Answer:
The answer is option A.
I hope this helps you.
Answer:
Macro Algae
Explanation:
probz
Classify each of the following word equations as a synthesis, decomposition, single displacement, or double displacement reaction.
Will give brainliest.
Answer:
12: This is decomposition because nitrogen triiodide is breaking apart.
13: This is double displacement because the elements in the compounds are "switching".
14: This is synthesis because the water and carbon dioxide are combining.
15: This is also synthesis because the hydrogen and oxygen are combining.
16: This is single displacement because the sodium is "switching" the element it's bonding with.
How many oxygen molecules are needed to make 10 carbon dioxide molecules according to the following balanced chemical equation? 2 CO + O2 → 2 CO2
five oxygen molecules
step by step explanation.
according to the equation,one molecule of oxygen is enough to react with two carbon molecules thus 10 carbon molecules need 5oxygen molecules
Which element's neutral atoms will have the electron configuration
1s22s22p3s23p'?
a. boron
b. carbon
c. silicon
d. aluminum
Answer:
Alumunium
Explanation:
Alumunium = [Ne] 3s² 3p¹
Ne = [He]2s²2p⁶
He = 1s
Alumunium = 1s 2s²2p⁶3s² 3p¹
Answer:
D
Explanation:
Use your trendline equation to determine the gas pressure at 200 K and 400 K. (notice the temperature units) How many times greater is the pressure at 400 K in comparison to 200 K? Is this what you’d expect? Why?
Answer:
The pressure will be twice the initial pressure
Explanation:
Gay-Lussac's law states that the pressure of a gas is directely proportional to absolute temperature under constant volume. That is because vibrations of a gas increase when temperature increases, increasing the pressure of the gas.
That means if the temperature of a gas is doubled, the pressure will be twice the initial pressure.
1. Natural gas is used as a cooking fuel in many restaurants and homes. The primary chemical components of natural gas are hydrocarbons known as alkanes. Research and list the primary alkanes found in natural gas, and explain why these compounds excel as fuel sources.
Answer:
The main component of natural gas is methane (CH4) at 60 to 90% followed by various combination of ethane, propane, and butane whose percentage can vary from 0 to 20% each. For each unit mass of alkanes, the combustion energy (energy released when the fuel reacts with oxygen) released is very high about 13 to 15 kcal/g, which is higher than even those generated by petrol or diesel. So, for heating or other energy generation purpose for household purposes, this source of energy is used.
The equation for combustion of methane is shown below. Upon combustion, carbondioxide and water is produced with simultaneous generation of heat which is the source of energy used for consumption.
CH4 + 2O2 --> CO2+ 2H2O + heat [ For methane, the combustion energy is ~ 6kcal/g]
As the CH2 units are increased in the alkanes, the combustion energy increases, for e.g., ethane has combustion energy of 7 kcal/g and propane has about 12 kcal/g.
Explanation:
What element forms an ion with an electronic configuration of 1s22s22p6 (or [Ne] ) and a −2 charge? Give the symbol for the element. g
Answer:
Mg²⁺
Explanation:
Electronic configuration = 1s22s22p6 (or [Ne] )
Charge = -2
This means the element has two extra electrons. So total electrons = 12.
The lement is Magnesium and the ion is Mg²⁺
What are the three types of combustion reactions
Answer:
Slow combustion
Spontaneous combustion
Explosive combustion
Explanation:
-Slow combustion reactions: Occurs at low temperatures. Cellular respiration is an example.
-Spontaneous combustion reactions: Occurs suddenly without an outside heat source. The heat source is the result of oxidation.
-Explosive combustion reactions: Involves an oxidizing agent.
hopefully this helped :3
Answer:
Three types are: Rapid Combustion, Complete Combustion, and Spontaneous Combustion.
Explanation:
Note: there are more types! This is just three random ones I picked to list. Hope this helps! :)
Considering the steps involved in dissolution, which of the following do you expect to speed up a dissolution process?
