Answer:
a) [tex]1.065\times 10^{-8} [/tex]fraction of carbon−14 in a piece of charcoal remains after 14.0 years.
b) [tex]0.000\times 10^{-3} [/tex]fraction of carbon−14 in a piece of charcoal remains after [tex]1.900\times 10^4 years[/tex]
c) [tex]0.0000\times 10^{-4} [/tex]fraction of carbon−14 in a piece of charcoal remains after [tex]1.0000\times 10^5 years[/tex].
Explanation:
The fraction of a radioactive isotope remaining at time t is given by:
[tex][A]=\frac{(\frac{1}{2})^t}{t_{\frac{1}{2}}}[/tex]
Taking log both sides:
[tex]\log [A]=t\log[\frac{1}{2}]-\log [t_{\frac{1}{2}}][/tex]
[A] = fraction at given time t
[tex]t_{\frac{1}{2}}[/tex] = half life of the carbon−14 =5,730 years
a)When , t = 14 years
[tex]\log [A]=t\log[\frac{1}{2}]-\log [t_{\frac{1}{2}}][/tex]
[tex]\log [A]= 14 years\times (-3010)-\log [5,730 years][/tex]
[tex][A]=1.065\times 10^{-8} [/tex]
b)When , t = [tex]1.900\times 10^4 years[/tex]
[tex]\log [A]=t\log[\frac{1}{2}]-\log [t_{\frac{1}{2}}][/tex]
[tex]\log [A]= 1.900\times 10^4 years\times (-3010)-\log [5,730 years][/tex]
[tex][A]=0.000\times 10^{-3} [/tex [/tex]
c)When , t = [tex]1.0000\times 10^5 years[/tex]
[tex]\log [A]=t\log[\frac{1}{2}]-\log [t_{\frac{1}{2}}][/tex]
[tex]\log [A]= 1.0000\times 10^5 years\times (-3010)-\log [5,730 years][/tex]
[tex][A]=0.0000\times 10^{-4} [/tex]
A student placed 13.0 g of glucose (C6H12O6) in a volumetric flask, added enough water to dissolve the glucose by swirling, then carefully added additional water until the 100.-mL mark on the neck of the flask was reached. The flask was then shaken until the solution was uniform. A 60.0-mL sample of this glucose solution was diluted to 0.500 L. How many grams of glucose are in 100. mL of the final solution?
Answer:
There are 1.5597 grams of glucose in 100 mL of 0.08658 M glucose solution.
Explanation:
13.0 grams of glucose was added to volumetric flask and solution was made up to 100 mL.
Moles of glucose = [tex]\frac{13.0 g}{180.156 g/mol}=0.07215 mol[/tex]
Volume of the solution = 100 mL = 0.1 L
Molarity of the solution = [tex]\frac{Moles}{\text{Volume of solution}}[/tex]
[tex]M=\frac{0.07215 mol}{0.1 L}=0.7215 M[/tex]
60 mL of 0.7215 M was diluted to 0.500 L, the molarity of the solution after dilution be [tex]M_2[/tex]
[tex]M_1=0.7215 M,V_2=60 mL=0.060 L[/tex]
[tex]M_2=?,V_2=0.500 L[/tex]
[tex]M_1V_1=M_2V_2[/tex] (Dilution)
[tex]M_2=\frac{0.7215 M\times 0.060 L}{0.500 L}[/tex]
[tex]M_2=0.08658 M[/tex]
Mass of glucose in 0.08658 M glucose solution:
In 1 L of solution = 0.08658 moles
In 1000 mL of solution = 0.08658 moles
Then in 100 mL of solution :
[tex]\frac{0.08658}{1000}\times 100 mol=0.008658 mol[/tex] of glucose
Mass of 0.008658 moles of glucose:
[tex]0.008658 mol\times 180.156 g/mol=1.5597 g[/tex]
There are 1.5597 grams of glucose in 100 mL of 0.08658 M glucose solution.
Answer:
m_{C_6H_{12}O_6}=1.56g
Explanation:
Hello,
After the dilution, the concentration of glucose is:
[tex]c=\frac{13.0g}{100mL}=0.13g/mL[/tex]
Now, we apply the dilution equation to compute the concentration after the dilution to 0.500L (500mL):
[tex]c_1*V_1=c_2*V_2\\c_2=\frac{c_1*V_1}{V_2}=\frac{0.13g/mL*60.0mL}{500mL} \\c_2=0.0156g/mL[/tex]
Finally, we compute the present grams in 100mL of the 0.0156g/mL glucose solution as shown below:
[tex]m_{C_6H_{12}O_6} =100mL*0.0156g/mL\\m_{C_6H_{12}O_6}=1.56g[/tex]
Best regards.
You heat 51 grams of magnesium over a Bunsen burner for several minutes until it reacts with oxygen in the air. Then you weigh the resulting product and see that it is now 53 grams. How does this happen without breaking the Law of Conservation of Mass?
Final answer:
The increase in mass of magnesium after burning is due to its combination with oxygen to form magnesium oxide, which is in accordance with the Law of Conservation of Mass. No mass is lost or created; the oxygen simply adds to the original mass of the magnesium metal.
Explanation:
The observation that a piece of magnesium gains mass when burnt can be explained without violating the Law of Conservation of Mass. This law states that mass is neither created nor destroyed in a chemical reaction. When magnesium burns in the presence of oxygen, a chemical change occurs, resulting in the formation of magnesium oxide, a white crumbly powder. The reaction can be depicted by the word equation magnesium + oxygen → magnesium oxide. The increase in mass from 51 grams to 53 grams is due to the addition of oxygen from the air. When magnesium (Mg) reacts with oxygen (O₂), the oxygen atoms combine with magnesium to form magnesium oxide (MgO), thus accounting for the increase in mass.
For example, if you heat 10.0 grams of calcium carbonate (CaCO₃) and produce 4.4 g of carbon dioxide (CO₂) and 5.6 g of calcium oxide (CaO), the total mass of the products equals the original mass of the reactants, which is in agreement with the Conservation of Mass. Similarly, in the magnesium burning experiment, the mass of the magnesium and the oxygen that combined during burning would equal the mass of the magnesium oxide produced, if all the reactants and products could be contained and measured.
1. Define the term disproportionation reaction. 2. Write the balanced equation for the conversion of 4-chlorobezaldehye into 4-chlorobenzyl alcohol and 4-chlorobenzoic acid. 3. What is the oxidation number of the aldehyde carbon in each product of the Cannizzaro reaction? Show all work for credit.
Answer:
3 is the answer
Explanation:
Ethanol (C2H5OH) and water (H2O) are both liquids at or near room temperature. When a sample of ethanol is poured into a beaker containing water, the two liquids. a. Are miscible because each can hydrogen bond with the other. b. Are immiscible because each has strong hydrogen bonds that cannot be broken. c. Are miscible because both are liquids and all liquids are miscible. d. Are immiscible because organic and non-organic compounds cannot mix.
Answer:
a. Are miscible because each can hydrogen bond with the other.
