Answer:
There are required 70.1 years for the activity of a sample of cesium-137 to fall to 20.0 percent of its original value
Explanation:
The radioactive decay follows always first-order kinetics where its general law is:
Ln[A] = -Kt + ln[A]₀
Where [A] is actual concentration of the atom, k is rate constant, t is time and [A]₀ is initial concentration.
We can find rate constant from half-life as follows:
Rate constant:
t(1/2) = ln 2 / K
As half-life of Cesium-137 is 30.2 years:
30.2 years = ln 2 / K
K = 0.02295 years⁻¹
Replacing this result and with the given data of the problem:
Ln[A] = -Kt + ln[A]₀
Ln[A] = -0.02295 years⁻¹* t + ln[A]₀
Ln ([A] / [A₀]) = -0.02295 years⁻¹* t
As you want time when [A] is 20% of [A]₀, [A] / [A]₀ = 0.2:
Ln (0.2) = -0.02295 years⁻¹* t
70.1 years = t
There are required 70.1 years for the activity of a sample of cesium-137 to fall to 20.0 percent of its original valueCompound A is an alkene that was treated with ozone to yield only (CH3CH2CH2)2C=O. Draw the major product that is expected when compound A is treated with a peroxy acid (RCO3H) followed by aqueous acid (H3O+).
Answer:
2,2,3,3-tetrapropyloxirane
Explanation:
In this case, we have to know first the alkene that will react with the peroxyacid. So:
What do we know about the unknown alkene?
We know the product of the ozonolysis reaction (see figure 1). This reaction is an oxidative rupture reaction. Therefore, the double bond will be broken and we have to replace the carbons on each side of the double bond by oxygens. If [tex](CH_3CH_2CH_2)_2C=O[/tex] is the only product we will have a symmetric molecule in this case 4,5-dipropyloct-4-ene.
What is the product with the peroxyacid?
This compound in the presence of alkenes will produce peroxides. Therefore we have to put a peroxide group in the carbons where the double bond was placed. So, we will have as product 2,2,3,3-tetrapropyloxirane. (see figure 2)
Which led to the formation of oceans after water on Earth's surface evaporated?
Answer:
condensation
Explanation:
What led to the formation of oceans after the water on the Earth's surface evaporated is a condensation reaction of water vapour.
The water present on the surface of the earth was able to evaporate as a result of the hot condition of the primitive earth. As the earth cools down the water vapour present in the atmosphere began to condense, gradually forming puddles of water and eventually leading to the formation of various oceans as we currently have it on the earth.
Answer:
as earth cooled, water in the atmosphere condensed.
Explanation:
2.Which of the alcohols listed below would you expect to react most rapidly with PBr3?A)CH3CH2CH2CH2CH2CH2OHB)(CH3CH2)2CH(OH)CH2CH3C)(CH3CH2)2CHOHCH3D)(CH3CH2)3COHE)(CH3CH2)2C(CH3)OH
Answer:
A) CH3CH2CH2CH2CH2CH2OH
Explanation:
For this question, we have the following answer options:
A) CH3CH2CH2CH2CH2CH2OH
B) (CH3CH2)2CH(OH)CH2CH3
C) (CH3CH2)2CHOHCH3
D) (CH3CH2)3COH
E) (CH3CH2)2C(CH3)OH
We have to remember the reaction mechanism of the substitution reaction with [tex]PBr_3[/tex]. The idea is to generate a better leaving group in order to add a "Br" atom.
The [tex]PBr_3[/tex] attacks the "OH" generation new a bond to P (O-P bonds are very strong), due to this new bond we will have a better leaving group that can remove the oxygen an allow the attack of the Br atom to generating a new C-Br bond. This is made by an Sn2 reaction. Therefore we will have a faster reaction with primary substrates. In this case, the only primary substrate is molecule A. So, "CH3CH2CH2CH2CH2CH2OH" will react faster.
See figure 1
I hope it helps!
