrium constant decreases
Answer:
The correct answers are:
c. The concentrations of the products decrease
d. The equilibrium constant decreases
Explanation:
Changes in temperature shift the equilibrium. In this problem, the reaction is endothermic, so it absorbs heat so heat is considered as a reactant:
Reactants + heat ⇒ Products
If the temperature is decreased, the heat is decreased, so reactants are removed from the reaction at equilibrium. According to Le Chaterlier's principle, the system will try to compensate the produced change. If reactants are removed, the systems will form reactants and the equilibrium will shift toward the left (formation of more reactants). In consequence, the amount of products will be decreased.
Thus, acorrect option is: c. The concentrations of the products decrease.
Since the equilibrium constant is given by the ratio of concentration of products over concentration of reactants, if the concentration of products decrease, the equilibrium constant also decreases. So, another correct option is: d. The equilibrium constant decreases.
A pharmaceutical company is making a large volume of nitrous oxide (NO). They predict they will be able to make a maximum amount of 4860 grams with the materials they have in stock. From the previous 10 volumes they have made, they know that the percent yield of this reaction is fairly low at 47%. How much will the actual yield be?
Answer:
2284.2 g.
Explanation:
The following data were obtained from the question:
Percentage yield = 47%
Theoretical yield = 4860 g
Actual yield =?
The percentage yield is simply defined as the ratio of actual yield to the theoretical yield multiplied by 100. Mathematically, it is expressed as:
Percentage yield = Actual yield /Theoretical yield x 100
With the above formula, we can obtain the actual yield as follow:
Percentage yield = Actual yield /Theoretical yield x 100
47% = Actual yield /4860
Cross multiply
Actual yield = 47% × 4860
Actual yield = 47/100 x 4860
Actual yield = 2284.2 g
Therefore, the actual yield is 2284.2 g.
Draw the Lewis structure of ethyne (C₂H₂) and then choose the appropriate pair of molecular geometries of the two central atoms. Your answer choice is independent of the orientation of your drawn structure.
A) linear / linear
B) trigonal/pyramidal
C) pyramidal/trigonal
D) trigonal pyramidal/trigonal pyramidal
E) planar / linear
Answer:
A) linear / linear
Explanation:
In this case, we have a triple bond beetween the atoms (See figure 1). If we have this triple bond we will have an Sp hybridization (in both carbons). We have to remember the relationship between the geometry and the hybridization:
-) Sp3 = Tetrahedral
-) Sp2 = Trigonal
-) Sp = Linear
Due to the hybridization, we will have a linear structure between the atoms. The angle between the atoms is 180º (See figure 2).
So, if we have a hybridization Sp for both carbons, we will have a linear geometry in each carbon. Therefore, the answer is A.
Considering that catalysts are not consumed in a reaction, how do you think increasing the amount of catalyst would affect the reaction rate for the decomposition of hydrogen peroxide?
a. increase
b. decrease
c. no effect
Answer:
a. increase
Explanation:
Catalysis is the process of increasing the rate of a chemical reaction by adding a substance known as a catalyst, which is not consumed in the catalyzed reaction.
By default, catalysts exists to speed up the rate of reactions. Increasing the amount of catalysts means that there would be an increase in the rate of reaction. The correct option is A.
Need help please! Theres three parts to this that I don't understand at all
Answer:
Chemical reaction B governs the process
Explanation:
The first part of the question asks to convert the mass of the calcium carbonate given to number of moles.
Mathematically;
Number of moles = mass/molar mass
Molar mass of CaCO3 = 100 g/mol
So the number of moles of CaCO3 will be 2.49/100 = 0.0249 moles
The second part of the question asks to convert the mass of carbon iv oxide to moles of carbon iv oxide
Mathematically;
That is same as ;
Number of moles = mass/molar mass
molar mass of CO2 is 44 g/mol
Number of moles of CO2 = 1.13/44 = 0.0256 moles
Now, if we compare the values of these number of moles, we can see that there are almost equal.
What this means is that the number of moles of calcium carbonate reacted is equal to the number of moles of carbon iv oxide produced.