Explanation:
Both ethanol and water are miscible. The reason why they can both mix freely is due to the hydrogen bonds that will form between their molecular structure.
Hydrogen bonds are special dipole-dipole attraction between polar molecules in which hydrogen atoms are directly joined to an electronegative atom.
Ethanol has an hydroxyl group which will bond to form an intermolecular bond with the oxygen and hydrogen on the water molecule. This attraction makes them miscible.
Final answer:
Ethanol and water are miscible because they can form hydrogen bonds with each other, resulting in a homogeneous solution. The polar nature of both substances facilitates this mixing.
Explanation:
When ethanol (C₂H₅OH) is poured into water (H₂O), the two liquids are miscible because each can form hydrogen bonds with the other. This mixing leads to the formation of a single homogeneous solution. This miscibility is due to the polar nature of both ethanol and water, which allows ethanol molecules to hydrogen bond with water molecules, resulting in a stable mixture. In contrast, substances that cannot form such interactions are immiscible, like oil and water, which do not mix to form a homogeneous solution.
Write down the equation called the modified Raoult's law
The carbon−14 decay rate of a sample obtained from a young tree is 0.266 disintegration per second per gram of the sample. Another wood sample prepared from an object recovered at an archaeological excavation gives a decay rate of 0.178 disintegration per second per gram of the sample. What is the age of the object? (The half-life of carbon−14 is 5715 years.) × 10 years Enter your answer in scientific notation.
Answer:
3.30 × 10³ yearsExplanation:
1) Carbon-14 disintegration rate of a young (alive) tree: 0.266 per second per gram.
2) Carbon-14 disintegration rate of a wood sample prepeared from an object recovered at an archaelogical excavation (dead matter): 0.178 per second per gram.
3) Ratio of decay: A/A₀ = 0.178 / 0.266 = 0.669
4) Half-life equation: A/A₀ = (1/2)ⁿ, where n is the number of half-lives since the object died.
5) Substitute and solve for n:
0.669 = (1/2)ⁿ[tex]nlog(1/2)=log(0.669)\\ \\n=\frac{log(0.669)}{log(1/2)}\\ \\n=0.578[/tex]
That means that 0.578 half-life has elapsed since the wood with which the object was created was dead matter.
6) Convert number of half-lives to years:
0.578 half-life × 5715 years/half-life = 3,303 years = 3.30 × 10³ years.Which of the following would you expect to cause the evolution of a gas when added to hydrogen peroxide? a. A piece of raw carrot. b. A piece of raw cabbage. c. A piece of raw radish. d. More than one answer is correct.
Answer:
D: More than one Answer is correct.
Explanation:
The enzyme at the heart of the question is peroxidase.The reaction of enzyme catalase with hydrogen peroxide produces oxygen gas. This catalase enzyme is present in all the above mentioned vegetables, i.e. radish, cabbage as well as carrot.
So more than one answer is correct in this case.
More than one of the listed items could cause the evolution of gas when added to hydrogen peroxide, as each can contain enzymes that act as catalysts for the decomposition of hydrogen peroxide. Hence option d More than one answer is correct.
The question is related to the decomposition of hydrogen peroxide (H₂O₂) and the role of catalysts in this chemical process. The decomposition reaction is known as the breakdown reaction happens when a single reactant splits into two or more products. Hydrogen peroxide has the ability to decompose into water and oxygen gas, but in the absence of a catalyst, this reaction is slow. When a catalyst like manganese (IV) oxide is added to a solution of hydrogen peroxide, it speeds up the reaction, leading to the rapid evolution of oxygen gas without being consumed in the process. As for the answer to the question, raw carrot, cabbage, or radish contain various enzymes that could act as potential catalysts. Therefore, more than one answer is correct as each of these foods could potentially contain enzymes that catalyze the decomposition of hydrogen peroxide, leading to the evolution of gas.
A 1.2 kg block of iron at 32 ∘C is rapidly heated by a torch such that 12 kJ is transferred to it. What temperature would the block of iron reach (assuming the complete transfer of heat and no loss to the surroundings)? If that same amount of heat (12 kJ ) was quickly transferred to a 850 g pellet of copper at 27 ∘C, what temperature would the copper reach before it begins losing heat to the surroundings? Use the equation for heat capacity and the following heat capacity values:
Answer:
For iron
Final temperature = 54,22°C
For copper
Final Temperature = 63.67 °C
Explanation
Hello,
You are using a torch to warm up a block of iron that has an initial temperature of 32°C.
The first you have to know is that the "heat capacity" could simply define as the heat required to go from an initial temperature to a final temperature.
So you need to use the heat capacity equation as follow in the paper.
The equation has to have all terms in the same units, so:
q = 12000 J
s = 0.450 J / g °C
m = 1200 g
Ti = 32 °C
2. The Food and Drug Administration (FDA) mandates that all vinegar sold in the United States must have a minimum concentration of 4% acetic acid by mass. Calculate the minimum molarity of acetic acid in vinegar according to this standard. Assume the density of vinegar is the same as for water.
Answer:
The minimum molarity of acetic acid in vinegar according to given standards is 0.6660 mol/L.
Explanation:
4% acetic acid by mass means that 4 gram of acetic acid in 100 g solution.
Given that density of the vinegar is same is that of water = 1 g/mL
Mass of the vinegar solution = 100 g
Volume of the vinegar solution = V
[tex]Density=\frac{Mass}{Volume}[/tex]
[tex]1 g/mL=\frac{100 g}{V}[/tex]
V = 100 mL = 0.1 L
Moles of acetic acid =[tex]\frac{4 g}{60.052 g/mol}=0.0666 mol[/tex]
[tex]Molarity=\frac{\text{Moles of solute}}{\text{Volume of solution(L)}}[/tex]
[tex]M=\frac{0.0666 mol}{0.1 L}=0.6660 mol/L[/tex]
The minimum molarity of acetic acid in vinegar according to given standards is 0.6660 mol/L.
consider the reaction between sulfite and a metal anion, X2-, to form the metal, X, and thiosulfate: 2 X2-(aq) + 2 SO32- + 3 H2O(l) → 2 X(s) + S2O32- (aq) + 6 OH- for which Eocell = 0.63. Given that the Eored for sulfite is -0.57 V, calculate Eored for X. Enter your answer to 2 decimal places
Answer:
[tex]E_{red}^{0}[/tex] for X is -1.20 V
Explanation:
Oxidation: [tex]2\times[/tex][[tex]X^{2-}(aq.)-2e^{-}\rightarrow X(s)[/tex]]
reduction: [tex]2SO_{3}^{2-}(aq.)+3H_{2}O(l)+4e^{-}\rightarrow S_{2}O_{3}^{2-}(aq.)+6OH^{-}(aq.)[/tex]
---------------------------------------------------------------------------------------------------
overall:[tex]2X^{2-}(aq.)+2SO_{3}^{2-}(aq.)+3H_{2}O(l)\rightarrow 2X(s)+S_{2}O_{3}^{2-}(aq.)+6OH^{-}(aq.)[/tex]
So, [tex]E_{cell}^{0}=E_{red}^{0}(SO_{3}^{2-}\mid S_{2}O_{3}^{2-})-E_{red}^{0}(X\mid X^{2-})[/tex]
or, [tex]0.63=-0.57-E_{red}^{0}(X\mid X^{2-})[/tex]
or, [tex]E_{red}^{0}(X\mid X^{2-})= -1.20[/tex]
So, [tex]E_{red}^{0}[/tex] for X is -1.20 V
Answer: The standard reduction potential of X is -1.20 V
Explanation:
For the given chemical equation:
[tex]2X^{2-}(aq.)+2SO_3^{2-}+3H_2O(l)\rightarrow 2X(s)+S_2O_3^{2-}(aq.)+6OH^-[/tex]
The half reaction follows:
Oxidation half reaction: [tex]X^{2-}(aq.)\rightarrow X(s)+2e^-[/tex] ( × 2 )
Reduction half reaction: [tex]2SO_{3}^{2-}(aq.)+3H_{2}O(l)+4e^{-}\rightarrow S_{2}O_{3}^{2-}(aq.)+6OH^{-}(aq.)E^o=-0.57V[/tex]
Substance getting oxidized always act as anode and the one getting reduced always act as cathode.