In the reaction 2 AgI + HgI 2 → Ag 2HgI 4, 2.00 g of AgI and 3.50 g of HgI 2 were used. What is the limiting reactant?
Answer:
AgI I the limiting reactant.
Explanation:
The balanced equation for the reaction is given below:
2AgI + HgI2 → Ag2HgI4
Next, we shall determine masses AgI and HgI2 that reacted from the balanced equation.
This is illustrated below:
Molar mass of Agl = 108 + 127 = 235 g/mol
Mass of AgI from the balanced equation = 2 x 235 = 470 g
Molar mass of HgI2 = 201 + (2x127) = 455 g/mol
Mass of HgI2 from the balanced equation = 1 x 455 = 455 g
From the balanced equation above,
470 g of AgI reacted with 455 g of HgI2.
Finally, we shall determine the limiting reactant as follow:
From the balanced equation above,
470 g of AgI reacted with 455 g of HgI2.
Therefore, 2 g of AgI will react with
= (2 x 455)/470 = 1.94 g of HgI2.
From the calculations made above, we can see that only 1.94 g out of 3.50 g of HgI2 given is needed to react completely with 2 g of AgI.
Therefore, AgI I the limiting reactant.
What are the conjugate acid-base pairs in the following chemical reaction? HBr(aq)+ CH3COOH(aq) ⇌ CH3C(OH)2+(aq) + Br-(aq)
Answer:
HBr, CH3C(OH)2 and CH3COOH, Br-
Explanation:
The conjugate acid-base pairs acid reacts with base to form a conjugate acid and conjugate base.
Conjugate acid is formed when a bases receives a proton (H+) and a conjugate base is formed when an acid losses a proton (H+).
From the given equation:
HBr, CH3C(OH)2 and CH3COOH, Br- are conjugate acid-base pair, where HBr is an acid and CH3C(OH)2 is a conjugate acid while CH3COOH and Br- is the conjugate base.
What is the atomic mass for Helium (He)? Question 5 options: 8 2 3 4
13. Arrange each group of units from smallest to
largest
a)km, mm, cm, m
b) mg, kg, g
C) L, mL
d) s, ms, min, h
Answer:
A. mm,cm,m,Km
B. mg, g, Kg
C. mL,L
D. ms, s, min, h
Explanation:
A piece of solid metal is put into an aqueous solution of . Write the net ionic equation for any single-replacement redox reaction. Assume that the oxidation state of in the resulting solution is 2 .
The question is incomplete,the complete question is as follows:
A piece of solid Fe metal is put into an aqueous solution of Cu(NO3)2. Write the net ionic equation for any single-replacement redox reaction that may be predicted. Assume that the oxidation state of in the resulted solution is 2 . (Use the lowest possible coefficients for the reaction. Use the pull-down boxes to specify states such as (aq) or (s). If a box is not needed, leave it blank. If no reaction occurs, leave all boxes blank and click on Submit.)
Answer:
Fe(s) + Cu^2+(aq) => Fe^2+(aq) + Cu(s)
Explanation:
An ionic equation is a chemical equation which shows clear image of reactions of the electrolytes in aqueous solution.
Molecular reaction equation for the reaction between iron and copper II nitrate is as follows:
Fe(s) + Cu(NO3)2(aq) => Fe(NO3)2(aq) +Cu(s)
The net ionic equation for any single-replacement redox reaction is as follows:
Fe(s) + Cu^2+(aq) => Fe^2+(aq) + Cu(s)
105/22 • (1.251 - 0.620)=
Answer:
105/22*(1.251-0.620)
105/22*0.631
4.772*0.631
3.011132
Hope it helps
Answer:
3.0
Explanation:
First, complete the operations inside the parenthesis according to the normal rules for significant figures. Because there are subsequent calculations, keep at least one extra significant figure when possible: (4.7727) × (0.631).
The final product will be rounded to two significant figures because it can’t be more precise than the least precise number in the problem, 22. The final product is 3.0.
