Answer:
specific heat
Explanation:
please like,rate and tag as brainliest
Which reaction is an example of heterogeneous catalysis?
Answer:
Explanation:
Industrial examples
Process Reactants, Product(s)
Ammonia synthesis (Haber–Bosch process) N2 + H2, NH3
Nitric acid synthesis (Ostwald process) NH3 + O2, HNO3
Hydrogen production by Steam reforming CH4 + H2O, H2 + CO2
Ethylene oxide synthesis C2H4 + O2, C2H4O
Answer:
The answer is A - Ethene gas reacts with hydrogen gas by using a nickel catalyst.
Explanation:
just did on Edge
Let A be the last two digits of your 8-digit student ID. Example: for 20245347, A = 47 Radio waves, from your favorite radio station has a frequency of (A + 88.3) MHz (megahertz). What is the corresponding wavelength for this frequency in meters?
Answer:
2.22 m
Explanation:
Step 1:
Data obtained from the question:
Frequency = (A + 88.3) MHz
We assume that the student ID is 20245347 as given in the question.
Therefore, A = 47 (last two digit of the 8-digit student ID)
Frequency = (47 + 88.3) MHz
Frequency = 135.3 MHz = 135.3x10^6 Hz
Wavelength =?
Recall:
Velocity of electromagnetic wave is 3x10^8 m/s2
Step 2:
Determination of the wavelength of the radio wave. This is illustrated below:
Velocity = wavelength x frequency
Wavelength = Velocity /frequency
Wavelength = 3x10^8 / 135.3x10^6
Wavelength = 2.22 m
How many moles of helium gas are contained in 4.0: flask at STP
To calculate the number of moles of helium in a 4.0-liter flask at STP, divide the volume of the gas by the molar volume of a gas at STP (22.4 L/mol). This yields approximately 0.17857 moles of helium gas.
First, we need to know the volume of the flask, but since the student has not specified the volume correctly, let's assume it is '4.0 liters'. At STP, one mole of any gas occupies 22.4 liters. To find the number of moles in the flask, we use the molar volume of a gas at STP.
Here's the calculation:
Divide the volume of the gas in the flask by the molar volume of a gas at STP.[tex]\frac{4.0\ L}{ 22.4\ L/mol} = number\ of\ moles\ of\ helium.[/tex]Calculate: [tex]\frac{4.0\ L}{22.4\ L/mol} = 0.17857\ mol.[/tex]Therefore, the flask contains approximately 0.17857 moles of helium gas.
16. Metals are good conductors of electricity because they
a. form crystal lattices.
b. contain positive ions.
c. contain mobile valence electrons.
d. form ionic bonds.
Metals are good conductors of electricity due to the mobile valence electrons which can move freely within the metallic crystal lattice, facilitating electric charge transfer.
Explanation:Metals are good conductors of electricity because they contain mobile valence electrons. In metallic bonds, these valence electrons are not associated with a particular atom or pair of atoms, but move freely within the crystal lattice of positively charged metal ions. They form what is often referred to as an 'electron sea'. These free moving electrons can carry charge from one place to another when a voltage (electric potential difference) is applied, making metals good conductors of electricity.
Learn more about Electric Conduction in Metals here:https://brainly.com/question/32879067
#SPJ3
A beaker holds 962 g of a brine solution that is 6.20 percent salt. If 123g of water are evaporated from the beaker, how much salt must be added to have an 8.60 percent brine solution? How many grams of the 8.6% brine solution will be produced?
To increase the salt concentration to 8.6%, you need to add 8794g of salt to the solution. The final mass of the 8.6% brine solution will be 961g.
Explanation:To have an 8.60 percent brine solution, you would need to add salt to compensate for the loss of water due to evaporation. First, calculate the mass of water after evaporation by subtracting 123g from the initial mass of the brine solution (962g - 123g = 839g).
Then, find the mass of the salt needed by multiplying the final mass of the solution by the desired percent of salt (839g / 0.0860 = 9756g). Subtract the initial mass of the brine solution to determine the amount of salt that must be added (9756g - 962g = 8794g).
To find the mass of the 8.6% brine solution that will be produced, subtract the mass of the added salt from the final mass of the solution (9756g - 8794g = 961g).
Learn more about Salt concentration in brine solution here:https://brainly.com/question/28668523
#SPJ3
Liquid nitrogen trichloride is heated in a 2.50−L closed reaction vessel until it decomposes completely to gaseous elements. The resulting mixture exerts a pressure of 818 mmHg at 95°C. What is the partial pressure of each gas in the container?
Answer:
1. Partial pressure of N2 is 204.5 mmHg
2. Partial pressure of Cl2 is 613.5 mmHg
Explanation:
Step 1:
The equation for the reaction. This is given below:
NCl3 —> N2 (g) + Cl2 (g)
Step 2:
Balancing the equation.
NCl3 (l) —> N2 (g) + Cl2 (g)
The above equation is balanced as follow:
There are 2 atoms of N on the right side and 1 atom on the left side. It can be balance by putting 2 in front of NCl3 as shown below:
2NCl3 (l) —> N2 (g) + Cl2 (g)
There are 6 atoms of Cl on the left side and 2 atoms on the right side. It can be balance by putting 3 in front of Cl2 as shown below:
2NCl3 (l) —> N2 (g) + 3Cl2 (g)
Now the equation is balanced.
Step 2:
Determination of the mole fraction of each gas.
From the balanced equation above, the resulting mixture of the gas contains:
Mole of N2 = 1
Mole of Cl2 = 3
Total mole = 4
Therefore, the mole fraction for each gas is:
Mole fraction of N2 = mole of N2/total mole
Mole fraction of N2 = 1/4
Mole fraction of Cl2 = mole of Cl2/total mole
Mole fraction of Cl2 = 3/4
Step 3:
Determination of the partial pressure of N2.
Partial pressure = mole fraction x total pressure
Total pressure = 818 mmHg
Mole fraction of N2 = 1/4
Partial pressure of N2 = 1/4 x 818
Partial pressure of N2 = 204.5 mmHg
Step 4:
Determination of the partial pressure of Cl2.
Partial pressure = mole fraction x total pressure
Total pressure = 818 mmHg
Mole fraction of Cl2 = 3/4
Partial pressure of Cl2 = 3/4 x 818
Partial pressure of Cl2 = 613.5 mmHg
Answer:
[tex]p_{N_2}=204.5mmHg\\p_{Cl_2}=613.5mmHg[/tex]
Explanation:
Hello,
In this case, the undergoing chemical reaction is:
[tex]2NCl_3(g)\rightarrow N_2(g)+3Cl_2(g)[/tex]
Thus, by knowing that the nitrogen trichloride is completely decomposed, one assumes there is one mole of nitrogen and three moles of chlorine (stoichiometric coefficients) as a basis to compute the partial pressures since they have the mole ratio from the nitrogen trichloride. Hence, the mole fractions result:
[tex]x_{N_2}=\frac{1}{1+3}=0.25\\ x_{Cl_2}=1-0.25=0.75[/tex]
In such a way, for the final pressure 818 mmHg, the partial pressures become:
[tex]p_{N_2}=x_{N_2}p_T=0.25*818mmHg=204.5mmHg\\p_{Cl_2}=x_{Cl_2}p_T=0.75*818mmHg=613.5mmHg[/tex]
Best regards.