To calculate the [tex]E^o_{cell}[/tex] of the reaction, we use the equation:
[tex]E^o_{cell}=E^o_{cathode}-E^o_{anode}[/tex]
We are given:
[tex]E^o_{cell}=0.63V[/tex]
Putting values in above equation, we get:
[tex]0.63=-0.57-E^o_{anode}\\\\E^o_{anode}=-0.57-0.63=-1.20V[/tex]
Hence, the standard reduction potential of X is -1.20 V
Are beverages usually acidic or basic? Why is this?
Answer:
ACIDIC
Explanation:
Usually beverages contain carbonated water. Carbonated water is the water containing dissolved carbon dioxide gas.
The carbon dioxide gas (CO₂) when dissolved in water produces carbonic acid (H₂CO₃), which makes the drinks acidic in nature.
H₂O(l) + CO₂(g) → H₂CO₃(aq)
This makes the pH of the drinks around 3 - 4.
Therefore, the beverages are usually acidic and not basic.
Final answer:
Beverages can either be acidic or basic based on their pH value. Orange juice is an example of a mildly acidic beverage, while baking soda is an example of a basic beverage.
Explanation:
Beverages can be either acidic or basic. The acidity or basicity of a beverage is determined by its pH value. On the pH scale, which ranges from 0 to 14, anything below 7.0 is considered acidic, and anything above 7.0 is considered basic. The pH of common beverages can vary - for example, orange juice has a pH of approximately 3.5, making it mildly acidic, while baking soda has a pH of 9.0, making it basic.
An ideal gas is compressed isothermally and reversibly at 400K from 1 m3 to 0.5 m3 9200 J heat is evolved during compression. What is the work done and how many moles of gas were compressed during this process?
Answer : The work done on the system is 9200 J and the number of moles of gas is 3.99 moles.
Explanation : Given,
Initial volume of gas = [tex]1m^3[/tex]
Final volume of the gas = [tex]0.5m^3[/tex]
Temperature of the gas = 400 K
Heat evolved = -9200 J
According to the question, this is the case of isothermal reversible compression of gas.
As per first law of thermodynamic,
[tex]\Delta U=q+w[/tex]
where,
[tex]\Delta U[/tex] = internal energy
q = heat
w = work done
As we know that, the term internal energy is the depend on the temperature and the process is isothermal that means at constant temperature.
So, at constant temperature the internal energy is equal to zero.
[tex]q=-w[/tex]
Thus, the work done on the system = -q = -(-9200J) = 9200 J
The expression used for work done will be,
[tex]w=-nRT\ln (\frac{V_2}{V_1})[/tex]
where,
w = work done on the system = 9200 J
n = number of moles of gas = ?
R = gas constant = 8.314 J/mole K
T = temperature of gas = 400 K
[tex]V_1[/tex] = initial volume of gas = [tex]1m^3[/tex]
[tex]V_2[/tex] = final volume of gas = [tex]0.5m^3[/tex]
Now put all the given values in the above formula, we get the number of moles of gas.
[tex]9200J=-n\times 8.314J/moleK\times 400K\times \ln (\frac{0.5m^3}{1m^3})[/tex]
[tex]n=3.99mole[/tex]
Therefore, the work done on the system is 9200 J and the number of moles of gas is 3.99 moles.
What is the limiting reactant when 1.50 g of lithium and 1.50 g of nitrogen combine to form lithium nitride, a component of advanced batteries, according to the following unbalanced equation? Li + N2 ⟶ Li3 N
Answer: The limiting reagent in the given equation is lithium metal.
Explanation:
Limiting reagent is defined as the reagent which is present in less amount and also it limits the formation of products.
Excess reagent is defined as the reagent which is present in large amount.
To calculate the number of moles, we use the equation:
[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex] ....(1)
For Lithium metal:Given mass of lithium metal = 1.50 g
Molar mass of lithium metal = 6.94 g/mol
Putting values in equation 1, we get:
[tex]\text{Moles of lithium metal}=\frac{1.50g}{6.94g/mol}=0.216mol[/tex]
For nitrogen gas:Given mass of nitrogen gas = 1.50 g
Molar mass of nitrogen gas = 28.01 g/mol
Putting values in equation 1, we get:
[tex]\text{Moles of nitrogen gas}=\frac{1.50g}{28.01g/mol}=0.053mol[/tex]
For the given chemical reaction:[tex]6Li+N_2\rightarrow 2Li_3N[/tex]
By stoichiometry of the reaction:
6 moles of lithium reacts with 1 mole of nitrogen gas
So, 0.216 moles of lithium will react with = [tex]\frac{1}{6}\times 0.216=0.036moles[/tex] of nitrogen gas.
As, the given amount of nitrogen gas is more than the required amount. Thus, it is considered as an excess reagent.
So, lithium metal is considered as a limiting reagent because it limits the formation of products.
In the reaction between lithium and nitrogen to form lithium nitride, lithium is the limiting reactant because we need more lithium than nitrogen for the reaction, but we have less available.
Explanation:To identify the limiting reactant in a chemical reaction, we need to compare the stoichiometric ratios of the reactants with the given amounts. The balanced equation for the reaction between lithium (Li) and nitrogen (N2) to form lithium nitride (Li3N) is:
6Li + N2 ⟶ 2Li3N
From this balanced equation, we see that six moles of lithium react with one mole of nitrogen. If we first convert the given masses to moles, we find that we have more moles of lithium (0.216 moles) than nitrogen (0.0535 moles). However, considering the stoichiometric ratio from the balanced equation, we need more moles of lithium than nitrogen for the reaction to occur. Therefore, lithium is the limiting reactant here as it will be completely used up before nitrogen is.
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Ideal gas (n 2.388 moles) is heated at constant volume from T1 299.5 K to final temperature T2 369.5 K. Calculate the work and heat during the process and the change of entropy of the gas. The isobaric heat capacity of the gas is Cp.m 28.253 J-K1-mol1.