Sulfur dioxide reacts with oxygen to form sulfur trioxide. What change in hybridization of the sulfur occurs in this reaction ? g
Answer:
PLEASE LOOK INN TO THE FILE YOU WILL GET ANSWER AND ALSO SUMMARY THANKS FOR ASKING QUESTION.
Explanation:
The argon atoms are excited into an excited state before emitting the 488.0 nm laser. It is known that the energy of the first ionization energy of argon is 1520 kJ mol-1. What is the energy level of the excited state (in unit eV) lies below the vacuum energy level (0 eV)
Answer:
Explanation:
Given that:
The argon atoms are excited into an excited state before emitting the 488.0 nm laser.
the energy of the first ionization energy of argon is 1520 kJ mol-1.
SInce 1 eV = 96.49 kJ/mol
Therefore, the energy of the first ionization energy of argon in eV is = ( 1520/ 96.49) eV
= 15.75 eV
To find where the energy level of the excited state lies below the vacuum energy level, let's first determine, the energy liberated by using planck expression.
[tex]E = \dfrac{hc}{\lambda}[/tex]
[tex]E = \dfrac{6.6 \times 10^{-34} \times 3 \times 10^8}{488 \times 10^{-9}}[/tex]
[tex]E = \dfrac{1.98 \times 10^{-25}}{488 \times 10^{-9}}[/tex]
[tex]E = \dfrac{1.98 \times 10^{-25}}{488 \times 10^{-9}}[/tex]
[tex]E =4.057 \times 10^{-19} \ J[/tex]
Converting Joules (J) to eV ; we get,
[tex]E =\dfrac{4.057 \times 10^{-19}}{1.6 \times 10^{-19}}[/tex]
E = 2.53 eV
The energy levels of the first exited state = -13.223 eV
What would happen to the measured cell potentials if 30 mL solution was used in each half-cell instead of 25 mL
Answer:
The answer is "[tex]\bold{\log \frac{[0] mole}{[R]mole}}[/tex]"
Explanation:
[tex]E_{cell} =E_{cell}^{\circ} - \frac{0.0591}{n}= \log\frac{[0]}{[R]}\\[/tex]
In the above-given equation, we can see from [tex]E_{ceu}[/tex], of both oxidant [tex]conc^n[/tex]as well as the reactant were connected. however, weight decreases oxidant and reduction component concentration only with volume and the both of the half cells by the very same factor and each other suspend
[tex]\to \log \frac{\frac{\text{oxidating moles}}{25 \ ml}}{\frac{\text{moles of reduction}}{25 ml}} \ \ = \ \ \log \frac{\frac{\text{oxidating moles}}{30 \ ml}}{\frac{\text{moles of reduction}}{30 ml}} \\\\\\[/tex]
[tex]\to {\log \frac{[0] mole}{[R]mole}}[/tex]
The cell potential of the electrochemical reaction has been the same when the volume has been reduced from 30 mL to 25 mL in each half cells.
The cell potential has been given as the difference in the potential of the two half cells in the electrochemical reaction.
The two cells has been set with the concentration of solutions in the oxidation and reduction half cells.
Cell potential changeThe cell potential has been changed when there has been a change in the potential of the half cells.
The volume of 30 mL to the solution has been, resulting in the cell potential difference of x.
With the volume of 25 mL, there has been the difference in the potential being similar to the 30 mL solution, i.e. x.
Thus, the cell potential of the electrochemical reaction has been the same when the volume has been reduced from 30 mL to 25 mL in each half cells.
Learn more about cell potential, here:
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what is the oxidation state of the oxygen atoms in co2,h2o and o2 and what does this information tell you about photosynthesis and respiration
Answer:
-2, -2 and 0.
- Respiration is a process in which energy is produced and photosynthesis is a process in which energy is used.