A hydrogen atom can be in the 1S state, whose energy we'll call 0, the 2S state, or any of 3 2P states. The 2S and 2P states have energies of 10.2 eV. There are other states with higher energy but we'll ignore them for simplicity. The 2P states have distinctive optical properties, so we're interested in how many are present even when it's a small fraction of the total?
Answer:
-- 5.8×10⁻⁹ of the H is in 2P states at T=5900 K, a typical Sun surface temperature.
-- 3.3×10⁻¹² of the H is in 2P states at T=4300 K, a typical Sunspot temperature.
Explanation:
Answer:
Explanation:
your question seems to be uncompleted , but however i have a solution to a similar question, check the attachment and follow the format to evaluate your question.
check the attachement
Why does pressure change in this way? Select all
that apply.
Answer:
A. Kinetic energies of molecules increase.
B. Speeds of molecules increase.
C. Number of collisions per second increase.
Final answer:
Pressure changes according to Le Chatelier's principle and can be influenced by mechanical and thermal mechanisms, as well as by the kinetic activity of gas molecules.
Explanation:
Pressure changes can be understood in terms of Le Chatelier's principle, which states that when a change is imposed on a system at equilibrium, the system adjusts to counteract that change.
In the case of gases, increasing the pressure (by decreasing volume) leads to a shift in the equilibrium to reduce the number of gas particles, if possible, thereby reducing the pressure. Conversely, decreasing the pressure (by increasing volume) would have the opposite effect.
Pressure also varies due to thermal and mechanical mechanisms. Heating air causes it to rise and reduce surface air pressure, while cooling air causes it to descend, increasing pressure.
Mechanical changes occur when airflow is blocked, causing a build-up of air and increased pressure. Additionally, the kinetic theory of gases suggests that gases exert pressure because of the continuous motion of their particles, colliding with container walls and exerting force.
Furthermore, variations in pressure at different points of a fluid are important for driving the phenomena of buoyancy, as well as affecting divers and airplane passengers through changes in ambient pressure.
At a certain temperature, 0.660 mol SO 3 is placed in a 4.00 L container. 2 SO 3 ( g ) − ⇀ ↽ − 2 SO 2 ( g ) + O 2 ( g ) At equilibrium, 0.110 mol O 2 is present. Calculate K c .
Answer:
[tex]Kc=6.875x10^{-3}[/tex]
Explanation:
Hello,
In this case, for the given chemical reaction at equilibrium:
[tex]2 SO_3 ( g ) \rightleftharpoons 2 SO_ 2 ( g ) + O_ 2 ( g )[/tex]
The initial concentration of sulfur trioxide is:
[tex][SO_3]_0=\frac{0.660mol}{4.00L}=0.165M[/tex]
Hence, the law of mass action to compute Kc results:
[tex]Kc=\frac{[SO_2]^2[O_2]}{[SO_3]^2}[/tex]
In such a way, in terms of the change [tex]x[/tex] due to the reaction extent, by using the ICE method, it is modified as:
[tex]Kc=\frac{(2x)^2*x}{(0.165-2x)^2}[/tex]
In that case, as at equilibrium 0.11 moles of oxygen are present, [tex]x[/tex] equals:
[tex]x=[O_2]=\frac{0.110mol}{4.00L}=0.0275M[/tex]
Therefore, the equilibrium constant finally turns out:
[tex]Kc=\frac{(2*0.0275)^2*0.0275}{(0.165-2*0.0275)^2} \\\\Kc=6.875x10^{-3}[/tex]
Best regards.
onsider two aqueous solutions of nitrous acid (HNO2). Solution A has a concentration of [HNO2]= 0.55 M and solution B has a concentration of [HNO2]= 1.55 M . You may want to reference (Page 743) Section 16.6 while completing this problem. Part A Which statement about the two solutions is true? Which statement about the two solutions is true? Solution A has the higher percent ionization and solution B has the higher pH. Solution B has the higher percent ionization and the higher pH. Solution B has the higher percent ionization and solution A has the higher pH. Solution A has the higher percent ionization and the higher pH.
Answer:
Solution A has the higher percent ionization and the higher pH.
Explanation:
Percent ionization depends on the concentration of acid in a solution. If the solution having more concentration of acid so the percent ionization will be lower while if the solution have low amount of acid i. e. dilute solution so the percent ionization will be higher. In solution A, the concentration of HNO2 is lower which is an acid so the percent ionization is higher and the pH of the solution is also higher as compared to solution B.
Answer:
Solution A has the higher percent ionization and the higher pH
Explanation:
The human body can get energy by metabolizing proteins, carbohydrates or fatty acids, depending on the circumstances. Roughly speaking, the energy it gets comes mostly from allowing all the carbon atoms in the food molecules to become oxidized to carbon dioxide by reaction with oxygen from the atmosphere. Hence the energy content of food is roughly proportional to the carbon content. Let's consider stearic acid , a fatty acid from which fats are made, and fructose , one of the simplest carbohydrates. Using the idea above about energy content, calculate the ratio of the energy the body gets metabolizing each gram of stearic acid to the energy the body gets metabolizing each gram of fructose. Round your answer to the correct number of significant digits.
The ratio of energy obtained from metabolizing stearic acid to fructose is 3:1.
Explanation:To calculate the ratio of the energy the body gets metabolizing stearic acid to the energy the body gets metabolizing fructose, we need to compare their carbon content. The molecular formula of stearic acid is C18H36O2, which means it has 18 carbon atoms. The molecular formula of fructose is C6H12O6, which means it has 6 carbon atoms. Since the energy content of food is roughly proportional to the carbon content, we can calculate the ratio by dividing the number of carbon atoms in stearic acid (18) by the number of carbon atoms in fructose (6). This gives us a ratio of 3:1.
Learn more about energy content of food here:https://brainly.com/question/32218552
#SPJ3
The energy ratio of metabolizing one gram of stearic acid (a fatty acid) to one gram of fructose (a carbohydrate) is 2.25, with stearic acid providing 9 kcal/g and fructose providing 4 kcal/g of energy.