Answer : The work, heat during the process and the change of entropy of the gas are, 0 J, 3333.003 J and -10 J respectively.
Explanation :
(a) At constant volume condition the entropy change of the gas is:
[tex]\Delta S=-n\times C_v\ln \frac{T_2}{T_1}[/tex]
We know that,
The relation between the [tex]C_p\text{ and }C_v[/tex] for an ideal gas are :
[tex]C_p-C_v=R[/tex]
As we are given :
[tex]C_p=28.253J/K.mole[/tex]
[tex]28.253J/K.mole-C_v=8.314J/K.mole[/tex]
[tex]C_v=19.939J/K.mole[/tex]
Now we have to calculate the entropy change of the gas.
[tex]\Delta S=-n\times C_v\ln \frac{T_2}{T_1}[/tex]
[tex]\Delta S=-2.388\times 19.939J/K.mole\ln \frac{369.5K}{299.5K}=-10J[/tex]
(b) As we know that, the work done for isochoric (constant volume) is equal to zero. [tex](w=-pdV)[/tex]
(C) Heat during the process will be,
[tex]q=n\times C_v\times (T_2-T_1)=2.388mole\times 19.939J/K.mole\times (369.5-299.5)K= 3333.003J[/tex]
Therefore, the work, heat during the process and the change of entropy of the gas are, 0 J, 3333.003 J and -10 J respectively.
A+B→2C When the reaction begins, the researcher records that the rate of reaction is such that 1 mole of A is consumed per minute. After making changes to the reaction, the researcher notes that 2 moles of A are consumed per minute. What change could the researcher have made to effect this change?
The researchers can do to increase the rate of consumption of mol A:
1. increase the concentration of substance B 2. raise the temperature 3. add catalyst 4. increase the pressureFurther explanationChemical reactions involve several factors, including factors such as reactants and products and external factors such as temperature and pressure
Factors that influence the speed of reaction in product formation.
Influencing factors include:
1. concentration of reactantsthe concentration of the reactants, the faster the reaction will be
2. surface areaThe form of reactant molecules in the form of powder or solids affects the speed of the reaction. For the same number of masses, the smaller particle size will make the reaction run faster because the reaction involves the larger surface area of the reactant
3. catalystCatalysts are added to help the reaction faster
Low activation energy is usually due to the addition of a catalyst. The catalyst makes activation energy more lace. The activation energy itself is a minimum of energy that must be possessed to a particle so that the collision produces a reaction
External factors that influence include:
1. PressurePressure will reduce the volume so that the reaction increases
High pressure also causes the reaction rate to be faster. The number of collisions between molecules is greater because of the distance between the molecules.
2. TemperatureHigh temperatures will speed up the reaction that occurs
the temperature is related to kinetic energy, heating makes kinetic energy increase and particles occur more often because particles move faster
For reaction
A + B ---> 2C
Reaction speed can be formulated:
[tex]\large{\boxed{\boxed{\bold{v~=~k.[A]^a[B]^b}}}[/tex]
where
v = reaction speed, M / s
k = constant, mol¹⁻⁽ᵃ⁺ᵇ⁾. L⁽ᵃ⁺ᵇ⁾⁻¹. S⁻¹
a = reaction order to A
b = reaction order to B
[A] = [B] = concentration of substances
So the researchers can do to increase the rate of consumption of mol A to 2 mol / min :
1. increase the concentration of substance B 2. raise the temperature 3. add catalyst 4. increase the pressureLearn morethe factor can decrease the rate of a chemical reaction
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Keywords: reaction rate, products, reactants , the researcher
Final answer:
To double the rate of a reaction from consuming 1 mole of A per minute to 2 moles per minute, the researcher could have increased the concentration of reactants, raised the temperature, or added a catalyst.
Explanation:
If a researcher observes that the consumption rate of a reactant A doubled from 1 mole per minute to 2 moles per minute in the reaction A + B → 2C, several changes could account for this increased reaction rate. One possibility is that the concentration of either reactant A or B was increased. According to the rate law, which must be determined experimentally, the rate of the reaction can be influenced by changing the concentrations of the reactants. For instance, if the reaction is first order with respect to both A and B, as represented by the rate equation rate = k [A] [B], then doubling the concentration of either reactant would indeed double the rate.
Another possibility is increasing the temperature, which generally increases reaction rates because it results in a greater proportion of molecules having the necessary activation energy to react. Additionally, the researcher could have added a catalyst to the reaction. A catalyst provides an alternative pathway for the reaction with a lower activation energy, thus increasing the rate without being consumed in the process.
One of the parameters monitored for in reactor water chemistry is boron. Which of the following is a reason for boron addition? to control reactivity to protect pumps to inhibit corrosion to preclude caustic stress corrosion
Answer:
To control reactivity.
Explanation:
One of the natural isotopes of boron (¹⁰B) with an abundance of 20%, has a high neutron cross section, wich means that it can absorb a high number of protons, thus helping in the leakage of protons to the outiside. Boron is usually added to the reactor water in form of boric acid (H₃BO₃).
Equal volumes of hydrogen and helium gas are at the same pressure. The atomic mass of helium is four times that of hydrogen. If the total mass of both gases is the same, the ratio of the temperature of helium (He) to that of hydrogen (H2) is:
Answer:
The ratio of the temperature of helium to that of hydrogen gas is 2:1.
Explanation:
Atomic mass of hydrogen = M
Temperature of hydrogen gas =T
Pressure of the hydrogen gas = P
Mass of the hydrogen gas = m
Moles of the hydrogen gas = [tex]n=\frac{m}{2M}[/tex]
Volume of the hydrogen gas = V
Using an ideal gas equation:
[tex]PV=nRT=PV=\frac{mRT}{M}[/tex]...(1)
Temperature of helium gas =T'
Pressure of the helium gas = P'= P
Mass of the helium gas = m' =m
Moles of the helium gas = [tex]n'=\frac{m}{M'}=\frac{m}{4M}[/tex]
Volume of the helium gas = V' = V
Using an ideal gas equation:
[tex]P'V'=n'RT'=\frac{mRT'}{4M}[/tex]...(2)
Divide (2) by (1)
[tex]\frac{P'V'}{PV}=\frac{\frac{mRT'}{4M}}{\frac{mRT}{2M}}[/tex]
[tex]\frac{T'}{T}=\frac{2}{1}[/tex]
The ratio of the temperature of helium to that of hydrogen gas is 2:1.
The equilibrium constant, Kc, is calculated using molar concentrations. For gaseous reactions another form of the equilibrium constant, Kp, is calculated from partial pressures instead of concentrations. These two equilibrium constants are related by the equation Kp=Kc(RT)Δn where R=0.08206 L⋅atm/(K⋅mol), T is the absolute temperature, and Δn is the change in the number of moles of gas (sum moles products - sum moles reactants). For example, consider the reaction:N2(g)+3H2(g)⇌2NH3(g)for which Δn=2−(1+3)=−2.