Explanation:
Hello,
In this case, oxygen is a substance that is used for animals and us to acquire the energy necessary for several functions by the cellular respiration (we also need glucose), besides, it is a product of the photosynthesis carried out by vegetable cells (plants). Moreover, carbon dioxide and water are used by the plants to produce oxygen we need as well as glucose via the aforementioned photosynthesis, thus, both chemical reactions are shown below:
[tex]C_6H_{12}O_6+6O_2\rightarrow 6CO_2+6H_2O\ \ \ respiration\\\\6CO_2+6H_2O\rightarrow C_6H_{12}O_6+6O_2\ \ \ photosynthesis[/tex]
In such a way, since the oxygen in carbon dioxide and water has an oxidation state of -2 (reduced form) we can say that the respiration is a process in which energy is produced and since the oxygen yielded during the photosynthesis has an oxydation state of 0, we can say that photosynthesis is a process in which energy is used.
Best regards.
Calculate the molality of a solution containing 141.5 g of glycine (NH2CH2COOH) dissolved in 4.456 kg of H2O
Answer:
0.423 m.
Explanation:
The following data were obtained from the question:
Mass of glycine (NH2CH2COOH) = 141.5 g
Mass of water = 4.456 kg
Molality =.?
Next, we shall determine the number of mole in 141.5 g of glycine (NH2CH2COOH.
This is illustrated below:
Mass of glycine (NH2CH2COOH) = 141.5 g
Molar mass of glycine (NH2CH2COOH) = 14 + (2x1) + 12 + (2x1) + 12 + 16 + 16 + 1 = 75 g/mol
Mole of glycine (NH2CH2COOH) =.?
Mole = mass /Molar mass
Mole of glycine (NH2CH2COOH) = 141.5/75
Mole of glycine (NH2CH2COOH) = 1.887 moles
Finally, we shall determine the molality of the solution as follow:
Molality is simply defined as the mole of solute per kilogram of water. Mathematically it is expressed as:
Molality = mole / mass (kg) of water
With the above formula, we can obtain the molality of the solution as follow:
Mole of glycine (NH2CH2COOH) = 1.887 moles
Mass of water = 4.456 kg
Molality =.?
Molality = mole /mass (kg) of water
Molality =1.887/4.456
Molality = 0.423 m
Therefore, the molality of the solution is 0.423 m
At a constant temperature, a sample of a gas in a balloon that originally had a volume of 5.00 L and pressure of 626 torr has its volume changed to 6.72 L. Calculate the new pressure in torr.
Answer:
466 torr
Explanation:
Step 1: Given data
Initial pressure (P₁): 626 torrInitial volume (V₁): 5.00 LFinal pressure (P₂): ?Final volume (V₂): 6.72 LConstant temperatureStep 2: Calculate the final pressure
Since we have a gas changing at a constant temperature, we can calculate the final pressure using Boyle's law.
P₁ × V₁ = P₂ × V₂
P₂ = P₁ × V₁ / V₂
P₂ = 626 torr × 5.00 L / 6.72 L
P₂ = 466 torr
Which of the following is required for the flow of current in all systems?
a) the presence of ions
b) an electrical potential ofo
c) a closed circuit
d) a short circuit
Answer:
I would say c) a closed circuit.
Hope I was right.
How has the work of chemists affected the environment over the years?
Answer:
Chemistry is one of the causes for global warming, and in some cases it can even cause certain illnesses.
Answer:
Chemists have both hurt the environment and helped the environment by their actions.
Explanation:
<3
Fireworks are chemical reactions that release energy. Which of these phenomena are caused by chemical reactions that release energy? If you’re not sure, make a guess.
Answer:
All chemical reactions involve energy. Energy is used to break bonds in reactants, and energy is released when new bonds form in products. Endothermic reactions absorb energy, and exothermic reactions release energy. The law of conservation of energy states that matter cannot be created or destroyed.