A. Sweeping all the solute particles into a pile within the solvent.
B. Stirring the solution vigorously.
C. Grinding the solute down into tiny particles.
D. Gently heating the solution.
Answer:
C. Grinding the solute down into tiny particles.
Explanation:
The dissolution of a solute has something to do with particle size. The size of solute particles usually determines how quickly a solute dissolves in a solvent. When large solute particles are introduced into the solvent, the large solute particles do not easily interact with solvent particles hence preventing easy dissolution in the solvent.
However, when the solute is ground into tiny particles, smaller solute particles interact more effectively with solvent particles hence dissolution is faster.
Therefore, tiny solute particles will dissolve faster in a solvent than a lump of solute. Summarily, small particle size enhances dissolution of a solute in the appropriate solvent.
Answer: stirring the solution vigorously
Grinding the solute down into tiny particles
gently heating the solution
Explanation:
A dissolution will proceed more readily when heated . Breaking up the solute as much as possible will aid in overcoming the solute-solute interaction, as will stirring the solution
A 25.0-mL sample of 0.150 M hydrochloric acid is titrated with a 0.150 M NaOH solution. The pH at the equivalence point is _____________.
Answer:
7 (neutral).
Explanation:
Hello,
In this case, for the chemical reaction:
[tex]HCl+NaOH\rightarrow NaCl+H_2O[/tex]
We can notice that since hydrochloric acid and sodium hydroxide are strong, they will fully dissociate during the titration, therefore, as they are in stoichiometric proportions in equal concentrations for the equivalence point, the pH will be 7 (neutral) since all the chloride ions are neutralized by the sodium ions.
Moreover, for the given acid solution, the required volume of sodium hydroxide will be:
[tex]V_{NaOH}=\frac{25.0mL*0.150M}{0.150M}=25.0mL[/tex]
To attain a complete titration until the equivalence point.
Best regards.
Calculate how many moles of NO2 form from 3.0 mol N2O5 2N2O5(g)→4NO2(g)+O2(g)Express your answer using two significant figures.
Answer:
6.0 moles NO2(g)
Explanation:
Based on the reaction every 2 moles N2O5(g) gives reaction with 4 moles NO2(g).Then when we have 3.0 mol N2O5(g),
2 moles N2O5(g) 4 moles NO2(g)
3 moles N2O5(g) ? moles NO2(g)
______________________________________
3 *4 / 2 = 6.00 moles NO2(g)
Answer:
There are 6 mol of NO2 with respect to 3 mol of N2O5
Explanation:
Approach 1 ( dimensional analysis ) :
3 Moles of N2O5 [tex]*[/tex] ( 4 moles of NO2 / 2 Moles of N2O5 ) - moles of N2O5 cancel out, leaving you with the moles of NO2 -
3 [tex]*[/tex] 4 / 2 = 12 / 2 = 6 moles of NO2
So as you can see in the formula there are 4 moles of NO2 present per 2 Moles of N2O5 - " 4NO2 and 2N2O5. " As we wanted the moles of N2O5 to cancel out, the 2 moles of N2O5 was kept as the denominator, and hence we received the fraction we needed.
Approach 2 :
There are 3 Moles of N2O5. The ratio of Moles of N2O5 to moles of NO2 is provided by the reaction -
Moles of N2O5 : Moles of NO2,
2 : 4,
1 : 2
Therefore the moles of NO2 will be two times as much as the given moles of N2O5, or 3 [tex]*[/tex] 2 = 6 moles of NO2
Vitamin c is known chemically by the name ascorbic acid determine the empirical formula of ascorbic acid if it is composed of 40.92% carbon, 4.58% hydrogen, and 54.50% oxygen.