The question asks to calculate the ratio of the energy the body gets metabolizing each gram of stearic acid to the energy the body gets metabolizing each gram of fructose. Stearic acid is a fatty acid, and fructose is a simple carbohydrate. According to the data provided, each gram of carbohydrates yields approximately 4 kcal of energy, while each gram of fat yields about 9 kcal. Therefore, the ratio of energy from fats to carbohydrates is 9 kcal/g to 4 kcal/g.
To calculate the ratio of energy from stearic acid to fructose:
Identify the energy values: 9 kcal/g for stearic acid and 4 kcal/g for fructose.
Divide the energy value of stearic acid by the energy value of fructose: 9 kcal/g \/ 4 kcal/g = 2.25.
This means that metabolizing one gram of stearic acid yields 2.25 times the energy compared to metabolizing one gram of fructose.
What is hydroponics
Answer:
Explanation:
Hydroponics is the process of growing crops using only water and liquid fertilizer. This process is great when your in the big city.
Hydroponics is a method of growing plants in a nutrient-rich water solution rather than soil, which allows precise control of nutrients and is used in research and commercial greenhouses for robust crop production.
Explanation:Hydroponics is a highly efficient farming technique where plants are grown in a water-nutrient solution, rather than in soil. This method allows for precise control over the nutritional environment of the plants, which is why it is favored in scientific research for studying plant nutrient deficiencies and for producing robust, healthy crops.
In hydroponic systems, the need for soil is eliminated, and plants are given the exact nutrients they require directly. Because of this, hydroponics is used not only in laboratories but also in commercial greenhouse environments to cultivate flowers, vegetables, and other crops.
These crops are often resilient to pests and harsh conditions, contributing to sustainable food production and agricultural development.
Greenhouse management and hydroponics go hand in hand, as many plants grown hydroponically are also cultivated under controlled climates within greenhouses. The elimination of soil in hydroponics also helps mitigate the ecological, economic, and health concerns associated with excessive pesticide use in traditional agriculture.
Tarnish on tin is the compound SnO. A tarnished tin plate is placed in an aluminum pan of boiling water. When enough salt is added so that the solution conducts electricity, the tarnish disappears. Imagine that the two halves of this redox reaction were separated and connected with a wire and a salt bridge. Part A Calculate the standard cell potential given the following standard reduction potentials: Al3++3e−→Al;E∘=−1.66 V Sn2++2e−→An;E∘=−0.140 V
Answer:
1.52V
Explanation:
Oxidation half equation:
2Al(s)−→2Al^3+(aq) + 6e
Reduction half equation
3Sn2^+(aq) + 6e−→3Sn(s)
E°cell= E°cathode - E°anode
E°cathode= −0.140 V
E°anode= −1.66 V
E°cell=-0.140-(-1.66)
E°cell= 1.52V
Final answer:
The overall cell potential for the redox reaction between tin and aluminum in a galvanic cell is calculated as +1.52 V, indicating a spontaneous reaction with tin being oxidized at the anode and aluminum reduced at the cathode.
Explanation:
To calculate the standard cell potential for a redox reaction involving tin (Sn) and aluminum (Al), we apply the reduction potentials of their respective half-reactions. The half-reaction for tin is as follows: Sn(s) ightarrow Sn2+(aq) + 2 e - , with an associated standard reduction potential (E & Ocirc ;) of - 0.140 V, however, its oxidation potential is actually +0.140 V.
For aluminum, the half-reaction is: Al3+(aq) + 3 e - ightarrow Al(s), with an E & Ocirc ; of - 1.66 V. In a galvanic cell, the aluminum will oxidize, and since it’s a reduction potential, for oxidation, we take the negative of this value, which would make it +1.66 V.
To find the overall cell potential, we use the equation Ecell = Ecathode - Eanode. In this reaction, Sn(s) is our anode, and Al(s) is our cathode. However, since aluminum's E & Ocirc ; is already negative (signifying oxidation), we reverse its sign for use in the cell potential equation.
Ecell = Ecathode - Eanode = (+1.66 V) - (+0.14 V) = +1.66 V - 0.14 V = +1.52 V
The standard cell potential is positive, indicating that the redox reaction is spontaneous. Tin is oxidized at the anode, and aluminum is reduced at the cathode, forming the basis for electric current flow in the cell.
An aqueous solution of Na2CrO4 at 25oC is slowly added to an aqueous solution containing 0.001 M Pb(NO3)2and 0.100 M Ba(NO3)2. Which solid will precipitate first? The Ksp of BaCrO4 is 1.17 × 10−10, and Ksp of PbCrO4 is 2.80 × 10−13.
Answer:
The one that will begin to precipitate first will be the lead chromate (PbCrO₄)
Explanation:
First of all, let's determine the equations involved:
Pb(NO₃)₂ → Pb²⁺ + 2NO₃⁻
0.001 0.001 0.002
Ba(NO₃)₂ → Ba²⁺ + 2NO₃⁻
0.100 0.100 0.200
Sodium chromate as a soluble salt, can be also dissociated in:
Na₂CrO₄ → 2Na⁺ + CrO₄²⁻
As the chromate can react to both cations of the aqueous solution, there will be formed 2 precipitates. When the saturation point is reached, which is determined by the Kps, everything that cannot be dissolved will precipitate.
The first to saturate the solution will precipitate first.
CrO₄²⁻ + Pb²⁺ ⇄ PbCrO₄
s s s² = Kps
Kps = s² ⇒ [CrO₄²⁻] . [Pb²⁺] = 2.80×10⁻¹³
[CrO₄²⁻] . 0.001 = 2.80×10⁻¹³
[CrO₄²⁻] = 2.80×10⁻¹³ / 0.001 = 2.80×10⁻¹⁰
This is the concentration for the chromate when the lead chromate starts to precpitate.
CrO₄²⁻ + Ba²⁺ ⇄ BaCrO₄
s s s² = Kps
Kps = [CrO₄²⁻] . [Ba²⁺]
1.17×10⁻¹⁰ = [CrO₄²⁻] . 0.100
[CrO₄²⁻] = 1.17×10⁻¹⁰ / 0.100 = 1.17×10⁻⁹
The first one that precipitates needs less chromate ion to start precipitating, in conclusion the one that will begin to precipitate first will be lead chromate.
The solid that will precipitate first is PbCrO₄.
To determine which solid will precipitate first, we need to compare the solubility product constants [tex](\( K_{\text{sp}} \))[/tex] for each possible precipitate. The compound with the lower [tex]\( K_{\text{sp}} \)[/tex] will precipitate first because it has lower solubility in water.
The solubility products given are:
- [tex]\( K_{\text{sp}} \) of BaCrO\(_4\) = \( 1.17 \times 10^{-10} \)[/tex]
- [tex]\( K_{\text{sp}} \) of PbCrO\(_4\) = \( 2.80 \times 10^{-13} \)[/tex]
We need to find the concentration of [tex]\(\text{CrO}_4^{2-}\) (\([ \text{CrO}_4^{2-} ]\))[/tex] at which each compound will begin to precipitate.
