Answer:
1) increases
2) smaller
Explanation:
Generally, as electron- electron repulsion increases and more electrons are added to the atom while Z is held constant, the electron cloud size is increased. The size of the anion formed is usually measured as the size of this extended electron cloud. Hence the larger electron cloud means a larger anion size compared to the size of the neutral atom.
For a cation, the converse is true and the cation is found to be smaller than the neutral atom.
This question is incomplete, I got the complete one from google as below:
I−>I>I+
1. As Z stays constant and the number of electrons increases, the electron-electron repulsions ______ (increases or decreases), and the anion becomes larger.
2. The reverse is true for the cation, which becomes ____ (smaller or larger) than the neutral atom.
Answer:
1. As Z stays constant and the number of electrons increases, the electron-electron repulsion increases, and the anion becomes larger.
2.The reverse is true for the cation, which becomes smaller than the neutral atom.
Explanation:
1. As Z stays constant and the number of electrons increases, the electron-electron repulsion increases, and the anion becomes larger. This is because in anions of the same atoms, the net force of attraction on electrons decreases.
2. The reverse is true for the cation, which becomes smaller than the neutral atom. This is because in cations of the same atoms, the net force of attraction on electrons increases.
A cell is set up with copper and lead electrodes in contact with CuSO4(aq) and Pb(NO3)2(aq), respectively, at 25°C. The standard reduction potentials are Pb2+ + 2e–? Pb E° = –0.13 V Cu2+ + 2e–? Cu E° = +0.34 V
If sulfuric acid is added to the Pb(NO3)2 solution, forming a precipitate of PbSO4, the cell potential
a. It is impossible to tell what will happen.
b. decreases.
c. is unchanged.
d. increases.
Answer:
increases
Explanation:
Now look at the matter closely. Given the values of electrode potential stated in the question, one can see that lead will function as the anode and copper as the cathode since lead has a more negative electrode potential.
This implies that at the anode, the half reaction going on is this;
Pb(s) -------> Pb^2+(aq) + 2e.
There will be a build up of Pb^2+ in the anode compartment. The addition of H2SO4 and formation of PbSO4 favours the removal of the Pb^2+ ions;
Pb^2+(aq) + SO4^2-(aq) -------> PbSO4(s)
As this continues, Pb^2+ concentration begins to decrease, in order to maintain equilibrium, more Pb^2+ is formed thereby increasing the current and voltage flowing in the cell as more electrons are transferred from anode to cathode(more current flows).
In the case when the sulfuric acid is added to the Pb(NO3)2 solution increases so the cell potential is increased.
Impact on cell potential:Here the anode and copper treated as the cathode because it contains more negative electrode potential. Due to this, it should be half reaction
Pb(s) -------> Pb^2+(aq) + 2e.
Also, there should be the addition of H2SO4 and creation of the PbSO4 that eliminates the Pb^2+ ions.
So,
Pb^2+(aq) + SO4^2-(aq) -------> PbSO4(s)
Here the concentration of Pb^2+ should reduced for maintaining the equilibrium.
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The reaction between sucrose and water to produce fructose and glucose is first order // overall. The data below shows the change in concentration of sucrose over time at 298 K. C12H22011 (aq)sucrose + H2O(l)艹C6H1206(aq)fructose + C6H1206(aq)oucose C12H2201l M Time (minutes) 1.002 0.808 0.630 0.0 60.0 130.0 a. Find the average rate ofdisappearance of C12H22011 from t = 0 min to t-60 min. b. Find the average rate of appearance of fructose from t 0 min to t 60 min. c. Calculate the rate constant, k, for the decomposition of sucrose at 25°C. Include units. d. How long will it take for the concentration of sucrose to drop from 1.002 to 0.212 M? e. What is the half-life for the decomposition of sucrose at 25°C?