Answer:
[tex]=C_3H_4O_3[/tex]
Explanation:
When percentage composition is given, and asked for the empirical formula, it is simplest to assume 100 g of material. Thus,
Mass C = 40.92 g. Moles C = 40.92 g x 1 mole/12 g = 3.41 moles C
Mass H = 4.58 g. Moles H = 4.58 g x 1 mole/1.0 g = 4.58 moles H
Mass O = 54.50 g. Moles O = 54.50 g x 1 mole/16 g = 3.41 moles O
Now, we want to get the moles into whole numbers, so we begin by dividing all by the smallest, i.e. divide all values by 3.41.
Moles C = 3.41/3.41 = 1
Moles H = 4.58/3.41 = 1.34
Moles O = 3.41/3.41 = 1
Now, in order to get 1.34 to be a whole number we multiply it (and all others) by 3
Moles C = 1x3 = 3
Moles H = 1.34x3 = 4
Moles O = 1x3 = 3
Empirical Formula [tex]=C_3H_4O_3[/tex]
In a study of the decomposition of the compound XX via the reaction
X(g)⇌Y(g)+Z(g)X(g)⇌Y(g)+Z(g)
the following concentration-time data were collected:
Time (min)(min) [X](M)[X](M)
0 0.467
1 0.267
2 0.187
3 0.144
4 0.117
5 0.099
6 0.085
7 0.075
Given that the rate constant for the decomposition of hypothetical compound XX is is 1.60 M^−1⋅min^−1. Calculate the concentration of XX after 18.0 minmin .
Answer:
( About ) 0.03232 M
Explanation:
Based on the units for this reaction it should be a second order reaction, and hence you would apply the integrated rate law equation "1 / [X] = kt + 1 / [[tex]X_o[/tex]]"
This formula would be true for the following information -
{ [tex]X_o[/tex] = the initial concentration of X, k = rate constant, [ X ] = the concentration after a certain time ( which is what you need to determine ), and t = time in minutes }
________
Therefore, all we have left to do is plug in the known values. The initial concentration of X is 0.467 at a time of 0 minutes, as you can tell from the given data. This is not relevant to the time needed in the formula, as we need to calculate the concentration of X after 18 minutes ( time = 18 minutes ). And of course k, the rate constant = 1.6
1 / [X] = ( 1.6 )( 18 minutes ) + 1 / ( 0.467 ) - Now let's solve for X
1 / [X] = 28.8 + 1 / ( 0.467 ),
1 / [X] = 28.8 + 2.1413...,
1 / [X] = 31,
[X] = 1 / 31 = ( About ) 0.03232 M
Now for this last bit here you probably are wondering why 1 / 31 is not 0.03232, rather 0.032258... Well, I did approximate one of the numbers along the way ( 2.1413... ) and took the precise value into account on my own and solved a bit more accurately. So that is your solution! The concentration of X after 18 minutes is about 0.03232 M
The concentration after 18 minutes is 1.45 × 10^-13 M.
We have the equation of the reaction as; X(g)⇌Y(g)+Z(g)X(g)⇌Y(g)+Z(g) and we are informed in the question that the reaction follows the first order kinetics.
From the first order kinetics; ln[A] = ln[A]o - kt
Where;
[A] = concentration at time t
[A]o = initial concentration
k = rate constant
t = time taken
Hence;
ln[A] = ln[0.467] - (1.60 M^−1⋅min^−1 × 18 min)
A = 1.45 × 10^-13 M
Learn more about rate of reaction: https://brainly.com/question/17960050
need this asap , help please
Answer:
Path A-B-D involves a catalyst and is slower than A-C-D
Explanation:
The diagram above illustrates both the catalyzed path and the uncatalyzed path of a chemical reaction.
The catalysed path is the path expressed with broken lines and the uncatalyzed path is the path expressed with thick small line as shown in the diagram above.
The catalyzed path has a higher activation energy than the uncatalyzed path.
Therefore, the catalyzed path will be slower that the uncatalyzed path because, the catalyzed path will require a higher energy to overcome the activation energy in order for the reaction to proceed to product.
On the other hand, the uncatalyzed path has a lower activation energy and a lesser amount of energy is needed to overcome it in order for the reaction to proceed to product.
Draw an aldohexose.
