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 %
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.
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.
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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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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Which of the following does not influence the effectiveness of a detergent
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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.
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.
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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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.
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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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.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
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:
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.
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).