g . Calculate the molar concentration for each of the following solutions. (a) 1.50 g NaCl in 100.0 mL of solution (b) 1.50 g K2Cr2O7 in 100.0 mL of solution (c) 5.55 g CaCl2 in 125 mL of solution (d) 5.55 g Na2SO4 in 125 mL of solution
Answer:
(a) [tex]M=0.257M[/tex]
(b) [tex]M=0.0510M[/tex]
(c) [tex]M =0.500M[/tex]
(d) [tex]M= 0.391M[/tex]
Explanation:
Hello,
In this case, since the molarity or molar concentration of a solution is computed by dividing the moles of solute by the volume of solution in liters, we proceed as follows:
(a) The molar mass of sodium chloride is 58.45 g/mol and the volume in liters is 0.100 L, therefore, the molarity is:
[tex]M=\frac{1.50gNaCl}{0.100L} *\frac{1molNaCl}{58.45gNaCl} =0.257M[/tex]
(b) The molar of potassium dichromate is 294.2 g/mol and the volume in liters is 0.100 L, therefore, the molarity is:
[tex]M=\frac{1.50gK_2Cr_2O_7}{0.100L} *\frac{1molK_2Cr_2O_7}{294.2gK_2Cr_2O_7} =0.0510M[/tex]
(c) The molar of calcium chloride is 111 g/mol and the volume in liters is 0.125 L, therefore, the molarity is:
[tex]M=\frac{5.55gCaCl_2}{0.100L} *\frac{1molCaCl_2}{111gCaCl_2} =0.500M[/tex]
(d) The molar of sodium sulfate is 142 g/mol and the volume in liters is 0.125 L, therefore, the molarity is:
[tex]M=\frac{5.55gNa_2SO_4}{0.100L} *\frac{1molNa_2SO_4}{142gNa_2SO_4} = 0.391M[/tex]
Best regards.
A solution containing a unknown ionic compound, vigorously bubbles when hydrochloric acid (HCl) is added to the solution. This might indicate that the solution contains which anion?
Answer:
CO3^2-
Explanation:
In qualitative analysis, we try to use chemical reactions to determine the composition of an unknown substance. The addition of certain reagents to the unknown solution gives certain results that show the presence or absence of certain species from the unknown sample.
When dilute HCl is added to an unknown sample and effervescence is observed, then the unknown sample must contain CO3^2- or HCO3^-. The presence of these species is confirmed if the gas evolved is passed through limewater and the gas turns limewater milky.
Draw a picture of what you imagine solid sodium chloride looks like at the atomic level. (Do NOT draw Lewis structures.) Make sure to include a key. Then describe what you've drawn and any assumptions you are making.
Answer:
Kindly check the explanation section.
Explanation:
PS: kindly check the attachment below for the required diagram that is the diagram showing solid sodium chloride looks like at the atomic level.
The chemical compound known as sodium chloride, NaCl has Molar mass: 58.44 g/mol, Melting point: 801 °C and
Boiling point: 1,465 °C. The structure of the solid sodium chloride is FACE CENTRED CUBIC STRUCTURE. Also, solid sodium chloride has a coordination number of 6: 6.
In the diagram below, the positive sign shows the sodium ion while the thick full stop sign represent the chlorine ion.
The NaCl has been the ionic structure with an equal number of sodium and chlorine ions bonded.
In the structure, there has been each Na ion bonded with the Cl ions. There has been the transfer of electrons between the structure in order to attain a stable configuration.
The expected structure of the NaCl would be the image attached below.
The image has been the cubic structure of NaCl. With the presence of Na ions at the vertex of the structure, there has been the presence of the Cl ion with every Na ion for the electron transfer.
For more information about the structure of NaCl, refer to the link:
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At 2000°C the equilibrium constant for the reaction 9_1.gif is 9_2.gif If the initial concentration of 9_3.gif is 0.200 M, what are the equilibrium concentrations of 9_4.gif and 9_5.gif?