Answer:
See explanation
Explanation:
We have to remember that theory behind the carbohydrates. Carbohydrates are molecules with several hydroxyl groups in which the main functional group can be an aldehyde or a ketone.
If we have an aldehyde as a main functional group we will have an "aldose". If we have a ketone as a main functional group we will have a "ketose".
We can also, classify the carbohydrates using the number of carbons. So, for example, if we have 5 carbons and a ketone as the main functional group we will have a "keto-pentose". If we have for example 4 carbons and an aldehyde as the main functional group we will have a "tetra-aldose".
In this case, we have an aldohexose, so we will have 6 carbons and an aldehyde as main functional group. So, we can draw a structure with 6 carbons, in carbon 1 we have to put the aldehyde group and in the other carbons we have to put "OH" groups.
See figure 1
I hope it helps!
How many formula units make up 36.0 g of magnesium chloride (MgCl2)?
Express the number of formula units numerically.
Answer: There are [tex]2.29\times 10^{23}[/tex] formula units
Explanation:
According to avogadro's law, 1 mole of every substance occupies 22.4 L at STP and contains avogadro's number [tex]6.023\times 10^{23}[/tex] of particles.
To calculate the moles, we use the equation:
[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text {Molar mass}}=\frac{36.0g}{95g/mol}=0.38moles[/tex]
1 mole of [tex]MgCl_2[/tex] contains = [tex]6.023\times 10^{23}[/tex] formula units
Thus 0.38 moles of [tex]MgCl_2[/tex] contains = [tex]\frac{6.023\times 10^{23}}{1}\times 0.38=2.29\times 10^{23}[/tex] formula units
Thus there are [tex]2.29\times 10^{23}[/tex] formula units
A naturally occurring oil co-distills with water to produce an oil/water distillate that is 20% oil by weight. If the molecular weight of the oil 100 g/mol, what was the partial pressure of the oil during distillation assuming atmospheric pressure is 760 mm Hg
Answer:
Explanation:
Partial pressure of oil = mole fraction of oil x total pressure
mole fraction of oil = mole of oil / mole of water + mole of oil
= mole of oil = mass of oil / molecular weight of oil
= 20 / 100 = .2
mole of water = 80 / 18
= 4.444
mole fraction of oil = .2 / .2 + 4.444
= .2 / 4.644
Partial pressure of oil = mole fraction of oil x total pressure
= (.2 / 4.644 ) x 760 mm
= 32.73 mm Hg .
Suppose you have a bucket of sand containing 5.5 billion grains of sand ( 5.5×109 grains). Most of the grains of sand are white, but there are some brown grains mixed evenly throughout.
If the concentration of brown sand is 6.0% , how many grains of brown sand are in the bucket?
brown grains
If the concentration of brown sand is 6.0 ppm, how many grains of brown sand are in the bucket?
brown grains
If the concentration of brown sand is 6.0 ppb, how many grains of brown sand are in the bucket?
brown grains
Explanation:
5.5 billion grains of sand ( 5.5×109 grains)
If the concentration of brown sand is 6.0% , how many grains of brown sand are in the bucket?
Number of grains = Concentration of brown side * Bucket of sand
Brown grains = 0.06 * 5.5×10^9 = 0.33 x 10^9 = 3.3 x 10^8 grains
If the concentration of brown sand is 6.0 ppm, how many grains of brown sand are in the bucket?
Number of grains = Concentration of brown side * Bucket of sand
6ppm = 6 / 1000000 = 0.000006
Brown grains = 0.000006 * 5.5×10^9 = 3.3 x 10^4 grains
If the concentration of brown sand is 6.0 ppb, how many grains of brown sand are in the bucket?
Number of grains = Concentration of brown side * Bucket of sand
6ppb = 6 / 1000000000 = 0.000000006
Brown grains = 0.000000006 * 5.5×10^9 = 3.3 x 10^1 = 33 grains
Identify the particle that must receive 2 electrons to acquire a charge of +1. a) K b) Fe2+ c) O2- d) Nee) Al3+ (URGENT) Needs to be done in 30 mins
Answer:
E) Al³⁺
Explanation:
A reaction involving a gain of electrons is known as a reduction reaction because the oxidation number of the species gaining the electron is reduced.