Answer:
(a)
Rate of appearance of sucrose = - d[C12H22O11] / dt = - ( 0.808 - 1.002 ) / ( 60.0 - 0.0) = 0.00323 M/s
(b)
Rate of appearance of fructose = d[C6H12O6] / dt = (1.002 - 0.808) / (60.0 - 0.0) = 0.00323 M/s
(c)
k = (1 / t ) * ln[A]/[A]t
k = ( 1 / 60.0 ) * ln[1.002 / 0.808]
k = 0.00359 min-1
(d)
0.00359 = ( 1 / t ) * ln[1.002 / 0.212]
t = 432.6 min
(e)
Half life time = 0.693 / k = 0.693 / 0.00359 = 193 min
Explanation:
First-order reactions are defined as the chemical reactions in which rate of the reaction is linearly dependent on the concentration of only one reactant.
The answers can be explained as:
(a) Rate of appearance of the sucrose from the chemical reaction is:
Rate = [tex]\dfrac{\text d [\text C_{12}\text H_{22}\text O_{11}]}{\text {dt}}[/tex] = [tex]\dfrac{0.808 - 1.002}{60.0 -0.0}[/tex]
Rate = 0.00323 m/s
(b) Rate of appearance of Fructose from the given chemical reaction is:
Rate = [tex]\dfrac{\text d [\text C_{6}\text H_{12}\text O_{6}]}{\text {dt}}[/tex] = [tex]\dfrac{1.002 - 0.808 }{60.0 -0.0}[/tex]
Rate = 0.00323 m/s
(C) Rate constant for the reaction is:
[tex]\text k &= \dfrac{1}{\text t}\times \dfrac {\text{ln [A]}}{\text {[A]} \text t}[/tex]
[tex]\text k &= \dfrac{1}{60}\times \dfrac {\text{ln} (1.002)}{(0.808)}[/tex]
k = 0.00359 minute⁻¹
(d) Time required for the concentration of sucrose to drop from 1.002 to 0.212 M is:
[tex]0.00359 &= \dfrac{1}{t} \times {\text{ln}\dfrac{[1.002]}{[0.212]}[/tex]
t = 432.6 minutes
(e) The half-life of the decomposition of sucrose at 25°C is:
Half-life = [tex]\dfrac{0.693}{\text k} = \dfrac{0.693}{0.00359}[/tex]
Half-life = 193 minutes.
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True False The colors emitted depends on the number of free electrons passing through the lamp. True False When a free electron hits an atom, the atom is always excited to the highest energy level possible. True False The kinetic energy of the free electron at the point of collision increases as the voltage of the battery increases. True False The kinetic energy of the free electron at the point of collision is higher if the atom is closer to the source of electrons. True False The only way to emit IR photons is if there are empty electronic energy levels really close to the ground state (lowest energy level). True False When atomic electrons are excited to a higher level, they always return to their lowest energy level by jumping down one level at a time.
Answer:
False
False
True
False
False
False
Explanation:
The colour emitted by an excited atom depends on the emission Spectra of the atom rather than on the number of free electrons passing through the lamp.
When a free electrons hits an atom, the atom can be excited to various intermediate states other than the highest energy level depending on the amount of energy it absorbed.
Increase in voltage of the battery also increases the kinetic energy of the electron
The kinetic energy of the electron depends on the voltage difference not on the distance from the atom.
The Paschen, Brackett, and Pfund series of lines are due to transitions from higher-energy orbits to orbits with n = 3, 4, and 5, respectively; these transitions release substantially less energy, corresponding to infrared radiation.
Excited atoms can jump from a higher level to the ground state in a series of steps or directly to the ground state
17
¿Cuál de las siguientes
propiedades distingue una solución
sobresaturada de una diluida?
Answer:
Which of the following
properties distinguishes a solution
oversaturated with a dilute?