Answer:
[tex][N_2]_{eq}=[H_2]_{eq}=0.09899M[/tex]
[tex][NO]_{eq}=0.00202M[/tex]
Explanation:
Hello,
In this case, for the given chemical reaction:
[tex]2NO\rightleftharpoons N_2+O_2[/tex]
We know the equilibrium constant and equilibrium expression:
[tex]Kc=2.4x10^3=\frac{[N_2][O_2]}{[NO]^2}[/tex]
That in terms of the reaction extent [tex]x[/tex] (ICE procedure) we can write:
[tex]2.4x10^3=\frac{x*x}{(0.2M-2*x)^2}[/tex]
In such a way, solving for [tex]x[/tex] by using a quadratic equation or solver, we obtain:
[tex]x_1=0.09899M\\x_2=0.1010M[/tex]
Clearly the solution is 0.09899M since the other value will result in a negative equilibrium concentration of NO. In such a way, the equilibrium concentrations of all the species are:
[tex][N_2]_{eq}=[H_2]_{eq}=x=0.09899M[/tex]
[tex][NO]_{eq}=0.2M-2*0.09899M=0.00202M[/tex]
Regards.
A piece of plastic sinks in oil but floats in water. Place these three substances in order from lowest density to greatest density.
Answer:
[tex]\rho _{oil}<\rho _{plastic}<\rho _{water}[/tex]
Explanation:
Hello,
In this case, since water and oil are immiscible due to the oil's nonpolarity and water's polarity, when mixed, the oil remains on the water since it is less dense than water. In such a way, for a plastic sunk in the oil and floating on the water (in middle of them) we can conclude that the plastic have a mid density, therefore, the required organization is:
[tex]\rho _{oil}<\rho _{plastic}<\rho _{water}[/tex]
Best regards.
According to the following reaction, how many moles of ammonia will be formed upon the complete reaction of 31.2 grams of nitrogen gas with excess hydrogen gas ? nitrogen(g) + hydrogen(g) ammonia(g)
Answer:
4.46 mol of NH3
Explanation:
The equation of he reaction is given as;
2N + 3H2 --> 2NH3
From the stochiometry of the reaction, 1 mol of Nitrogen produces 2 mol of Ammonia.
Mass of Nitrogen = 31.2g
Molar mass of Nitrogen = 14g/mol
Number of moles = Mass / Molar mass = 31.2 / 14 = 2.23 mol
Since 1 mol of N = 2 mol of NH3;
2.23 mol of N2 would produce x
x = 2.23 * 2 = 4.46 mol of NH3
4. A 0.100 M solution of NaOH is used to titrate an HCl solution of unknown concentration. To neutralize the solution, an average volume of
the titrant was 38.2 mL. The starting volume of the HCI solution was 20 ml. What's the concentration of the HCI?
O A.0.284 M
B. 3.34 M
C. 0.191 M
D. 0.788 M
Answer:
C. 0.191 M
Explanation:
Our goal for this question, is to calculate the concentration of the HCl solution. For this, in the experiment, a solution of NaOH was used to find the moles of HCl. Therefore, our first step is to know the reaction between HCl and NaOH:
[tex]HCl~+~NaOH~->~NaCl~+~H_2O[/tex]
The "titrant" in this case is the NaOH solution. If we know the concentration of NaOH (0.100M) and the volume of NaOH (38.2 mL=0.0382 L), we can calculate the moles using the molarity equation:
[tex]M=\frac{mol}{L}[/tex]
[tex]0.100~M=\frac{mol}{0.0382~L}[/tex]
[tex]mol=0.100~M*0.0382~L=0.0382~mol~of~NaOH[/tex]
Now, in the reaction, we have a 1:1 molar ratio between HCl and NaOH (1 mol of HCl is consumed for each mole of NaOH added). Therefore we will have the same amount of moles of HCl in the solution:
[tex]0.0382~mol~of~NaOH\frac{1~mol~HCl}{1~mol~NaOH}=0.0382~mol~HCl[/tex]
If we want to calculate the molarity of the HCl solution we have to divide by the litters of HCl used in the experiment (20 mL= 0.02 L):
[tex]\frac{0.0382~mol~HCl}{0.02~L}~=~0.191~M[/tex]
The concentration of the HCl solution is 0.191 M
I hope it helps!
Calculate the entropy change in the surroundings associated with this reaction occurring at 25∘C. Express the entropy change to three significant figures and include the appropriate units.