Calculation for BaCrO₄:
The reaction for the precipitation of BaCrO₄ is:
[tex]\[ \text{Ba}^{2+} (aq) + \text{CrO}_4^{2-} (aq) \rightarrow \text{BaCrO}_4 (s) \][/tex]
The [tex]\( K_{\text{sp}} \)[/tex] expression is:
[tex]\[ K_{\text{sp}} = [\text{Ba}^{2+}] [\text{CrO}_4^{2-}] \][/tex]
Given:
[tex]\[ K_{\text{sp}} (\text{BaCrO}_4) = 1.17 \times 10^{-10} \][/tex]
[tex]\[ [\text{Ba}^{2+}] = 0.100 \, \text{M} \][/tex]
We can solve for [tex]\([ \text{CrO}_4^{2-} ]\)[/tex]:
[tex]\[ 1.17 \times 10^{-10} = (0.100) [\text{CrO}_4^{2-}] \][/tex]
[tex]\[ [\text{CrO}_4^{2-}] = \frac{1.17 \times 10^{-10}}{0.100} \][/tex]
[tex]\[ [\text{CrO}_4^{2-}] = 1.17 \times 10^{-9} \, \text{M} \][/tex]
Calculation for PbCrO₄:
The reaction for the precipitation of PbCrO₄ is:
[tex]\[ \text{Pb}^{2+} (aq) + \text{CrO}_4^{2-} (aq) \rightarrow \text{PbCrO}_4 (s) \][/tex]
The [tex]\( K_{\text{sp}} \)[/tex] expression is:
[tex]\[ K_{\text{sp}} = [\text{Pb}^{2+}] [\text{CrO}_4^{2-}] \][/tex]
Given:
[tex]\[ K_{\text{sp}} (\text{PbCrO}_4) = 2.80 \times 10^{-13} \][/tex]
[tex]\[ [\text{Pb}^{2+}] = 0.001 \, \text{M} \][/tex]
We can solve for [tex]\([ \text{CrO}_4^{2-} ]\)[/tex]:
[tex]\[ 2.80 \times 10^{-13} = (0.001) [\text{CrO}_4^{2-}] \][/tex]
[tex]\[ [\text{CrO}_4^{2-}] = \frac{2.80 \times 10^{-13}}{0.001} \][/tex]
[tex]\[ [\text{CrO}_4^{2-}] = 2.80 \times 10^{-10} \, \text{M} \][/tex]
Comparison:
- The concentration of [tex]\(\text{CrO}_4^{2-}\)[/tex] needed to precipitate BaCrO₄ is [tex]\( 1.17 \times 10^{-9} \, \text{M} \)[/tex].
- The concentration of [tex]\(\text{CrO}_4^{2-}\)[/tex] needed to precipitate PbCrO₄ is [tex]\( 2.80 \times 10^{-10} \, \text{M} \)[/tex].
Since [tex]\( 2.80 \times 10^{-10} \, \text{M} \)[/tex] is smaller than[tex]\( 1.17 \times 10^{-9} \, \text{M} \), PbCrO\(_4\)[/tex] will precipitate first.
Compound A has a solubility of 0.2 g/mL in toluene at toluene's boiling point and a solubility of 0.05 g/mL at 0 ºC. How much toluene would be necessary to recrystallize 3.2 g of A. What would be the maximum amount of A that could be recovered if the saturated solution was allowed to cool to 0 ºC. How much A would be recovered, if you accidentally used twice as much toluene as was necessary?
Answer:
1) 16 mL of toluene is necessary to recrystallize 3.2 g of compound A
2) 2.4 g of A is the maximum amount that could be recovered.
3) 1.6 g of compound A can be recovered if you accidentally use twice as much toluene as was necessary
Explanation:
1)
The volume of the solvent ( toluene) = [tex]\frac{starting \ amount }{solubility \ at \boiling \ point }[/tex]
= [tex]\frac{3.2 \ g}{0.2 \ g/mL}[/tex]
= 16 mL
∴ 16 mL of toluene is necessary to recrystallize 3.2 g of compound A
2)
The maximum amount A that could be recovered if the saturated solution was allowed to cool to 0 ºC is determined by the difference of the starting amount and amount left in the solution at 0 ºC
i.e
maximum amount of A = 3.2 - ( 16 mL × 0.05 g/mL)
= 3.2 g - 0.8 g
= 2.4 g
∴ 2.4 g of A is the maximum amount that could be recovered.
3)
Amount of A that would be recovered, if you accidentally used twice as much toluene as was necessary is calculated as follows;
amount of A = 3.2g - (32 mL × 0.05 g/mL)
= 3.2 g - 1.6 g
= 1.6 g
Thus; 1.6 g of compound A can be recovered if you accidentally use twice as much toluene as was necessary
Which of the following statements is TRUE? Question 1 options: There is a "heat tax" for every energy transaction. A spontaneous reaction is always a fast reaction. The entropy of a system always decreases for a spontaneous process. Perpetual motion machines are a possibility in the near future. None of these are true.
Question:
Which of the following statements is TRUE?
A. Perpetual motion machines are a possibility in the near future.
B. The entropy of a system always decreases for a spontaneous process.
C. A spontaneous reaction is always a fast reaction.
D. There is a "heat tax" for every energy transaction.
E. None of the above are true.
Answer:
The correct answer is D)
There is a "heat tax" for every energy transaction.
Explanation:
Heat and work are two different ways in which energy is moved from one device to another. In the field of thermodynamics the distinction between Heat and Work is significant. The transfer of thermal energy between systems is heat. This is what is referred to as "heat tax".
No other statement in the question above is correct.
Cheers!
his mechanism has been proposed for the reaction between chloroform and chlorine. Step 1: Cl2(g) 2Cl(g) fast Step 2: CHCl3(g) Cl(g) CCl3(g) HCl(g) slow Step 3: CCl3(g) Cl(g) CCl4(g) fast a. Write the stoichiometric equation for the overall reaction. b. Identify any reaction intermediates in this mechanism. c. Write the rate equation for the rate determining step. d. Show how the rate equation in c. can be used to obtain the rate law for the overall reaction. e. If the concentrations of the reactants are doubled, by what ratio does the reaction rate change
Answer:
a) Balanced Overall Stoichiometric Equation
CHCl₃(g) + Cl₂(g) → CCl₄(g) + HCl(g)
b) The reaction intermediates include Cl(g) and CCl₃(g)
c) The rate of reaction of the rate determining step is:
Rate = k [CHCl₃] [Cl]
d) The rate of overall reaction is given as
Rate = K [CHCl₃] √[Cl₂]
e) The reaction rate increases by a multiple of 2√2 when the reactants' concentrations are doubled.