The supersaturated solution is one in which the solvent has dissolved more solute than it can dissolve in the saturation equilibrium. The solute can be a solid, or a gas. The molecules of the solvent surround those of the solute and seek to open space between themselves to be able to harbor more amount of solute.A dilute solution is a solution that has not reached the maximum concentration of solute dissolved in a solvent. The additional solute will dissolve when added in a dilute solution and will not appear in the aqueous phase. It is considered a state of dynamic equilibrium where the speeds in which the solvent dissolves the solute are greater than the recrystallization rate.A geochemist in the field takes a sample of water from a rock pool lined with crystals of a certain mineral compound . He notes the temperature of the pool, , and caps the sample carefully. Back in the lab, the geochemist first dilutes the sample with distilled water to . Then he filters it and evaporates all the water under vacuum. Crystals of are left behind. The researcher washes, dries and weighs the crystals. They weigh 3.00 g.
a. Using only the information above, can you calculate the solubility of X in water at 26 degrees Celsius.?
b. If yes than calculate the solubility of X. Round your answer to 3 significant digits
Answer:
Check the explanation
Explanation:
Kindly check the attached image below to see the step by step explanation to the question above.
Radiation detectors use which of the following properties to detect radioactivity?
○ charge
○ ionizing
○ mass
○ energy
Answer:
ionizing
I hope I helped :)
Which of these statements indicates an experiment is systematic?
A. The experiment is repeated many times.
B. The experiment is about a testable question.
C. The experiment is very complicated and uses lots of chemicals.
D. The experiment includes a good plan for what measurements to take.
Answer:
D. The experiment includes a good plan for what measurements to take.
Explanation:
a gas occupies 34.5 liters at a pressure of 345 torr what will be the volume of this gas if the pressure becomes 760 torr
Answer:
THE NEW PRESSURE IS 15.5 L.
Explanation:
Using Boyle's law which states that the pressure of a given mass of gas is inversely proportional to the volume of the gas at constant temperature.
Mathematically,
P1V1 =P2V2
P1 = 345 torr = 345/760 atm = 0.45 atm
P2 = 760 torr = 760/760 atm = 1 atm
V1 = 34.5 L
V2 = ?
Rearranging the variables, making V2 the subject of the formula, we obtain:
V2 = P2 V1 / P1
V2 = 1 * 34.5 / 0.45
V2 = 15.525 Litres
The volume of the gas when the pressure becomes 760 torr or 1 atm is 15.5 Litres
Determine the amount of energy required to boil 50 g of
ethanol.
Answer:
42050 J.
Explanation:
Data obtained from the question:
Mass (M) of ethanol = 50g
Heat of vaporisation (ΔHv) of ethanol = 841 J/g
Heat (Q) =.?
The heat required to boil 50g of ethanol can be obtained as follow:
Q = MΔHv
Q = 50 x 841
Q = 42050 J.
Therefore, the heat required to boil 50g of ethanol is 42050 J.
The isothermal, first-order reaction of gaseous A occurs within the pores of a spherical catalyst pellet. The reactant concentration halfway between the external surface and the center of the pellet is equal to one-fourth the concentration at the external surface. (a) What is the relative concentration of A near the center of the pellet? (b) By what fraction should the pellet diameter be reduced to give an effectiveness factor of 0.77
Answer:
Check the explanation
Explanation:
Kindly check the attached image below to see the step by step explanation to the question above.
A balloon at 30.0°C has a volume of 222 mL. If the temperature is
increased to 53.1°C and the pressure remains constant, what will the new volume be, in ml?
Answer : The new volume will be, 238.9 mL
Explanation :
Charles' Law : It states that volume of the gas is directly proportional to the temperature of the gas at constant pressure and number of moles of gas.
Mathematically,
[tex]\frac{V_1}{T_1}=\frac{V_2}{T_2}[/tex]
where,
[tex]V_1\text{ and }T_1[/tex] are the initial volume and temperature of the gas.
[tex]V_2\text{ and }T_2[/tex] are the final volume and temperature of the gas.
We are given:
[tex]V_1=222mL\\T_1=30.0^oC=(30.0+273)K=303K\\V_2=?\\T_2=53.1^oC=(53.1+273)K=326.1K[/tex]
Putting values in above equation, we get:
[tex]\frac{222mL}{303K}=\frac{V_2}{326.1K}\\\\V_2=238.9mL[/tex]
Therefore, the new volume will be, 238.9 mL
At pH0 the more thermodynamically stable species is ______ and the most strongly oxidizing species is _______ . At pH14 the more thermodynamically stable species is _______ and the most strongly oxidizing species is ______ . At pH0 ______ will tend to disproportionate and pH14 ________will tend to disproportionate. Under acidic conditions mixing Mn3+(aq) and MnO42- (aq) will result in the formation of______.