Answer:
That means that if you are calculating entropy change, you must multiply the enthalpy change value by 1000. So if, say, you have an enthalpy change of -92.2 kJ mol-1, the value you must put into the equation is -92200 J mol-1
The entropy change in the surroundings associated with this reaction occurring at 25 degree C is calculated as ΔS = -ΔH/T J/K.
What is entropy?Entropy is a quantity which gives idea about the randomness or arrangement of atoms or molecules present in any sample.
Entropy change will be calculated as:
ΔS = -ΔH/T, where
ΔH = chnage in enthalpy (J/mole)
T = temperature (K)
So to calculate the entropy change first we have to know about the value of enthalpy in joules and then divide it by the temperature.
Hence the unit of entropy is joule per kelvin.
To know more about entropy, visit the below link:
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What is the edge length of a face-centered cubic unit cell that is made of of atoms, each with a radius of 154 pm
Answer:
The edge length of a face-centered cubic unit cell is 435.6 pm.
Explanation:
In a face-centered cubic unit cell, each of the eight corners is occupied by one atom and each of the six faces is occupied by a single atom.
Hence, the number of atoms in an FCC unit cell is:
[tex] 8*\frac{1}{8} + 6*\frac{1}{2} = 4 atoms [/tex]
In a face-centered cubic unit cell, to find the edge length we need to use Pythagorean Theorem:
[tex] a^{2} + a^{2} = (4R)^{2} [/tex] (1)
Where:
a: is the edge length
R: is the radius of each atom = 154 pm
By solving equation (1) for "a" we have:
[tex] a = 2R\sqrt{2} = 2*154 pm*\sqrt{2} = 435.6 pm [/tex]
Therefore, the edge length of a face-centered cubic unit cell is 435.6 pm.
I hope it helps you!
A student is using a coffee-cup calorimeter to determine the enthalpy change of the endothermic reaction of two aqueous solutions. After both solutions are added to the cup, the student neglects to put the lid on the cup. This would cause the magnitude of the calculated ΔH° value to be: the answer is: too small, since the solution will absorb heat from the room. But why? Wouldn't depend on if the reaction releases or absorbs heat. Wouldn't it be too large because heat escapes the cup? I'm so confused
Answer:
Explanation:
In all calorimetric experiment , the calorimeter must be isolated from the surrounding . Otherwise the heat change in the experiment can not be determined with precision .
The reaction is endothermic . Hence, there is lowering of temperature due to absorption of heat in the reaction equal to ΔH°. The value of ΔH° can be calculated by measuring fall in the temperature of the content . The fall in the temperature will be less when heat is allowed to come from the surrounding . Less fall of temperature will result in less ΔH° to be calculated .
Hence in the given experiment , if the student neglects to put lid on the cup , the experiment will give less value of ΔH°.
An atom of 122In has a mass of 121.910280 amu. Calculate the binding energy in MeV per NUCLEON. Use the masses: mass of 1H atom
The given question is incomplete, the complete question is:
An atom of 122In has a mass of 121.910280 amu. Calculate the binding energy in MeV per atom. Use the masses: mass of 1H atom = 1.007825 amu mass of a neutron = 1.008665 amu 1 amu = 931.5 MeV
Answer:
The correct answer is 1029.95 MeV.
Explanation:
Based on the given information,
The mass of proton is 1.007825, the mass of neutron is 1.008665.
The atomic number of Indium is 49, therefore, the number of neutrons will be, 122-49 = 73.
Now the calculated mass will be,
= 49 * 1.007825 + 73 * 1.008665
= 49.383425 + 73.632545
= 123.01597
The mass defect is calculated by subtracting the actual mas from calculated mass,
Mass defect = Actual mass - Calculated mass
= 121.910280 - 123.01597
= -1.10569 amu
Now the binding energy will be,
Binding energy = 1.10569 * 931.5 MeV
= 1029.95 MeV
What is the mass number of an element
Answer:
A (Atomic mass number or Nucleon number)
Explanation:
The mass number is the total number of protons and nucleons in an atomic nucleus.
Hope this helps.
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