Part A:For the reaction 2A(g)+2B(g)⇌C(g) Kc = 55.6 at a temperature of 313 ∘C .Calculate the value of Kp.Express your answer numerically.
Part B:For the reaction X(g)+2Y(g)⇌3Z(g)Kp = 3.83×10−2 at a temperature of 119 ∘C .Calculate the value of Kc.
Answer :
(a) The value of [tex]K_p[/tex] is, [tex]5.0\times 10^{-4}[/tex]
(b) The value of [tex]K_c[/tex] is, [tex]3.83\times 10^{-2}[/tex]
Explanation:
(a) We have to determine the value of [tex]K_p[/tex].
The given balanced reaction is,
[tex]2A(g)+2B(g)\rightleftharpoons C(g)[/tex]
The relation between [tex]K_p[/tex] and [tex]K_c[/tex] are :
[tex]K_p=K_c\times (RT)^{\Delta n}[/tex]
where,
[tex]K_p[/tex] = equilibrium constant at constant pressure = ?
[tex]K_c[/tex] = equilibrium concentration constant = 55.6
R = gas constant = 0.08206 L⋅atm/(K⋅mol)
T = temperature = [tex]313^oC=273+313=586K[/tex]
[tex]\Delta n[/tex] = change in the number of moles of gas = [1 - (2 + 2)] = -3
Now put all the given values in the above relation, we get:
[tex]K_p=55.6\times (0.08206L.atm/K.mol\times 586K)^{-3}[/tex]
[tex]K_p=0.00050=5.0\times 10^{-4}[/tex]
The value of [tex]K_p[/tex] is, [tex]5.0\times 10^{-4}[/tex]
(b) We have to determine the value of [tex]K_c[/tex].
The given balanced reaction is,
[tex]X(g)+2Y(g)\rightleftharpoons 3Z(g)[/tex]
The relation between [tex]K_p[/tex] and [tex]K_c[/tex] are :
[tex]K_p=K_c\times (RT)^{\Delta n}[/tex]
where,
[tex]K_p[/tex] = equilibrium constant at constant pressure = [tex]3.83\times 10^{-2}[/tex]
[tex]K_c[/tex] = equilibrium concentration constant = ?
R = gas constant = 0.08206 L⋅atm/(K⋅mol)
T = temperature = [tex]119^oC=273+119=392K[/tex]
[tex]\Delta n[/tex] = change in the number of moles of gas = [3 - (2 + 1)] = 0
Now put all the given values in the above relation, we get:
[tex]3.83\times 10^{-2}=K_c\times (0.08206L.atm/K.mol\times 392K)^{0}[/tex]
[tex]K_c=3.83\times 10^{-2}[/tex]
The value of [tex]K_c[/tex] is, [tex]3.83\times 10^{-2}[/tex]
To calculate Kp for the reaction 2A(g) + 2B(g) ⇌ C(g) with Kc = 55.6 at 313 °C, convert the temperature to Kelvin, use Δn = -3, and apply the formula Kp = Kc(RT)Δn. Calculating Kp gives its value. To find Kc for the reaction X(g) + 2Y(g) ⇌ 3Z(g) where Kp is given at 119 °C, convert temperature to Kelvin, use Δn = 0, and Kp = Kc, giving Kc = 3.83×10⁻².
Explanation:To calculate the equilibrium constant Kp for the reaction 2A(g) + 2B(g) ⇌ C(g), where Kc = 55.6 at a temperature of 313 °C. First, we convert the temperature to Kelvin by adding 273.15 to the Celsius temperature. The temperature in Kelvin is T = 313 + 273.15 = 586.15 K. We use the equation Kp = Kc(RT)Δn, where Δn is the change in moles of gas, R is the ideal gas constant 0.08206 L·atm/(K·mol), and T is the absolute temperature in Kelvin.
For the reaction, Δn = 1 - (2 + 2) = -3. Plugging values into the equation: Kp = 55.6 x (0.08206 L·atm/(K·mol) x 586.15 K)^-3. Calculating Kp gives the value of Kp for this reaction.
To find Kc from Kp for the reaction X(g) + 2Y(g) ⇌ 3Z(g), where Kp = 3.83×10⁻² at a temperature of 119 °C, we again convert to Kelvin (119 + 273.15 = 392.15 K), and this time Δn = 3 - (1 + 2) = 0. With Δn = 0, Kp = Kc, so Kc = 3.83×10⁻² for this reaction.
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If the fugacity of a pure component at the conditions of an ideal solution is 40 bar and its mole fraction is 0.4, what is its fugacity in the solution according to the Lewis-Randall rule?
Answer : The fugacity in the solution is, 16 bar.
Explanation : Given,
Fugacity of a pure component = 40 bar
Mole fraction of component = 0.4
Lewis-Randall rule : It states that in an ideal solution, the fugacity of a component is directly proportional to the mole fraction of the component in the solution.
Now we have to calculate the fugacity in the solution.
Formula used :
[tex]f_i=X_i\times f_i^o[/tex]
where,
[tex]f_i[/tex] = fugacity in the solution
[tex]f_i^o[/tex] = fugacity of a pure component
[tex]X_1[/tex] = mole fraction of component
Now put all the give values in the above formula, we get:
[tex]f_i=0.4\times 40\text{ bar}[/tex]
[tex]f_i=16\text{ bar}[/tex]
Therefore, the fugacity in the solution is, 16 bar.
If the pressure at a depth of 8 ft below the surface of a stagnant fluid is 3.5 psi, what will the pressure of the liquid be at a depth of 48 ft below the surface?
Answer : The pressure of the liquid at depth of 48 ft below the surface of fluid is, 21 psi.
Explanation : Given,
The pressure of the liquid at depth 8 ft = 3.5 psi
Now we have to calculate the pressure of the liquid at depth 28 ft.
According to the question,
As, the depth of 8 ft below the surface of a stagnant fluid has pressure = 3.5 psi
So, the depth of 48 ft below the surface of a stagnant fluid has pressure = [tex]\frac{3.5\text{ psi}}{8\text{ ft}}\times 48\text{ ft}=21\text{ psi}[/tex]
Therefore, the pressure of the liquid at depth of 48 ft below the surface of fluid is, 21 psi.
Red mercury (II) oxide decomposes to form mercury metal and oxygen gas according to the following equation: 2HgO (s) 2Hg (l) + O2 (g). If 4.00 moles of HgO decompose to form 1.50 moles of O2 and 603 g of Hg, what is the percent yield of this reaction?
hey there!:
2HgO (s) => 2Hg (l) + O2 (g)
2 moles of HgO decompose to form 2 moles of Hg and 1 mole of O2 according to the reaction mentioned in the question.
So 4.00 moles of HgO must give 4 moles of Hg and 2 moles of O2 theoretically.
603 g of Hg = 603 / 200.6 = 3 moles
Percent yield = ( actual yield / theoretical yield) * 100
= ( 3/4) * 100
= 75 %
Hope this helps!