Answer:
At pH0 the more thermodynamically stable species is lower and the most strongly oxidizing species is higher. At pH14 the more thermodynamically stable species is higher and the most strongly oxidizing species is lower . At pH0 convex curve will tend to disproportionate and pH14 concave curce will tend to disproportionate. Under acidic conditions mixing Mn3+(aq) and MnO42- (aq) will result in the formation of redox oxidation-reduction reaction.
1. A solution is a mixture. The part of
the solution in the greater amount is
called the
Answer:
Solvent
Explanation:
Many people use the solute to describe the solid being dissolved and the solvent to describe the thing doing the dissolving, but really solvent means the part of the solution with a greater amount/concentration.
For example, if you have 1 gram of salt in 10 liters of water, the water is the solvent.
Final answer:
A solution is a homogeneous mixture where the component with the largest quantity is known as the solvent, and the substance in the smaller quantity is the solute.
Explanation:
A solution is a homogeneous mixture of two or more pure substances. When we talk about a solution, we refer to its components as the solvent and the solute. The solvent is the part of the solution that is present in the greater amount and acts as the medium in which the other substance, or substances, are dispersed.
The solute, on the other hand, is any component of a solution that is not the solvent. In other words, it is the material present in the smaller amounts within the solution. When a solute is dissolved in a solvent, the resulting mixture is uniform on a microscopic level, ensuring that the composition is consistent throughout.
Solutions can come in various forms and are not limited to liquids; for instance, air is a solution that is in a gaseous state. It is important to note that the largest component by mole fraction is always considered the solvent, and this remains the same whether the components are in solid, liquid, or gaseous form.
A 100g sample of Carbon-14 has a half-life of 5 years. How much Carbon-14 is left after 10 years?
Answer: The carbon-14 left after 10 years is 25 g
Explanation:
Expression for rate law for first order kinetics is given by:
[tex]t=\frac{2.303}{k}\log\frac{a}{a-x}[/tex]
where,
k = rate constant
t = age of sample
a = let initial amount of the reactant
a - x = amount left after decay process
a) for rate constant
[tex]t_{\frac{1}{2}}=\frac{0.693}{k}[/tex]
[tex]k=\frac{0.693}{5years}=0.139years^{-1}[/tex]
b) for amount left after 10 years
[tex]t=\frac{2.303}{k}\log\frac{a}{(a-x)}[/tex]
[tex]10=\frac{2.303}{0.139}\log\frac{100}{(a-x)}[/tex]
[tex](a-x)=25g[/tex]
Thus carbon-14 left after 10 years is 25 g
Given the following reaction:
\ce{2KClO3 -> 2KCl + 3O2}2KClOX
3
2KCl+3OX
2
How many moles of \ce{KCl}KClK, C, l will be produced from 15.0 \text{ g}15.0 g15, point, 0, start text, space, g, end text of \ce{KClO3}KClOX
3
?
moles (round to three significant figures)
Answer:
0.122 moles
Explanation:
Final answer:
0.122 moles of KCl will be produced from the reaction of 15.0 g of KClO3, using stoichiometry and the molar mass of KClO3 which is 122.55 g/mol.
Explanation:
To determine the amount of KCl produced from 15.0 g of KClO3, we first need to know the molar mass of KClO3, which is 122.55 g/mol according to LibreTexts. By using stoichiometry, we can find out how many moles of KClO3 we have and, subsequently, how many moles of KCl will be produced according to the balanced chemical equation.
First, calculate the number of moles of KClO3:
15.0 g KClO3 × (1 mol KClO3 / 122.55 g) = 0.122 mol KClO3
From the balanced equation Î{2KClO3 -> 2KCl + 3O2}, we can see that 2 moles of KClO3 produce 2 moles of KCl. Therefore, the number of moles of KCl produced is equal to the number of moles of KClO3 available:
0.122 mol KClO3 × (2 mol KCl / 2 mol KClO3) = 0.122 mol KCl
Thus, 0.122 mol of KCl are produced from the reaction of 15.0 g of KClO3.