So what we conclude here is that we have an equal mole ratio between the two compounds.
So the reaction that would be the correct answer will present equal number of moles of carbon iv oxide and calcium carbonate
Thus, we can see that reaction B is the one that governs this process as it is the only reaction out of the three options that present the two compounds with equal number of moles.
Determine the half-life of a nuclide that loses 38.0% of its mass in 407 hours. Determine the half-life of a nuclide that loses 38.0% of its mass in 407 hours. 204 hours 568 hour 590 hours 291 hours 281 hours
Answer:
Use 62% - the equation is for the amount present at a given time. 0.62 = (1) e-kt -> ln(0.62)=-kt -> k = -ln(0.62)/t. I get k = .00117 hr-1 t(half) = 0.693/k = 590 hr.
HOPE THIS HELPS AND PLSSS MARK AS BRAINLIEST AND THNXX :)
The half-life is the time at which the substance's concentration is reduced by half of its initial amount. The half-life of a nuclide that lost its 38.0% mass is 590 hr. Thus, option C is correct.
What is half-life?Half-life is the time required by a substance to get reduced to half of its initial concentration. The half-life of the substance can be determined by the rate constant.
Given,
The initial quantity of substance (A₀) = 100
Remaining quantity (At) = 10 - 38 = 62
Time elapse (t) = 407 hours
The rate constant (k) is calculated as:
ln (At ÷ A₀) = - kt
ln (62 ÷ 100) ÷ 407 hour = - k
-0.47803580094 ÷ 407 = - k
k = 0.00117453513
Now, half-life from rate constant (k) is calculated as:
[tex]\rm t ^{\frac{1}{2}}[/tex] = 0.693 ÷ k
[tex]\rm t ^{\frac{1}{2}}[/tex] = 0.693 ÷ 0.00117453513
[tex]\rm t ^{\frac{1}{2}}[/tex] = 590 hours
Therefore, option C. 590 hours is the half-life of the substance.
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At 850 K, the equilibrium constant for the reaction
2SO2(g)+O2(g)↽−−⇀2SO3(g)
is Kc=15. If the given concentrations of the three gases are mixed, predict in which direction the net reaction will proceed toward equilibrium.
Left No net reaction Right
Answer:
Answers are in the explanation.
Explanation:
Given concentrations are:
SO₂ = 0.20M O₂ = 0.60M SO₃ = 0.60MSO₂ = 0.14M O₂ = 0.10M SO₃ = 0.40M And SO₂ = 0.90M O₂ = 0.50M SO₃ = 0.10MIn the reaction:
2SO₂(g) + O₂(g) ⇄ 2SO₃(g)
Kc is defined as:
Kc = 15 = [SO₃]² / [O₂] [SO₂]²
Where concentrations of each species are equilbrium concentrations.
Also, you can define Q (Reaction quotient) as:
Q = [SO₃]² / [O₂] [SO₂]²
Where concentrations of each species are ACTUAL concentrations.
If Q > Kc, the reaction will shift to the left until Q = Kc;
If Q < Kc, the reaction will shift to the right until Q = Kc
If Q = Kc, there is no net reaction because reaction would be en equilibrium.
Replacing with given concentrations:
Q = [0.60M]² / [0.60M] [0.20M]² = 15; Q = Kc → No net reactionQ = [0.40M]² / [0.10M] [0.14M]² = 82; Q > Kc, → Reaction will shift to the leftQ = [0.10M]² / [0.50M] [0.90M]² = 0.015; Q < Kc → Reaction will shift to the right
Ammonia is oxidized with air to form nitric oxide in the first step of the production of nitric acid. Two principal gas-phase reactions occur:
Answer:
4NH₃(g) + 5O₂(g) → 4NO(g) + 6H₂O
2NO(g) + O₂(g) → 2 NO₂
Explanation:
First of all, we need to consider the reaction for production of ammonia. In this reaction we have as reactants, nitrogen and hydroge.