In the given question, the oxidation number (charge) of particle accepting two electrons will decrease by 2. From the given options;
A. K is a neutral atom with oxidation number of 0. If is accepts two electrons, its oxidation number becomes -2.
K + 2e⁻ ----> K⁻²
B) Fe²⁺ has a charge of +2. If it accepts two electrons, its charge comes 0.
Fe⁺ + 2e⁻ ----> Fe
C) O²⁻ has a charge of -2. if it accepts two electrons, it will have a charge of -4.
O²⁻ + 2e⁻ ----> O⁴⁻
D) Ne has a charge of zero. If it accepts two electrons, its charge becomes -2.
Ne + 2e⁻ ----> Ne²⁻
E) Al³⁺ has a charge of +3. If it gains two electrons, its charge becomes +1.
Al³⁺ + 2e⁻ ----> Al⁺
A gas particle of mass 5.31 × 10^-23 kg has a velocity of 1.00 102 m/s. What is the kinetic energy of the molecule
Answer:
Kinetic energy = 1/2mv²
where m is the mass
v = velocity
m = 5.31 × 10^-23 kg
v = 1.00 × 10^2 m/s
Kinetic energy = 1/2 × 5.31 × 10^-23 × ( 1.00 × 10^2)²
= 2.655 × 10^-19 Joules
Hope this helps
A balloon filled with helium has a volume of 4.5 × 103 L at 25°C. What volume will the balloon occupy at 50°C if the pressure surrounding the balloon remains constant?
Answer:
[tex]V_2 = 4.87 * 10^3[/tex]
Explanation:
This question is an illustration of ideal Gas Law;
The given parameters are as follows;
Initial Temperature = 25C
Initial Volume = 4.5 * 10³L
Required
Calculate the volume when temperature is 50C
NB: Pressure remains constant;
Ideal Gas Law states that;
[tex]PV = nRT[/tex]
The question states that the pressure is constant; this implies that the constant in the above formula are P, R and n
Divide both sides by PT
[tex]\frac{PV}{PT} = \frac{nRT}{PT}[/tex]
[tex]\frac{V}{T} = \frac{nR}{P}[/tex]
Represent [tex]\frac{nR}{P}[/tex] with k
[tex]\frac{V}{T} = k[/tex]
[tex]k = \frac{V_1}{T_1} = \frac{V_2}{T_2}[/tex]
At this point, we can solve for the required parameter using the following;
[tex]\frac{V_1}{T_1} = \frac{V_2}{T_2}[/tex]
Where V1 and V2 represent the initial & final volume and T1 and T2 represent the initial and final temperature;
From the given parameters;
V1 = 4.5 * 10³L
T1 = 25C
T2 = 50C
Convert temperatures to degree kelvin
V1 = 4.5 * 10³L
T1 = 25 +273 = 298K
T2 = 50 + 273 = 323K
Substitute values for V1, T1 and T2 in [tex]\frac{V_1}{T_1} = \frac{V_2}{T_2}[/tex]
[tex]\frac{4.5 * 10^3}{298} = \frac{V_2}{323}[/tex]
Multiply both sides by 323
[tex]323 * \frac{4.5 * 10^3}{298} = \frac{V_2}{323} * 323[/tex]
[tex]323 * \frac{4.5 * 10^3}{298} = V_2[/tex]
[tex]V_2 = 323 * \frac{4.5 * 10^3}{298}[/tex]
[tex]V_2 = \frac{323 * 4.5 * 10^3}{298}[/tex]
[tex]V_2 = \frac{1453.5 * 10^3}{298}[/tex]
[tex]V_2 = 4.87 * 10^3[/tex]
Hence, the final volume at 50C is [tex]V_2 = 4.87 * 10^3[/tex]
What information does the geologic time scale provide?