Which of these substances is the least acidic? solution A, pOH = 1.5 solution B, pOH = 7.0 solution C, pOH = 13.5
1. A cylinder of gas has a pressure of 4.40 atm at 25°C. At what temperature in °C
will it reach a pressure of 6.50 atm? Answer 167°C.
Answer: The final temperature in °C is 167
Explanation:
To calculate the final temperature of the system, we use the equation given by Gay lussac's Law. This law states that pressure of the gas is directly proportional to the temperature of the gas at constant volume.
Mathematically,
[tex]\frac{P_1}{T_1}=\frac{P_2}{T_2}[/tex]
where,
[tex]P_1\text{ and }T_1[/tex] are the initial pressure and temperature of the gas.
[tex]P_2\text{ and }T_2[/tex] are the final pressure and temperature of the gas.
We are given:
[tex]P_1=4.40atm\\T_1=25^oC=(25+273)K=298K\\P_2=6.50atm\\T_2=?[/tex]
Putting values in above equation, we get:
[tex]\frac{4.40}{298}=\frac{6.50}{T_2}\\\\T_2=440K=(440-273)^0C=167^0C[/tex]
Thus the final temperature in °C is 167
To determine at what temperature a gas cylinder will reach 6.50 atm pressure from 4.40 atm at 25°C, we use Gay-Lussac's Law and find that the temperature is 167°C.
We can solve this problem using Gay-Lussac's Law, which states that for a given mass and constant volume of an ideal gas, the pressure is directly proportional to its temperature.
The formula for Gay-Lussac's Law is:
P1/T1 = P2/T2
Where:
P1 is the initial pressureT1 is the initial temperature in KelvinP2 is the final pressureT2 is the final temperature in KelvinFirst, we need to convert the initial temperature into Kelvin:
T1 (in Kelvin) = 25°C + 273 = 298 K
Now, we rearrange the equation to solve for T2:
T2 = (P2 × T1) / P1
Substitute the known values into the formula:
T2 = (6.50 atm × 298 K) / 4.40 atm
T2 ≈ 440 K
Finally, convert the temperature back to Celsius:
T2 (in °C) = 440 K - 273 = 167°C
Therefore, the temperature at which the gas will reach a pressure of 6.50 atm is 167°C.
Which of the following is a property of a base?
A
taste sour
(B
slippery feel
(c produces hydrogen gas when reacting with a metal
D
produces carbon dioxide when reacting with carbonates
Answer:
produces carbon dioxide when reacting with carbonates
The magnesium and calcium ions present in seawater ([Mg2+] = 0.059 M and [Ca2+] = 0.011 M) can be separated by selective precipitation with KOH.
If the concentration of Mg2+ in the solution were 0.039 M, what minimum [OH−] triggers precipitation of the Mg2+ ion? (Ksp=2.06×10−13.)
Answer:
the minimum [OH−] triggers precipitation = 2.3x10^-6 M
Explanation:
Mg2+ + 2OH- --> Mg(OH)2(s)
Ksp = [Mg2+][OH-]^2
(2.06x10^-13) = (0.039*x^2)
x = [OH-] = 2.3*10^-6 M
the minimum [OH−] triggers precipitation = 2.3x10^-6 M
Final answer:
The minimum [OH-] that triggers the precipitation of Mg2+ ion can be calculated using the Ksp value.
Explanation:
The minimum [OH-] that triggers the precipitation of Mg2+ ion can be calculated using the Ksp value. Given that the concentration of Mg2+ in the solution is 0.039 M, and the Ksp value is 2.06×10⁻¹³, the minimum [OH-] can be determined to initiate precipitation.
The elementary gas-phase reaction takes place isobarically and isothermally in a PFR where 63.2% conversion is achieved. The feed is pure A. It is proposed to put a CSTR of equal volume upstream of the PFR. Based on the entering molar flow rate to A to the first reactor, what will be the intermediate from the CSTR, X 1 , and exit conversion from the PFR, X 2 , based on the feed to first reactor?