Answer: The percent yield of the reaction is 75 %
Explanation:
We are given:
Moles of HgO decomposed = 4.00 moles
The given chemical reaction follows:
[tex]2HgO(s)\rightarrow 2Hg(l)+O_2(g)[/tex]
By Stoichiometry of the reaction:
2 moles of HgO produces 1 moles of oxygen gas
So, 4.00 moles of HgO will produce = [tex]\frac{1}{2}\times 4.00=2mol[/tex] of oxygen gas
To calculate the percentage yield of the reaction, we use the equation:
[tex]\%\text{ yield}=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100[/tex]
Experimental yield of oxygen gas = 1.50 moles
Theoretical yield of oxygen gas = 2.00 moles
Putting values in above equation, we get:
[tex]\%\text{ yield of oxygen gas}=\frac{1.50}{2.00}\times 100\\\\\% \text{yield of oxygen gas}=75\%[/tex]
Hence, the percent yield of the reaction is 75 %
You need to produce a buffer solution that has a pH of 5.31. You already have a solution that contains 10. mmol (millimoles) of acetic acid. How many millimoles of acetate (the conjugate base of acetic acid) will you need to add to this solution? The pKa of acetic acid is 4.74.
Answer:
37 mmol of acetate need to add to this solution.
Explanation:
Acetic acid is an weak acid. According to Henderson-Hasselbalch equation for a buffer consist of weak acid (acetic acid) and its conjugate base (acetate)-
[tex]pH=pK_{a}(acetic acid)+log[\frac{mmol of CH_{3}COO^{-}}{mmol of CH_{3}COOH }][/tex]
Here pH is 5.31, [tex]pK_{a}[/tex] (acetic acid) is 4.74 and number of mmol of acetic acid is 10 mmol.
Plug in all the values in the above equation:
[tex]5.31=4.74+log[\frac{mmol of CH_{3}COO^{-}}{10}][/tex]
or, mmol of [tex]CH_{3}COO^{-}[/tex] = 37
So 37 mmol of acetate need to add to this solution.
Calculate the vapor pressure at 25 °C of an aqueous solution that is 5.50% NaCl by mass. (Assume complete dissociation of the solute.) Tro, Nivaldo J.. Chemistry (p. 617). Pearson Education. Kindle Edition.
Answer:
23.38 mmHg is the vapor pressure at 25 °C of an aqueous solution that is 5.50% NaCl by mass.
Explanation:
Vapor pressure of water at 25 °C ,[tex]p^o= 23.8 mmHg[/tex]
Vapor pressure of the solution = [tex]p_s[/tex]
Number moles of water in 5.50% NaCl solution.In 100 gram of solution, 94.5 g of water is present.
[tex]n_1=\frac{94.5 g}{18 g/mol}=5.25 mol[/tex]
Number moles of NaCl in 5.50% NaCl solution.5.50 g of NaCl in 100 grams of solution.
[tex]n_2=\frac{5.50 g}{58.5 g/mol}=0.09401 mol [/tex]
Mole fraction of the solute = [tex]\chi_2=\frac{n_2}{n_1+n_2}[/tex]
[tex]\chi_2=\frac{0.09401 mol}{0.09401 mol+5.25 mol}=0.01759[/tex]
The relative lowering in vapor pressure of the solution with non volatile solute is equal to the mole fraction of solute in the solution:
[tex]\frac{p^o-p_s}{p^o}=\chi_2=\frac{n_2}{n_1+n_2}[/tex]
[tex]\frac{23.8 mmHg - p_s}{23.8 mmHg}=0.01759[/tex]
[tex]p_s=23.38 mmHg[/tex]
23.38 mmHg is the vapor pressure at 25 °C of an aqueous solution that is 5.50% NaCl by mass.
23.38 mmHg is the vapour pressure.
The pressure enforced by the vapours when in the thermodynamical equilibrium state on the system is called the vapour pressure.
How to calculate the vapour pressure?Vapour pressure [tex](p^{\circ})[/tex] of water at [tex]25 ^{\circ} \rm C[/tex] = 23.8 mmHgVapour pressure of aqueous solution = [tex](p_{s})[/tex]Step 1: Moles of water in 5.50% NaCl solution when 100 grams of solution = 94.5 g of water
[tex]\begin{aligned}\rm n &= \dfrac{94.5 \;\rm g}{18\;\rm g/mol}\\\\\\&= 5.25\;\rm mol\end{aligned}[/tex]
Step 2: Moles of NaCl in 5.50% solution when 5.50 g NaCl in 100 grams solution then,
[tex]\begin{aligned}\rm n &= \dfrac{5.50\;\rm g}{58.5\;\rm g/mol}\\\\\\&= 0.09401\;\rm mol\end{aligned}[/tex]
Step 3: Calculate the mole fraction of the solute:
[tex]\begin{aligned}\rm X_{2} &= \dfrac{\rm n_{2}}{\rm n_{1}+n_{2}}\\\\\rm X_{2} &= \dfrac{0.09401\;\rm mol}{0.09401 + 5.25\;\rm mol}\\\\&= 0.0175\end{aligned}[/tex]
Step 4: Calculate the vapour pressure
[tex]\begin{aligned}\rm X_{2} &= \dfrac{\rm n_{2}}{\rm n_{1}+n_{2}} = \dfrac{p^{\circ}-p_{s}}{p^{\circ}}\\\\0.0175 &= \dfrac{23.38 \;\text{mmHg} -p_{s}}{23.38 \;\rm mmHg}\\\\&= 23.38\;\rm mmHg\end{aligned}[/tex]
Therefore, 23.38 mmHg is the vapour pressure.
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Under identical conditions, separate samples of O2 and an unknown gas were allowed to effuse through identical membranes simultaneously. After a certain amount of time, it was found that 4.644.64 mL of O2 had passed through the membrane, but only 3.123.12 mL of of the unknown gas had passed through. What is the molar mass of the unknown gas? unknown molar mass: g/mol
Answer:
70.77 g/mol is the molar mass of the unknown gas.
Explanation:
Effusion is defined as rate of change of volume with respect to time.
Rate of Effusion=[tex]\frac{Volume}{Time}[/tex]
Effusion rate of oxygen gas after time t = [tex]E=\frac{4.64 mL}{t}[/tex]
Molar mass of oxygen gas = M = 32 g/mol
Effusion rate of unknown gas after time t = [tex]E'=\frac{3.12 mL}{t}[/tex]
Molar mass of unknown gas = M'
The rate of diffusion of gas, we use Graham's Law.
This law states that the rate of effusion or diffusion of gas is inversely proportional to the square root of the molar mass of the gas. The equation given by this law follows:
[tex]\text{Rate of effusion}\propto \frac{1}{\sqrt{\text{Molar mass of the gas}}}[/tex]
[tex]\frac{E}{E'}=\sqrt{\frac{M'}{M}}[/tex]
[tex]\frac{\frac{4.64 mL}{t}}{\frac{3.12 mL}{t}}=\sqrt{\frac{M'}{32 g/mol}}[/tex]
M' = 70.77 g/mol
70.77 g/mol is the molar mass of the unknown gas.
Under identical conditions 4.64 mL of O₂ and 3.12 mL of an unknown gas effuse for the same time. The molar mass of the unknown gas is 70.1 g/mol.