Answer:
The answer is D or information about when plants first appeared.
Explanation:
1. Corrosion in metals is an example of what?
Answer:
In the most common use of the word, this means electrochemical oxidation of metal in reaction with an oxidant such as oxygen or sulfates. Rusting, the formation of iron oxides, is a well-known example of electrochemical corrosion.The percent yield (isolation yield) of guaifenesin isolated from a 650 mg tablet containing 400 mg dose of drug, can be expressed as: Group of answer choices
Answer:
61.54%
Explanation:
Hello,
To calculate the percent yield of a product, we express it as ratio between the actual yield to the theoretical yield multiplied by 100.
Percent yield = (actual yield / theoretical yield) × 100
Actual yield = 400mg
Theoretical yield = 650mg
Percent yield = (400 / 650) × 100
Percent yield = 0.6154 × 100
Percent yield = 61.54%
Percent yield of guaifenesin in the drug is 61.54%
How many moles of carbon atoms are there in 0.500 mol of C2H6?
The number of moles of carbon atoms in 0.500 mol of ethane (C₂H₆) is equal to one mole.
What is a mole?A mole can be defined as a scientific unit that is utilized to calculate the quantities such as atoms, molecules, ions, or other particular particles. The mass of one mole of a given chemical element is atomic mass and that of 1 mole of a chemical compound is molar mass.
The number of entities found in one mole is equal to 6.023 × 10 ²³ which is known as Avogadro’s constant.
Given, the number of moles of C₂H₆ = 0.500 mole
One molecule of ethane has carbons = 2
One mole of ethane has moles of carbons = 2 moles
0.500 mol of ethane has moles of carbon atoms = 0.500×2 = 1 mol
Therefore, one mole of carbon atoms is present in 0.500 mol of ethane C₂H₆.
Learn more about the mole, here:
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Identify a reaction of sugars, starch, or cellulose to produce ethanol and carbon dioxide halogenation polymerization
Answer:
Fermentation
Explanation:
Fermentation is the general term used to describe the process by which sugars such as glucose, starch or cellulose are converted to ethanol and carbon (iv) oxide. It is anaerobic process meaning that it occurs in the absence of air or in very low oxygen concentrations.
Yeast and other microorganisms ferment glucose into ethanol and carbon (iv) oxide with the help of the enzyme zymase. Polysaccharides such as starch and cellulose are first broken down into glucose by enzymes such as diastases, maltase and cellulase, before it is then converted into ethanol and carbon (iv) oxide.
The equation for the conversion of glucose to ethanol and carbon (iv) oxide is as follows:
C₆H₁₂O₆(aq) -----> 2C₂H₅OH(aq) + 2CO₂(g)
An unknown gas Q requires 2.67 times as long to effuse under the same conditions as the same amount of nitrogen gas. What is the molar mass (g/mol) of Q
Answer:
The correct answer is 199.66 grams per mole.
Explanation:
Based on law of effusion given by Graham, a gas rate of effusion is contrariwise proportionate to the square root of molecular mass, that is, rate of effusion of gas is inversely proportional to the square root of mass. Therefore,
R1/R2 = √ M2/√ M1
Here rate is the rate of effusion of the gas expressed in terms of number of mole per uni time or volume, and M is the molecular mass of the gas.
Rate Q/Rate N2 = √M of N2/ √M of Q
The molecular mass of N2 or nitrogen gas is 28 grams per mole and M of Q is molecular mass of Q and based on the question Q needs 2.67 times more to effuse in comparison to nitrogen gas, therefore, rate of Q = rate of N2/2.67
Now putting the values we get,
rate of N2/2.67/rate of N2 = √28/ √M of Q
√M of Q = √ 28 × 2.67
M of Q = (√ 28 × 2.67)²
M of Q = 199.66 grams per mole
g Suppose you are titrating an acid of unknown concentration with a standardized base. At the beginning of the titration, you read the base titrant volume as 1.94 mL. After running the titration and reaching the endpoint, you read the base titrant volume as 23.82 mL. What volume of base was required for the titration
Answer:
21.88mL is the volume of base required for the titration.