Answer:
Explanation:
check below for explanation in the attached files.
when 1564 J of heat energy is added to a sample of gold at 25.0°C, the temperature of the gold increases to 449°C. what is the means of the gold sample? (specific heat of gold= 0.129 J/g-°C
Answer:
The mass of the gold sample is 28.59 g
Explanation:
Calorimetry is the measurement and calculation of the amounts of heat exchanged by a body or a system. The amount of heat received or transferred by a body when it undergoes a temperature variation (Δt) without changing its physical state (solid, liquid or gaseous) is calculated by:
Q = c * m * ΔT
Where Q is the heat exchanged by a body of mass m, made up of a specific heat substance c and where ΔT is the temperature variation.
In this case:
Q= 1564 Jc= 0.129 [tex]\frac{J}{g*C}[/tex]m=?ΔT= Tfinal - Tinitial= 449°C - 25°C= 424 °CReplacing:
[tex]1564 J=0.129\frac{J}{g*C}*m*424 C[/tex]
Solving:
[tex]m=\frac{1564 J}{0.129\frac{J}{g*C}*424 C}[/tex]
m=28.59 g
The mass of the gold sample is 28.59 g
Calculate the percent oxygen in sodium oxide.
Answer:
% is 51.6% of Oxygen
Explanation:
According to the working in the photo
51.6% is the percent of oxygen in sodium oxide. It represents the amount of solute in a given amount of solution.
What is mass percentage?A concentration is expressed as a mass percent. Furthermore, it defines the component of a particular blend. The solution component can be expressed in terms of mass percentage. It represents the amount of solute in a given amount of solution.
The amount of solute is measured in mass or moles. We shall study the mass percent equation using numerous solved numerical examples in this post. Mass is often represented in grams, although any unit of measurement is acceptable as long as an equivalent unit is used both for the component or solute mass and thus the total or solution mass.
mass percentage= (mass of oxygen / mass of sodium oxide)×100
=(32/62)×100
= 51.6%
Therefore, 51.6% is the percent of oxygen in sodium oxide.
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Write a balanced chemical equation for the standard formation reaction of solid sodium carbonate
Answer:
2Na(s) + C(s) + 3/2 O2(g) -----> Na2CO3(s).
Explanation:
Balancing and writring chemical reactions in Chemistry is a very important aspect of chemistry and,
writting and balancing of chemical reaction is often regarded as the heart of chemistry. A balanced Chemical reaction shows the number of each species in a reaction.
Hence, the balanced chemical equation for the standard formation reaction of solid sodium carbonate is given below;
(1). The reaction(unbalanced Reaction):
Na(s) + C(s) + O2(g) -----> Na2CO3(s).
Where solid sodium carbonate = Na2CO3(s).
(2). Balanced Reaction;
2Na(s) + C(s) + 3/2 O2(g) -----> Na2CO3(s)
The balanced chemical equation for the standard formation reaction of
solid sodium carbonate is shown below:
2Na(s) + C(s) + 3/2 O2(g) -----> Na2CO3(s).
The reaction involves elements which include Sodium(Na) , Carbon(C) and
Oxygen(O2). They react under certain conditions to form solid sodium
carbonate.
The reaction involves specific number of atoms involved in the reaction
which is why the equation must be balanced. The number of atoms of an
element on the left hand side should always be equal to those on the right
hand side.
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A 3-column table with 4 rows. The first column labeled Scenario has entries A, B, C, D. The second column labeled Object 1 has entries lotion at 26 degrees Celsius, lasagna at 170 degrees Celsius, ocean water at 25 degrees Celsius, eggs at 2 degrees Celsius. The third column labeled Object 2 has entries skin at 37 degrees Celsius, plate at 20 degrees Celsius, lava 950 degrees Celsius, sidewalk at 80 degrees Celsius.
Indicate the direction of heat flow in each scenario.
Scenario A:
Scenario B:
Scenario C:
Scenario D:
The direction of heat flow in each scenario will be:
Scenario A = toward object 1Scenario B = toward object 2Scenario C = toward object 1Scenario D = toward object 1What is heat transfer?It should be noted that heat transfer means the transfer of heat from a location to another.