What does Graham's law state?Graham's law states that the rate of effusion of a gas is inversely proportional to the square root of its molar mass.
r ∝ 1/√M
where,
r is the rate of effusion.M is the molar mass.Under identical conditions and after the same amount of time (t), 4.64 mL of O₂ and 3.12 mL of an unknown gas X effuse. The ratio of their rates of effusion is:
[tex]\frac{rO_2}{rX} = \frac{\frac{vO_2}{t} }{\frac{vX}{t} } = \frac{vO_2}{vX} = \frac{4.64 mL}{3.12 mL} = 1.48[/tex]
We can calculate the molar mass of the unknown gas by applying Graham's law.
[tex]\frac{rO_2}{rX} = 1.48 = \sqrt{\frac{M(X)}{M(O_2)} } = \sqrt{\frac{M(X)}{32.00 g/mol} }\\\\M(X) = 70.1 g/mol[/tex]
Under identical conditions 4.64 mL of O₂ and 3.12 mL of an unknown gas effuse for the same time. The molar mass of the unknown gas is 70.1 g/mol.
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An ideal gas is enclosed in a cylinder with a movable piston on top of it. The piston has a mass of 7 650 g and an area of 5.10 cm2 and is free to slide up and down, keeping the pressure of the gas constant. How much work is done on the gas as the temperature of 0.180 mol of the gas is raised from 18.5°C to 340°C?
Explanation:
According to the ideal gas equation, PV = nRT.
So, V = [tex]\frac{nRT}{P}[/tex] ......... (1)
Since, it is given that volume of the gas is increases. So, change in volume will be as follows.
[tex]\Delta V = V_{f} - V_{i}[/tex]
Hence, equation (1) will become as follows.
[tex]\Delta V[/tex] = [tex]\frac{nRT_{f}}{P} - \frac{nRT_{i}}{P}[/tex]
[tex]P \Delta V[/tex] = nR[tex](T_{f} - T_{i})[/tex]
Therefore, work done on the gas will be given as follows.
W = - P [tex]\Delta V[/tex]
= [tex]-n \times R(T_{f} - T_{i})[/tex]
= [tex]- 0.0180 mol \times 8.31 atm L/mol K (340^{o}C - 18.5^{o}C)[/tex]
= 48.09 J
Thus, we can conclude that work done on the gas is 48.09 J.
A serving fish contains 50 g of protein and 4 gof fat. If protein has a caloric calue of 4.0 kcal/g and fat has 9.1 kcal/g, how many kcal are in the serving?
Answer:
there are 236.4 kcal in the serving fish
Explanation:
1 g of protein has 4.0 kcal thus 50 g will have 50 x 4.0 = 200 kcal
1 g of fat has 9.1 kcal thus 4 g will have 4 x 9.1 = 36.4 kcal
the total kcal present in the serving will be the sum of kcal from protein and fat
200 kcal + 36.4 kcal = 236.4 kcal
Write a chemical equation for the reaction that occurs in the following cell: Cu|Cu2+(aq)||Ag+(aq)|Ag Express your answer as a balanced net ionic equation. Identify all of the phases in your answer.
Answer:
Explanation:
The cell reaction properly written is shown below:
Cu|Cu²⁺[tex]_{aq}[/tex] || Ag⁺[tex]_{aq}[/tex] | Ag
From this cell reaction, to get the net ionic equation, we have to split the reaction into their proper oxidation and reduction halves. This way, we can know that is happening at the electrodes and derive the overall net equation.
Oxidation half:
Cu[tex]_{s}[/tex] ⇄ Cu²⁺[tex]_{aq}[/tex] + 2e⁻
At the anode, oxidation occurs.
Reduction half:
Ag⁺[tex]_{aq}[/tex] + 2e⁻ ⇄ Ag[tex]_{s}[/tex]
At the cathode, reduction occurs.
To derive the overall reaction, we must balance the atoms and charges:
Cu[tex]_{s}[/tex] ⇄ Cu²⁺[tex]_{aq}[/tex] + 2e⁻
Ag⁺[tex]_{aq}[/tex] + e⁻ ⇄ Ag[tex]_{s}[/tex]
we multiply the second reaction by 2 to balance up:
2Ag⁺[tex]_{aq}[/tex] + 2e⁻ ⇄ 2Ag[tex]_{s}[/tex]
The net reaction equation:
Cu[tex]_{s}[/tex] + 2Ag⁺[tex]_{aq}[/tex] + 2e⁻⇄ Cu²⁺[tex]_{aq}[/tex] + 2e⁻ + 2Ag[tex]_{s}[/tex]
We then cancel out the electrons from both sides since they appear on both the reactant and product side:
Cu[tex]_{s}[/tex] + 2Ag⁺[tex]_{aq}[/tex] ⇄ Cu²⁺[tex]_{aq}[/tex] + 2Ag[tex]_{s}[/tex]
The net ionic equation for the cell - Cu|Cu2+(aq)||Ag+(aq)|Ag is Cu(s) + 2Ag+(aq) --> Cu2+(aq) + 2Ag(s). This reaction involves the transfer of electrons from copper to silver ions, converting copper metal to copper ions and silver ions to silver metal.
Explanation:The chemical equation for the reaction in the given cell, Cu|Cu2+(aq)||Ag+(aq)|Ag, can be expressed as follows:
Cu(s) --> Cu2+(aq) + 2e−
2Ag+(aq) + 2e− --> 2Ag(s)
Combining these two half-reactions gives the net ionic equation:
Cu(s) + 2Ag+(aq) --> Cu2+(aq) + 2Ag(s)
In this reaction, solid copper (Cu) reacts with silver ions (Ag+) in the aqueous solution to produce copper ions (Cu2+) and solid silver (Ag). All the phases for each species are defined in the equation. The copper metal is solid, represented by (s), both ionic species are in aqueous solution denoted by (aq).
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Sulfur dioxide and oxygen react to form sulfur trioxide during one of the key steps in sulfuric acid synthesis. An industrial chemist studying this reaction fills a 25.0L tank with 4.5 mol of sulfur dioxide gas and 4.5 mol of oxygen gas at 30.°C. He then raises the temperature, and when the mixture has come to equilibrium measures the amount of sulfur trioxide gas to be 1.4 mol. Calculate the concentration equilibrium constant for the reaction of sulfur dioxide and oxygen at the final temperature of the mixture. Round your answer to 2 significant digits.
Answer:
1.3 is the concentration equilibrium constant for the reaction of sulfur dioxide and oxygen at the final temperature of the mixture.