Explanation:
For an acid-base titration trying to find the concentration of an acid, you must add a known quantity of the acid and titrate it with an standarized base.
If you know the moles of base you add to the acid solution, these moles are equal to moles of acid.
In the buret of the titration, initial volume is 1.94mL and final volume is 23.82mL. The volume you are adding is the difference between initial and final volume, that is:
23.82mL - 1.94mL
21.88mL is the volume of base required for the titration.A cylindrical rod of length 1 m and radius 1 cm is submerged in water. The rod has a non-uniform mass distribution such that one half of the rod is much more massive than the other half. At which point in the rod can the buoyant force be considered to be acting?
A) at the rod's geometrical center
B) two of the given choices are correct
C) none of these
D) at the rod's center of mass
E) at the rod's center of gravity
Answer:
A) at the rod's geometrical center
Explanation:
Let us assume that the rod is replaced by water. And now this water volume is in translational and in rotational equilibrium.
Therefore, a net upward force must have been exerted by the surrounding liquid which acts at the center of mass of the water volume.
This force determines through the geometric center of the column of the cylindrical water
Moreover, the force is also independent of submerged body into it
Hence, the first option is correct
For the reaction below, initially the partial pressure of all 3 gases is 1.0atm. . 2NH3(g)--> N2(g) + 3H2(g) K, 0.83 1. When the reaction reach equilibrium the partial pressure of N2 will be greater than 1atm The reaction would shift toward the reactants The reaction would shift toward the products 2. When the reaction reach equilibrium the partial pressure of NH3 will be greater than 1atm 3. When the reaction reach equilibrium the partial pressure of H2 will be greater than 1atm
Answer:
The reaction would shift toward the reactants
When the reaction reach equilibrium the partial pressure of NH3 will be greater than 1atm
Explanation:
For the reaction:
2NH₃(g) ⇄ N₂(g) + 3H₂(g)
Where K is defined as:
[tex]K = \frac{P_{N_{2}}*P_{H_2}^3}{P_{NH_3}^2} = 0.83[/tex]
As initial pressures of all 3 gases is 1.0atm, reaction quotient, Q, is:
[tex]Q = \frac{1atm*{1atm}^3}{1atm^2} = 1[/tex]
As Q > K, the reaction will produce more NH₃ until Q = K consuming N₂ and H₂.
Thus, there are true:
The reaction would shift toward the reactantsWhen the reaction reach equilibrium the partial pressure of NH3 will be greater than 1atm
For the reaction: PCl5(g) ⇌ PCl3(g) + Cl2(g) at 600.0 K, the equilibrium constant is 11.5. Suppose that 1.500 g of PCl5 (MW=208.22 g/mol) is placed in an evacuated 500.0 mL bulb, which is then heated to 600.0 K. What is the total pressure (in atm) in the bulb at equilibrium?
Answer:
1.418688 atm
Explanation:
(a) Moles of PCl5 = mass / molar mass
=1.5 g / 208.22 g/mol
= 0.0072 moles
Also given,
T = 600 K
V = 0.500 L
Pressure of PCl5, P = nRT / V
= 0.0072 mol×0.0821 L-atm / (mol.K)×600 K / 0.500 L
= 0.709344 atm
(b) PCl5(g) ⇄ PCl3(g) + Cl2(g)
Initial 0.965 0 0
Change -x +x +x
Equilibrium (0.709344 -x) x x
K_p = 11.5 = x×x / (0.965 -x)
solving, we get x = 0.67027
So partial pressure of PCl5 at equilibrium = 0.709344 - 0.67027 = 0.039074 atm
(c) Partial pressure of PCl3 = Cl2 = 0.709344 atm
So total pressure = 0.709344+0.039074+ 0.67027= 1.418688 atm