Explanation:
Step 1: Cl₂(g) → 2Cl(g) fast
Step 2: CHCl₃(g) + Cl(g) → CCl₃(g) + HCl(g) slow
Step 3: CCl₃(g) + Cl(g) → CCl₄(g) fast
a) The overall reaction is a reaction between Chloroform (CHCl₃) and Chlorine (Cl₂). It can be obtained by summing all the elementary equations and eliminating the reaction intermediates (the species that appear on both sides upon summing up all the elementary equations).
Cl₂(g) + CHCl₃(g) + Cl(g) + CCl₃(g) + Cl(g) → 2Cl(g) + CCl₃(g) + HCl(g) + CCl₄(g)
Now, eliminating the reaction intermediates (the species that appear on both sides upon summing up all the elementary equations), we are left with
Cl₂(g) + CHCl₃(g) → HCl(g) + CCl₄(g)
CHCl₃(g) + Cl₂(g) → CCl₄(g) + HCl(g)
b) Like I mentioned in (a), the reaction intermediates are the species that appear on both sides upon summing up all the elementary equations. They include:
Cl(g) and CCl₃(g)
c) The rate determining step is usually the slowest step among the elementary equations.
The rate of reaction expression is usually written from the slowest step.
The slowest step is step 2.
CHCl₃(g) + Cl(g) → CCl₃(g) + HCl(g) slow
The rate of reaction is then given as
Rate = k [CHCl₃] [Cl]
d) The rate of reaction for the overall reaction is obtained by substituting for any intermediates that appear in the rate of reaction for the rate determining step
The rate of reaction of the rate determining step is:
Rate = k [CHCl₃] [Cl]
But we can substitute for [Cl] by obtaining an expression for it from the step 1.
Step 1: Cl₂(g) → 2Cl(g)
K₁ = [Cl]²/[Cl₂]
[Cl]² = K₁ [Cl₂]
[Cl] = √{K₁ [Cl₂]}
We then substitute this into the rate determining step
Rate = k [CHCl₃] [Cl]
Rate = k [CHCl₃] √{K₁ [Cl₂]}
Rate = (k)(√K₁) [CHCl₃] √[Cl₂]
Let (k)(√K₁) = K (the overall rate constant)
Rate = K [CHCl₃] √[Cl₂]
e) If the concentrations of the reactants are doubled, by what ratio does the reaction rate change
Old Rate = K [CHCl₃] √[Cl₂]
If the concentrations of the reactants are doubled, the new rate would be
New Rate = K × 2[CHCl₃] × √{2 × [Cl₂]}
New Rate = K × 2[CHCl₃] × √2 × √[Cl₂]
New Rate = 2√2 K [CHCl₃] √[Cl₂]
Old Rate = K [CHCl₃] √[Cl₂]
New Rate = 2√2 × (Old Rate)
The reaction rate increases by a multiple of 2√2 when the reactants' concentrations are doubled.
Hope this Helps!!!!
The standard free energy ( Δ G ∘ ′ ) (ΔG∘′) of the creatine kinase reaction is − 12.6 kJ ⋅ mol − 1 . −12.6 kJ⋅mol−1. The Δ G ΔG value of an in vitro creatine kinase reaction is − 0.1 kJ ⋅ mol − 1 . −0.1 kJ⋅mol−1. At the start of the reaction, the concentration of ATP is 5 mM, 5 mM, the concentration of creatine is 17 mM, 17 mM, and the concentration of creatine phosphate is 25 mM. 25 mM. Using the values given, calculate the starting concentration of ADP in micromolar.
To calculate the starting concentration of ADP in micromolar, use the equilibrium constant and the concentrations of ATP, creatine, and creatine phosphate at the start.
Explanation:To calculate the starting concentration of ADP in micromolar, we need to use the equilibrium constant and the concentrations of ATP, creatine, and creatine phosphate at the start. The equation for the creatine kinase reaction is:
ATP + Creatine → ADP + Creatine Phosphate
Given that the standard free energy change (ΔG°) is -12.6 kJ/mol and the ΔG is -0.1 kJ/mol, we can calculate the equilibrium constant (K) using the equation: ΔG = -RT ln K.
Using the given values, we can substitute them into the equation to solve for K and then use the concentrations of ATP, creatine, and creatine phosphate to calculate the starting concentration of ADP in micromolar.
Learn more about Starting Concentration of ADP in Micromolar here:https://brainly.com/question/28590870
#SPJ2
Using the values given the ADP concentration came as approximately 527.12 μM.
To find the starting concentration of ADP in micromolar, we can use the Gibbs free energy equation:
ΔG = ΔG° + RT ln(Q)
where ΔG is the Gibbs free energy change under cellular conditions, ΔG° is the standard Gibbs free energy change, R is the universal gas constant (8.314 J/mol·K), T is the temperature in Kelvin, and Q is the reaction quotient.
The creatine kinase reaction is:
Creatine + ATP <=> Creatine phosphate + ADP
Given:
ΔG°' = -12.6 kJ/mol = -12600 J/molΔG = -0.1 kJ/mol = -100 J/mol[ATP] = 5 mM = 5 × 10⁻³ M[Creatine] = 17 mM = 1.7 × 10⁻² M[Creatine phosphate] = 25 mM = 2.5 × 10⁻² MWe need to find [ADP]. First, rearrange the Gibbs free energy equation to solve for Q:
ΔG - ΔG°' = RT ln(Q)
Q = e^{(ΔG - ΔG°') / RT}
Substitute the known values (assuming T = 298 K):
Q = e^{(-100 - (-12600)) / (8.314 × 298)}
Q = e^{12500 / 2479.87}
Q = e^{5.04} ≈ 155.50
Substitute Q into the reaction quotient expression:
Q = [Creatine phosphate] [ADP] / [Creatine][ATP]
155.50 = (2.5 × 10⁻²) [ADP] / ((1.7 × 10⁻²) (5 × 10⁻³))
155.50 = (2.5 × 10⁻²) [ADP] / (8.5 × 10⁻⁵)
155.50 = (2.5 / 8.5) × 10³ [ADP]
155.50 = 0.295 [ADP] × 10³
ADP ≈ 155.50 / 0.295 ≈ 527.12 μM
The starting concentration of ADP is approximately 527.12 μM.
Propose a mechanism for the formation of the monobrominated product. Draw all missing reactants and/or products in the appropriate boxes by placing atoms on the canvas and connecting them with bonds. Add charges where needed. Electron flow arrows should start on the electron(s) of an
Answer:
The mechanism is SN2
Explanation:
See mechanism of monobromination of alkane attached
Answer:
Explanation:
find the solution below
if you have 3.0 moles of argon gas at STP, how much volume will the argon take up?