3H₂ (g) + N₂(g) → 2NH₃ (g)
Afterwards, ammonia reacts to oxygen, to produce NO and H₂O
The equation for the process will be:
4NH₃(g) + 5O₂(g) → 4NO(g) + 6H₂O
Then, we take the nitric oxide to make it react, to produce NO₂, in order to produce nitric acid, for the final reaction:
2NO(g) + O₂(g) → 2 NO₂
3NO₂(g) + H₂O(g) → 2 HNO₃ (g) + NO(g)
Which statement is true according to the kinetic theory? Will mark brainliest
Answer:
[tex]\boxed{\sf Option \ E}[/tex]
Explanation:
All the gases at the same temperature and mass have the same average kinetic energy.
If the masses were different, then the different gases will have different velocities. If the temperature was higher then there would be a greater motion, if the temperature was lower, then there would be less motion.
Answer:
option E
Explanation:
what is the IUPAC name of KNO3
Using the bond energy data from your text (or the internet), determine (show calculations for) the approximate enthalpy change , ∆H, for each of the following reactions: (a) Cl2 (g) + 3F2 (g) ⟶ 2ClF3 (g)
Answer:
∆H= 438 KJ/mol
Explanation:
First, we have to find the energy bond values for each compound:
-) Cl-Cl = 243 KJ/mol
-) F-F = 159 KJ/mol
-) F-Cl = 193 KJ/mol
If we check the reaction we can calculate the number of bonds:
[tex]Cl_2_(_g_)~+~3F_2_(_g_)~->~2ClF_3_(_g_)[/tex]
In total we will have:
-) Cl-Cl = 1
-) F-F = 3
-) F-Cl = 6
With this in mind. we can calculate the total energy for each bond:
-) Cl-Cl = (1*243 KJ/mol) = 243 KJ/mol
-) F-F = (3*159 KJ/mol) = 477 KJ/mol
-) F-Cl = (6*193 KJ/mol) = 1158 KJ/mol
Now, we can calculate the total energy of the products and the reagents:
Reagents = 243 KJ/mol + 477 KJ/mol = 720 KJ/mol
Products = 1158 KJ/mol
Finally, to calculate the total enthalpy change we have to do a subtraction between products and reagents:
∆H= 1158 KJ/mol-720 KJ/mol = 438 KJ/mol
I hope it helps!
The approximate enthalpy change is:
∆H= 438 KJ/mol
Calculation for enthalpy change:First, we have to find the energy bond values for each compound:
Cl-Cl = 243 KJ/mol
F-F = 159 KJ/mol
F-Cl = 193 KJ/mol
Balanced chemical reaction:
Cl₂ (g) + 3F₂ (g) ⟶ 2ClF₃ (g)
Total number of bond for each:
Cl-Cl = 1
F-F = 3
F-Cl = 6
Total bond energy will be:
Cl-Cl = (1*243 KJ/mol) = 243 KJ/mol
F-F = (3*159 KJ/mol) = 477 KJ/mol
F-Cl = (6*193 KJ/mol) = 1158 KJ/mol
Now, we can calculate the total energy of the products and the reactants:
Reactants = 243 KJ/mol + 477 KJ/mol = 720 KJ/mol
Products = 1158 KJ/mol
Finally, to calculate the total enthalpy change we have to do a subtraction between products and reagents:
∆H= 1158 KJ/mol-720 KJ/mol = 438 KJ/mol
Find more information about enthalpy change here:
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Phosphorus pentafluoride, PF5, acts as a __________ during the formation of the anion PF−6. Select the correct answer below: A. Lewis acid B. Lewis base C. catalyst D. drying agent
Answer:
Lewis acid
Explanation:
In chemistry, a Lewis acid is any chemical specie that accepts a lone pair of electrons while a Lewis base is any chemical specie that donates a lone pair of electrons.
If we look at the formation of PF6^-, the process is as follows;
PF5 + F^- -----> PF6^-
We can see that PF5 accepted a lone pair of electrons from F^- making PF5 a lewis acid according to our definition above.
Hence in the formation of PF6^-, PF5 acts a Lewis acid.