It is the process in which the molecules are moved from a region of higher temperature to that of a lower temperature.
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Answer:
It Is Scenario D.
Explanation:
what kind of solution would have a Ka value that is much less than 1
Answer:
A weakly acidic solution.
Explanation:
Final answer:
A solution with a Ka value much less than 1 refers to a weak acid, where only a small portion of the acid is ionized, resulting in relatively low concentrations of hydronium ions and the conjugate base.
Explanation:
The type of solution that would have a Ka value much less than 1 is a weak acid solution. This is because the Ka (acid dissociation constant) value indicates the strength of the acid in solution; the lower the Ka, the weaker the acid. For a Ka value significantly less than 1, this means that only a small fraction of the acid molecules donate protons (H+) to the water, resulting in few hydronium ions (H₋₃O⁺) and the conjugate base (A-).
In contrast, strong acids have a Ka value that approaches infinity since they are almost completely ionized in solution. Acetic acid, with a Ka value of 1.75 x 10⁻⁵, is an example of a weak acid, indicating that in equilibrium, the concentration of the undissociated acetic acid is much greater than the concentrations of the acetate and hydronium ions.
A 0.0200 gram piece of unknown alkaline earth metal, M, is reacted with excess 0.500 M H 2 SO 4 , and the hydrogen gas produced is collected over water. The total gas pressure inside the collecting tube is 1.01 atm., the temperature 24.0 o C. The volume of gas collected is 19.6 mL. The gas in the tube contains water vapor, at a pressure of 0.029 atm. M(s) + H 2 SO 4 (aq) → MSO 4 (aq) + H 2 (g). What is the partial pressure of the dry hydrogen gas collected in the tube?
Answer:
0.981atm
Explanation:
According ot Dalton's law total pressure of a mixture of non-reactive gas is equal to sum of partial pressures of individual gases.
total pressure= 1.01at
Number of gases=2
Gases: water vapor and hydrogen
partial pressure of water vapor= 0.029atm
1.01= partial pressure of water vapor+ partial pressure of hydrogen
1.01= 0.029 + partial pressure of hydrogen
partial pressure of hydrogen = 0.981atm
Measurements show that the enthalpy of a mixture of gaseous reactants decreases by 162. kJ during a certain chemical reaction, which is carried out at a constant pressure. Furthermore, by carefully monitoring the volume change it is determined that -194. kJ of work is done on the mixture during the reaction. a.Calculate the change in energy of the gas mixture during the reaction. Round your answer to 3 significant digits b. Is the reaction exothermic or endothermic?
Answer:
= -356KJ
therefore, the reaction where heat is released is exothermic reaction since theΔH is negative
Explanation:
given that enthalpy of gaseous reactants decreases by 162KJ and workdone is -194KJ
then,
change in enthalpy (ΔH) = -162( released energy)
work(w) = -194KJ
change in enthalpy is said to be negative if the heat is evolved during the reaction while heat change(ΔH) is said to be positive if the heat required for the reaction occurs.
At constant pressure the change in enthalpy is given as
ΔH = ΔU + PΔV
ΔU = change in energy
ΔV = change in volume
P = pressure
w = -pΔV
therefore,
ΔH = ΔU -W
to evaluate energy change we have,
ΔU =ΔH + W
ΔU = -162+ (-194KJ)
= -356KJ
therefore, the reaction where heat is released is exothermic reaction since theΔH is negative
Final answer:
The change in energy of the gas mixture during the reaction is 32 kJ. The reaction is exothermic because it releases heat to the surroundings as indicated by the negative enthalpy change.
Explanation:
To calculate the change in energy of the gas mixture during the reaction, we use the formula for change in internal energy (ΔU) at constant pressure:
ΔU = ΔH - PΔV
where ΔH is the change in enthalpy and PΔV is the work done on or by the system. Given that the enthalpy of the system decreases by 162 kJ and the work done on the system is -194 kJ, we can substitute these values into the equation:
ΔU = -162 kJ - (-194 kJ)
ΔU = -162 kJ + 194 kJ
ΔU = 32 kJ
Therefore, the change in energy of the gas mixture during the reaction is 32 kJ. The reaction can be classified as exothermic since the enthalpy decreases (ΔH is negative), which means heat is released by the system to the surroundings.