Heat capacity is the amount of heat needed to raise the temperature of a substance 1 ∘ ∘ C or 1 K. Open Odyssey. In the Molecular Explorer, choose Measuring Specific Heat (16). Follow the directions for water only. What variable is plotted on the y - y- axis? total energy What variable is plotted on the x - x- axis? temperature What is the molar heat capacity? molar heat capacity = J ⋅ K − 1 ⋅ mol − 1 J⋅K−1 ⋅mol−1 What is the specific heat capacity?
Answer:J.K^-1. kg^-1
Explanation:
Heat capacity= H
Mass=m
Specific heat capacity= c
H=mc
J.K^-1 = c ×kg
c= J.K^-1/kg
=J.K^-1.kg^-1
Pure platinum is too soft to be used in jewelry because it scratches easily. To increase the hardness so that it can be used in jewelry, platinum is combined with other metals to form an alloy. To determine the amount of platinum in an alloy, a 8.528 g sample of an alloy containing platinum and cobalt is reacted with excess nitric acid to form 2.49 g of cobalt(II) nitrate. Calculate the mass percent of platinum in the alloy.
Answer:
percentage mass of platinum in the alloy ≈ 90.60 %
Explanation:
The alloy is 8.528 g sample of an alloy containing platinum and cobalt . The alloy react with excess nitric acid to form cobalt(ii) nitrate . Platinum is resistant to acid so it will definitely not react with the acid only the cobalt metal in the alloy will react with the acid.
The chemical reaction can be represented as follows:
Co (s) + HNO₃ (aq) → Co(NO₃)₂ (aq) + NO₂ (l) + H₂O (l)
The balanced equation
Co (s) + 4HNO₃ (aq) → Co(NO₃)₂(aq) + 2NO₂ (l) + 2H₂O (l)
Cobalt is the limiting reactant
atomic mass of cobalt = 58.933 g/mol
Molar mass of Co(NO₃)₂ = 58.933 + 14 × 2 + 16 × 6 = 58.933 + 28 + 96 = 182.933 g
58.933 g of cobalt produce 182.933 g of Co(NO₃)₂
? gram of cobalt will produce 2.49 g of Co(NO₃)₂
cross multiply
grams of cobalt that will react = (58.933 × 2.49)/182.933
grams of cobalt that will react = 146.74317000/182.933
grams of cobalt that will react= 0.8021689362 g
grams of cobalt that will react = 0.802 g
mass of platinum in the alloy = 8.528 g - 0.802 g = 7.726 g
percentage mass of platinum in the alloy = 7.726/8.528 × 100 = 772.600/8.528 = 90 .595 %
percentage mass of platinum in the alloy ≈ 90.60 %
A 25.888 g sample of aqueous waste leaving a fertilizer manufacturer contains ammonia. The sample is diluted with 73.464 g of water. A 10.762 g aliquot of this solution is then titrated with 0.1039 M HCl . It required 31.89 mL of the HCl solution to reach the methyl red endpoint. Calculate the weight percent NH3 in the aqueous waste.
Answer:
Weight % of NH₃ in the aqueous waste = 2.001 %
Explanation:
The chemical equation for the reaction
[tex]\\\\NH_3} + HCl -----> NH_4Cl[/tex]
Moles of HCl = Molarity × Volume
= 0.1039 × 31.89 mL × [tex]\frac{1 \ L}{1000 \ mL}[/tex]
= 0.0033 mole
Total mass of original sample = 25.888 g + 73.464 g
= 99.352 g
Total HCl taken for assay = [tex]\frac{10.762 \ g}{99.352 \ g}[/tex]
= 0.1083 g
Moles of NH₃ = [tex]\frac{0.0033 \ mol}{0.1083}[/tex]
= 0.03047 moles
Mass of NH₃ = number of moles × molar mass
Mass of NH₃ = 0.03047 moles × 17 g
Mass of NH₃ = 0.51799
Weight % of NH₃ = [tex]\frac{0.51799 \ g}{25.888 \ g} * 100%[/tex]%
Weight % of NH₃ in the aqueous waste = 2.001 %
A sample of NaOH (sodium hydroxide) contains a small amount of Na2CO3 (sodium carbonate). For titration to the phenolphthalein endpoint, 0.200 g of this sample requires 23.98 mL of 0.100 M HCl. An additional 0.700 mL of 0.100 M HCl is required to reach the methyl orange endpoint. What is the percentage of Na2CO3 by mass in the sample?
Answer:
3.71%
Explanation:
The phenolphthalein endpoint refers to the reactions:
OH⁻ + H⁺ → H₂O
CO₃⁻² + H⁺ → HCO₃⁻
While the methyl orange endpoint to:
HCO₃⁻ + H⁺ → H₂CO₃
So the additional volume required for the second endpoint tells us the amount of HCO₃⁻ species, which in turn is the total amount of Na₂CO₃ in the sample:
0.700 mL * 0.100 M * [tex]\frac{1mmolHCO_{3}^{-}}{1mmolHCl}[/tex] = 0.07 mmol HCO₃⁻
Now we calculate the mass of Na₂CO₃, using its molecular weight:
0.07 mmol HCO₃⁻ = 0.07 mmol Na₂CO₃
0.07 mmol Na₂CO₃ * 106 mg/mmol = 7.42 mg Na₂CO₃
No calculations using the volume of the first equivalence point are required because the problem already tells us the mass of the sample is 0.200 g.
0.200 g ⇒ 0.200 * 1000 = 200 mg
%Na₂CO₃ = 7.42 mg/200 mg * 100 = 3.71%
3. Calculate the theoretical value for the number of moles of CO2 that should have been produced in each balloon assuming that 1.45 g of NaHCO3 is present in an antacid tablet. Use stoichiometry (a mole ratio conversion must be present) to find your answers (there should be three: one answer for each balloon). (6 pts)
Answer:
the theoretical value for the number of moles of [tex]CO_{2(aq)}[/tex] is 0.0173 moles
Explanation:
The balanced chemical equation for the reaction is represented by:
[tex]H_3C_6H_5O_{7(aq)} + 3NaHCO_{3(aq)} ------>3CO_{2(g)}+3H_2O+Na_3C_6H_5O_{7(aq)}[/tex]
From above equation; we would realize that 3 moles of [tex]NaHCO_{3(aq)}[/tex] reacts with [tex]H_3C_6H_5O_{7(aq)}[/tex] to produce 3 moles of [tex]CO_{2(aq)}[/tex]
However ; the molar mass of [tex]NaHCO_{3(aq)}[/tex] = 84 g/mol
mass given for [tex]NaHCO_{3(aq)}[/tex] = 1.45 g
therefore , we can calculate the number of moles of [tex]NaHCO_{3(aq)}[/tex] by using the expression :
number of moles of [tex]NaHCO_{3(aq)}[/tex] = [tex]\frac{mass \ given}{ molar \ mass}[/tex]
number of moles of [tex]NaHCO_{3(aq)}[/tex] = [tex]\frac{1.45}{84}[/tex]
number of moles of [tex]NaHCO_{3(aq)}[/tex] = 0.0173 mole
Since the ratio of [tex]NaHCO_{3(aq)}[/tex] to [tex]CO_{2(aq)}[/tex] is 1:1; that implies that number of moles of [tex]NaHCO_{3(aq)}[/tex] is equal to number of moles of [tex]CO_{2(aq)}[/tex] produced.