Explanation:
Initial Concentration of sulfur dioxide = [tex][SO_2]=\frac{4.5 mol}{25 L}=0.18 M[/tex]
Initial Concentration of oxygen= [tex][O_2]=\frac{4.5 mol}{25 L}=0.18 M[/tex]
[tex]2SO_2+O_2\rightleftharpoons 2SO_3[/tex]
Initially (0.18 M) (0.18 M) 0
Eq'm (0.18 -2x) (0.18 -x) 2x
Equilibrium concentration of sulfur trioxide =[tex][SO_3]=2x=\frac{1.4 mol}{25 L}=0.056 M[/tex]
x = 0.028 M
Equilibrium concentration of sulfur dioxide =[tex][SO_2]'=(0.18 -2x)=0.18 - 0.056 =0.124 M[/tex]
Equilibrium concentration of oxygen=[tex][O_2]'=(0.18 -x)=0.18 - 0.028 =0.152 M[/tex]
The expression for an equilibrium constant will be :
[tex]K_c=\frac{[SO_3]^2}{[SO_2]'^2[O_2]'}[/tex]
[tex]K_c=\frac{(0.056 M)^2}{(0.124 M)^2(0.152 M)}=1.3418\approx 1.3[/tex]
1.3 is the concentration equilibrium constant for the reaction of sulfur dioxide and oxygen at the final temperature of the mixture.
Final answer:
To calculate the concentration equilibrium constant (Kc) for the reaction of SO2 and O2 to form SO3, find the equilibrium concentrations from the initial amounts and amount at equilibrium, then apply the equilibrium expression. The resulting Kc for the reaction at the final temperature is 0.020 when rounded to two significant digits.
Explanation:
The calculation of the concentration equilibrium constant for the reaction between sulfur dioxide and oxygen to form sulfur trioxide at a certain temperature involves using the equilibrium concentrations of reactants and products. The balanced chemical equation for the reaction is:
2 SO2(g) + O2(g) = 2 SO3(g)
Given 4.5 mol of SO2 and 4.5 mol of O2 initially in a 25.0 L tank, and 1.4 mol of SO3 at equilibrium, we can calculate the change in moles during the reaction (δ) and thus the equilibrium concentrations ([SO2], [O2], and [SO3]). The concentration equilibrium constant (Kc) is then found using the expression:
Kc = ([SO3]2)/([SO2]2 × [O2])
Through stoichiometry and equilibrium concentration calculations, the concentrations are:
[SO3] = 1.4 mol / 25.0 L = 0.056 M
[SO2] = (4.5 mol - 1.4 mol) / 25.0 L = 0.124 M (since 1 mol of SO2 is consumed for every mol of SO3 produced)
[O2] = (4.5 mol - 0.7 mol) / 25.0 L = 0.152 M (since 0.5 mol of O2 is consumed for every mol of SO3 produced)
Plugging these values into the Kc expression gives:
Kc = (0.0562) / (0.1242 × 0.152) = 0.0197
The calculated Kc at the final temperature is 0.020 (rounded to two significant digits).
At the base of a frictionless icy hill that rises at 25.0∘ above the horizontal, a toboggan has a speed of 11.5 m/s toward the hill. How high vertically above the base will it go before stopping?
The toboggan, having initial kinetic energy, gains potential energy as it goes up the hill until it stops. The kinetic energy's conversion into potential energy is described by the equation 0.5 * m * v^2 = m * g * h. TThe toboggan will stop approximately 2.82 m high above the base.
Explanation:This is a question about kinematics, more specifically potential and kinetic energy conversion in the presence of gravity. In this scenario, the kinetic energy of the toboggan is being converted into potential energy as it climbs up the hill.
The main formula we will use is the energy conservation law: Kinetic Energy = Potential Energy, or 0.5*m*v^2 = m*g*h where:
m is the mass, v is the velocity, g is the acceleration due to gravity (9.8 m/s^2), h is the height.Even though we do not know the mass of the toboggan, it cancels out in the equation. So the actual height could be evaluated as h = v^2 / (2*g), substituting the given v and g into this formula, h = (11.5)^2 / (2*9.8) ≈ 6.75 m.
The actual vertical height above the base is h' = h * sinθ, where θ is the angle of the hill. Substituting into the formula, h' = 6.75 * sin(25) ≈ 2.82 m. So, the toboggan will go approximately 2.82m high vertically above the base before stopping.
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The toboggan will ascend the hill to a height of approximately 6.75 meters before stopping. This solution was derived using the principles of conservation of energy and kinematics.
Explanation:This question can be solved using the principles of conservation of energy and kinematics in physics. The toboggan, at the bottom of the hill, possesses kinetic energy (KE) which can be calculated using the formula KE = 1/2 m×v², where 'm' is the mass of the toboggan and 'v' is its speed.
As the toboggan moves up the icy, frictionless hill, this kinetic energy is gradually converted to gravitational potential energy (PE = m×g×h, where 'm' is the mass, 'g' is the acceleration due to gravity, and 'h' is the height). At the point when the toboggan stops, all the kinetic energy has been converted into potential energy.
It's important to note that we don't actually need the mass of the toboggan for this calculation because it cancels out in the equation KE = PE. Rearranging the equation to solve for 'h' gives us h = v² / (2*g)
Substituting the given values (v = 11.5 m/s and g = 9.81 m/s²), we get a height of approximately 6.75 meters. Therefore the toboggan will go up the hill to a height of approximately 6.75 meters above the base before stopping.
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The common titanium alloy known as T-64 has a composition of 90 weight% titanium 6 wt% aluminum and 4 wt% vanadium. Calculate the concentrations as atomic percents.
Explanation:
Suppose in 100 g of alloy contains 90% titanium 6% aluminum and 4% vanadium.
Mass of titanium = 90 g
Moles of titanium = [tex]\frac{90 g}{47.87 g/mol}=1.8800 mol[/tex]
Total number of atoms of titanium ,[tex]a_t=1.8800 mol\times N_A[/tex]
Mass of aluminum = 6 g
Moles of aluminium = [tex]\frac{6 g}{26.98 g/mol}=0.2223 mol[/tex]
Total number of atoms of aluminium,[tex]a_a=0.2223 mol\times N_A[/tex]
Mass of vanadium = 4 g
Moles of vanadium= [tex]\frac{4 g}{50.94 g/mol}=0.0785 mol[/tex]
Total number of atoms of vanadium[tex]a_v=0.0785 mol\times N_A[/tex]
Total number of atoms in an alloy = [tex]a_t+a_a+a_v[/tex]
Atomic percentage:
[tex]Atomic\%=\frac{\text{total atoms of element}}{\text{Total atoms in alloy}}\times 100[/tex]
Atomic percentage of titanium:
:[tex]\frac{a_t}{a_t+a_a+a_v}\times 100=\frac{1.8800 mol\times N_A}{1.8800 mol\times N_A+0.2223 mol\times N_A+0.0785 mol\times N_A}\times 100=86.20\%[/tex]
Atomic percentage of Aluminium:
:[tex]\frac{a_a}{a_t+a_a+a_v}\times 100=\frac{0.2223 mol\times N_A}{1.8800 mol\times N_A+0.2223 mol\times N_A+0.0785 mol\times N_A}\times 100=10.19\%[/tex]
Atomic percentage of vanadium
:[tex]\frac{a_v}{a_t+a_a+a_v}\times 100=\frac{0.0785 mol\times N_A}{1.8800 mol\times N_A+0.2223 mol\times N_A+0.0785 mol\times N_A}\times 100=3.59\%[/tex]