What will the concentration of LiOH be if 27ml of a 4.5 M LiOH solution is diluted to 435ml? *
Answer: 0.28 M
Explanation:
Molarity of a solution is defined as the number of moles of solute dissolved per Liter of the solution.
According to the neutralization law,
[tex]M_1V_1=M_2V_2[/tex]
where,
[tex]M_1[/tex] = molarity of stock [tex]LiOH[/tex] solution = 4.5 ml
[tex]V_1[/tex] = volume of stock [tex]LiOH[/tex] solution = 27 ml
[tex]M_2[/tex] = molarity of dilute c solution = ?
[tex]V_2[/tex] = volume of dilute [tex]LiOH[/tex] solution = 435 ml
[tex]4.5\times 27=M_2\times 435[/tex]
[tex]M_2=0.28M[/tex]
Thus the concentration of LiOH be if 27 ml of a 4.5 M LiOH solution is diluted to 435 ml is 0.28 M
Complete the reaction, which is part of the electron transport chain. The abbreviation FMN represents flavin mononucleotide. Use the appropriate abbreviation for the product.
NADH + H^+ + FMN ⟶
The reactant that is reduced is_________ .
In complex III, electrons are transferred from coenzyme Q to cytochrome c, which contains iron.
QH2 + 2cyt c ( Fe3+) ⟶ Q + 2cyt c(Fex) + 2H^+
Determine the oxidation number for iron on the right side of the reaction arrow.
Answer:
1. NADH + H⁺ + FMN + Q ⟶ NAD⁺ + FMN + QH₂
2. The reactant that is reduced is Q
3. The charge on iron on the right side is +2, Fe²⁺
Explanation:
NADH + H⁺ + FMN + Q ⟶ NAD⁺ + FMN + QH₂
The reaction above is catalysed by NADH:ubiquinone oxidoreductase (complex 1), which transfers a hydride ion from NADH to FMN, from which two electrons pass through a series of of Fe-S centers to the iron-sulfur protein N-2. Electron transfer from N-2 to Ubiquinone forms QH₂
The species in a reaction which gains hydrogen irons is reduced, Therefore, the reactant that is reduced is Q, ubiquinone to form QH₂, ubiquinol.
To determine the oxidation number of iron on the right side of the reaction below,
QH2 + 2cyt c ( Fe3+) ⟶ Q + 2cyt c(Fex) + 2H^+
Sum of charges on the left side = Sum of charges on the right side
Sum of charges on the left side = 2 *+3 = +6
Therefore 2 * x + 2= 6
2x = 6 -2 = 4
x = 4/
x = 2
Therefore the charge on iron on the right side is +2, Fe²⁺
On January 2, 2020, Nash Company purchases a call option for $290 on Merchant common stock. The call option gives Nash the option to buy 1,050 shares of Merchant at a strike price of $51 per share. The market price of a Merchant share is $51 on January 2, 2020 (the intrinsic value is therefore $0). On March 31, 2020, the market price for Merchant stock is $54 per share, and the time value of the option is $210.
Final answer:
A call option is a financial contract that gives the buyer the right to purchase shares of a stock at a predetermined price. The net profit from this call option transaction can be calculated by determining the difference between the market price and the strike price, multiplied by the number of shares.
Explanation:
A call option is a type of financial contract that gives the buyer the right, but not the obligation, to purchase a specified number of shares of a stock at a predetermined price within a certain time frame. In this case, Nash Company purchased a call option to buy 1,050 shares of Merchant stock at a strike price of $51 per share. On March 31, 2020, the market price for Merchant stock is $54 per share, and the time value of the option is $210.
The net profit from this stock transaction can be calculated by determining the difference between the market price of the stock and the strike price, multiplied by the number of shares:
Net Profit = (Market Price - Strike Price) x Number of Shares
Using the given information, the net profit from this call option transaction would be:
Net Profit = ($54 - $51) x 1,050 = $3 x 1,050 = $3,150