number of moles of [tex]CO_{2(aq)}[/tex] = [tex]\frac{mass \ given}{ molar \ mass}[/tex]
0.0173 = [tex]\frac{mass \ given}{ 44 \ g/mol}[/tex]
mass of [tex]CO_{2(aq)}[/tex] = 0.0173 × 44
mass of [tex]CO_{2(aq)}[/tex] = 0.7612 g
Thus; the theoretical value for the number of moles of [tex]CO_{2(aq)}[/tex] is 0.0173 moles
A 215-g sample of copper metal at some temperature is added to 26.6 g of water. The initial water temperature is 22.22 oC, and the final temperature is 24.44 oC. If the specific heat of copper is 0.385 Jg-1oC-1, what was the initial temperature of the copper? Any additional constants needed can be found in your textbook.
The initial temperature of the copper metal was 27.38 degrees.
Explanation:
Data given:
mass of the copper metal sample = 215 gram
mass of water = 26.6 grams
Initial temperature of water = 22.22 Degrees
Final temperature of water = 24.44 degrees
Specific heat capacity of water = 0.385 J/g°C
initial temperature of copper material , Ti=?
specific heat capacity of water = 4.186 joule/gram °C
from the principle of:
heat lost = heat gained
heat gained by water is given by:
q water = mcΔT
Putting the values in the equation:
qwater = 26.6 x 4.186 x (2.22)
qwater = 247.19 J
qcopper = 215 x 0.385 x (Ti-24.4)
= 82.77Ti - 2019.71
Now heat lost by metal = heat gained by water
82.77Ti - 2019.71 = 247.19
Ti = 27.38 degrees
7. If you fill a balloon with 5.2 moles of gas and it creates a balloon with a volume of 23.5 liters, how many moles are in a balloon at the same temperature and pressure that has a volume of 14.9 liters
To solve this problem, we can use the ideal gas law equation PV=nRT. We can find the number of moles in the first balloon using the given information, and then use that value to find the volume of the second balloon.
Explanation:To solve this problem, we can use the ideal gas law, which states that PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature in Kelvin.
First, we can use the given information to find the number of moles in the first balloon. Rearranging the ideal gas law equation, we have n = PV / RT. Plugging in the values, we get n = (5.2 mol)(23.5 L) / (0.0821 atm L/mol K)(T in Kelvin).
Once we have the number of moles for the first balloon, we can use this value to find the volume of the second balloon. Rearranging the ideal gas law equation, we have V = nRT / P. Plugging in the values and solving for V, we get V = (5.2 mol)(0.0821 atm L/mol K)(T in Kelvin) / (P)
A 40.0-mL sample of 0.100 M HNO2 (Ka = 4.6 x 10-4 .) is titrated with 0.200 M KOH. Calculate: a. the pH when no base is added b. the volume of KOH required to reach the equivalence point. c. the pH after adding 5.00 mL of KOH d .the pH at one-half the equivalence point e. the pH at the equivalence point f. the pH after 30 mL of the base is added
The initial pH of 0.100 M HNO2 is approximately 2.17. It takes 20.0 mL of 0.200 M KOH to reach the equivalence point. The subsequent pH values at various points in the titration reflect the changing concentrations of HNO2 and OH-.
A 40.0-mL sample of 0.100 M HNO2 (Ka = 4.6 x 10-4) is titrated with 0.200 M KOH.
a. The pH when no base is added
To find the initial pH, we first need to calculate the concentration of H3O+ using the equilibrium expression for the weak acid:
HNO2 ⇌ H+ + NO2-
Using Ka, we get:
Ka = [H+][NO2-] / [HNO2]
4.6 x 10-4 = x2 / 0.100
Solving for x, we get:
x = √(4.6 x 10-4 * 0.100) = 0.00678 M
pH = -log(0.00678) ≈ 2.17
b. The volume of KOH required to reach the equivalence point
The moles of HNO2 are:
0.040 L * 0.100 M = 0.004 mol
Since KOH and HNO2 react in a 1:1 molar ratio, the volume of 0.200 M KOH required is:
0.004 mol / 0.200 M = 0.020 L = 20.0 mL
c. The pH after adding 5.00 mL of KOH
Moles of KOH added:
0.005 L * 0.200 M = 0.001 mol
Moles of HNO2 remaining:
0.004 mol - 0.001 mol = 0.003 mol
Concentration of HNO2 remaining = 0.003 mol/0.045 L = 0.0667 M
Concentration of NO2- formed = 0.001 mol/0.045 L = 0.0222 M
Using the Henderson-Hasselbalch equation:
pH = pKa + log([A-]/[HA])
pH = -log(4.6 x 10-4) + log(0.0222/0.0667) ≈ 3.35
d. The pH at one-half the equivalence point
At one-half the equivalence point, the concentration of [A-] = [HA], so:
pH = pKa = -log(4.6 x 10-4) ≈ 3.34
e. The pH at the equivalence point
At the equivalence point, all HNO2 has been converted to NO2-:
NO2- will hydrolyze to produce OH-:
NO2- + H2O ⇌ HNO2 + OH-
Using Kb for NO2-:
Kb = Kw/Ka = 1.0 x 10-14 / 4.6 x 10-4 = 2.17 x 10-11
Setting up the equation:
Kb = [OH-][HNO2] / [NO2-]
2.17 x 10-11 = x2 / 0.100
Solving for x:
x = √(2.17 x 10-11 * 0.100) = 1.47 x 10-6
pOH = -log(1.47 x 10-6) ≈ 5.83
pH = 14 - 5.83 = 8.17
f. The pH after 30 mL of the base is added
Moles of KOH added:
0.030 L * 0.200 M = 0.006 mol
Excess moles of OH-:
0.006 mol - 0.004 mol = 0.002 mol
Concentration of OH- in the total volume:
0.002 mol / 0.070 L = 0.02857 M
pOH = -log(0.02857) ≈ 1.54
pH = 14 - 1.54 = 12.46
The correct answers are: (a). [H⁺] = 2.17; (b). [tex]\text{Volume} = 20.0\ mL[/tex]; (c). [tex][NO_2^-] = \text{0.0222 M}[/tex]; (d). pH = 3.34; (e). [tex]\text{pH} = 8.02[/tex]; (f). pH = 12.46.
Let's solve the different parts of the titration problem step-by-step:
a. pH when no base is added:
First, we need to find the pH of a 0.100 M HNO₂ solution. HNO₂ is a weak acid and it partially ionizes in water:
HNO₂ ⇌ H⁺ + NO₂⁻The expression for the acid dissociation constant Ka is:
[tex]K_a = 4.6 \times 10^{-4} = \frac{[H^+][NO_2^-]}{[HNO_2]}[/tex]Assuming that the initial concentration of HNO₂ is C0 = 0.100 M and the change in concentration is x:
[tex]4.6 \times 10^{-4} = \frac{(x \times x)}{(0.100 - x)}[/tex]Assuming x is small relative to 0.100 M:
[tex]4.6 \times 10^{-4} \approx \frac{x^2} {0.100}[/tex]x² = 4.6 x 10⁻⁵x = 6.78 x 10⁻³ M[H⁺] = 6.78 x 10⁻³ M, pH = -log[H⁺] = -log(6.78 x 10⁻³) = 2.17b. Volume of KOH required to reach the equivalence point:
At the equivalence point, moles of HNO₂ = moles of KOH.Moles of HNO₂ = 0.100 M × 0.040 L = 0.00400 molFor KOH: 0.00400 mol = volume × 0.200 M[tex]\text{Volume} = \frac{\text{0.00400 mol}} {\text{0.200 M}} = 0.0200\ L = 20.0\ mL[/tex]c. pH after adding 5.00 mL of KOH:
Moles of KOH added = 0.200 M × 0.00500 L = 0.00100 molRemaining moles of HNO₂ = 0.00400 mol - 0.00100 mol = 0.00300 molTotal volume = 40.0 mL + 5.0 mL = 45.0 mL = 0.0450 L[tex][HNO_2] = \frac{\text{0.00300 mol}} {\text{0.0450 L}} = \text{0.0667 M}[/tex][tex][NO_2^-] = \frac{\text{0.00100 mol}} {\text{0.0450 L}} = \text{0.0222 M}[/tex]Using the Henderson-Hasselbalch equation:
[tex]pH = pKa + \log(\frac{[\text{NO}_2^-]}{[\text{HNO}_2]}) \\[/tex][tex]pK_a = -\log(4.6 \times 10^{-4}) = 3.34 \\[/tex][tex]pH = 3.34 + \log(\frac{0.0222}{0.0667}) = 3.34 - 0.477 = 2.86[/tex]d. pH at one-half the equivalence point:
At one-half the equivalence point, [HNO₂] = [NO₂⁻], so pH = pKa.
pH = pKa = 3.34e. pH at the equivalence point:
At the equivalence point, all HNO₂ has reacted to form NO₂⁻. The solution contains NO₂⁻ ions, which hydrolyze:
NO₂⁻ + H₂O ⇌ HNO₂ + OH⁻[tex]Kb = \frac{K_w}{K_a} = \frac{1.0 \times 10^{-14}}{4.6 \times 10^{-4}} = 2.17 \times 10^{-11}[/tex]Let x be the concentration of OH⁻:
[tex]K_b = \frac{x^2}{0.0500\text{ M}} \\[/tex][tex]2.17 \times 10^{-11} = \frac{x^2}{0.0500} \\[/tex][tex]x^2 = 1.085 \times 10^{-12} \quad \Rightarrow \quad x = \sqrt{1.085 \times 10^{-12}} = 1.04 \times 10^{-6}\text{ M} \\[/tex][tex][\text{OH}^-] = 1.04 \times 10^{-6}\text{ M} \\[/tex][tex]\text{pOH} = -\log[\text{OH}^-] = -\log(1.04 \times 10^{-6}) = 5.98 \\[/tex][tex]\text{pH} = 14 - \text{pOH} = 14 - 5.98 = 8.02[/tex]f. pH after 30 mL of the base is added:
Moles of KOH added = 0.200 M × 0.030 L = 0.00600 molExcess moles of KOH = 0.00600 mol - 0.00400 mol = 0.00200 molTotal volume = 40 mL + 30 mL = 70 mL = 0.070 L[tex][OH^-] = \frac{\text{0.00200 mol}}{\text{0.070 L}} = \text{0.0286 M}[/tex]pOH = -log(0.0286) = 1.54pH = 14 - pOH = 14 - 1.54 = 12.46Nuclear power plants produce energy using fission. One common fuel, uranium-235, produces energy through the fission reaction 235 92U+10n→fission fragments+neutrons+3.20×10−11 J/atom 92235U+01n→fission fragments+neutrons+3.20×10−11 J/atom What mass of uranium-235 is needed to produce the same amount of energy as the fusion reaction in Part A? Express your answer to three significant figures and include the appropriate units.
Answer:
mass of U-235 = 15.9 g (3 sig. figures)
Explanation:
1 atom can produce -------------------------> 3.20 x 10^-11 J energy
x atoms can produce ----------------------> 1.30 x 10^12 J energy
x = 1.30 x 10^12 / 3.20 x 10^-11
x = 4.06 x 10^22 atoms
1 mol ----------------------> 6.023 x 10^23 atoms
y mol ----------------------> 4.06 x 10^22 atoms
y = 0.0675 moles
mass of U-235 = 0.0675 x 235 = 15.8625
mass of U-235 = 15.9 g (3 sig. figures)
The half-equivalence point of a titration occurs halfway to the equivalence point, where half of the analyte has reacted to form its conjugate, and the other half still remains unreacted. If 0.300 moles of a monoprotic weak acid ( Ka=3.6×10−5 M) is titrated with NaOH , what is the pH of the solution at the half-equivalence point?
Answer:
pH=pKa
pH=4.44
Explanation:
Since the titration occur between a weak acid and a strong base.
then at half -equivalence point, the pH of the solution is equals to the pKa of the weak acid.
Therefore, pH=pKa
Ka of weak acid=3.6×10^−5
To calculate the pKa of the weak acid using the express below;
pKa =- log(Ka)
pKa=−log(3.6×10−5)=4.44
From the question, the pKa of the solution is at half -equivalence point
Then,
pH=pKa
pH=4.44
The question says that the titration occurred between a weak acid and a
strong base at half-equivalence point. Then we can deduce that the pH of
the solution is equal to the pKa of the weak acid.
pH=pKa
Ka of monoprotic weak acid=3.6×10⁻⁵
The pKa of the monoprotic weak acid will be calculated by :
pKa = - log(Ka)
pKa = −log(3.6×10⁻⁵) = 4.44
Since the pKa of the solution is at half -equivalence point
pH=pKa
pH=4.44
Read more on https://brainly.com/question/15867090