Electrolysis is the production of a chemical reaction by passing an electric current to liquid containing ions.
Calculate the second ionization energy of the metal M (?Hion2� in kJ/mol) using the following data:
Lattice enthalpy of MO(s), ?Hl� = -2383 kJ/mol
Bond dissociation enthalpy of O2(g) = +498 kJ/mol
First electron affinity of O = -141 kJ/mol
Second electron affinity of O = +744 kJ/mol
Enthalpy of sublimation of M = + 130 kJ/mol
First ionization energy of M = + 267 kJ/mol
Standard enthalpy of formation of MO(s), ?Hf� = -307 kJ/mol
From the information provided in the question, the second ionization energy of the metal is 578 kJ/mol.
From the question, we have the following information;
Lattice enthalpy of MO(s) = -2383 kJ/mol
Bond dissociation enthalpy of O2(g) = +498 kJ/mol
First electron affinity of O = -141 kJ/mol
Second electron affinity of O = +744 kJ/mol
First ionization energy of M = + 267 kJ/mol
Heat of sublimation of M = + 130 kJ/mol
Standard enthalpy of formation of MO(s) = -307 kJ/mol
Using Hess law of constant heat summation;
ΔHf = ΔHs + BE + ∑IE + ∑EA + U
ΔHs = Heat of sublimation of metal
ΔHf = Heat of formation MO
BE = Bond energy of O2
∑EA = sum of electron affinities of Oxygen
∑IE = Sum of the ionization energies of M
U = Lattice energy of MO
Let the second ionization energy be x
Substituting values;
(-307) = 130 + 498 + (267 + x) + 603 + (-2383)
(-307) = -885 + x
-x = -885 + 307
-x = -578
x = 578 kJ/mol
The second ionization energy of the metal is 578 kJ/mol.
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Student A uses 3N of force to move a cart 5 meters in 10 seconds. Student B uses 6N of force to move the same cart the same distance in 5 seconds. Which student did more work? Which student used more power? Use evidence and explain your answersss
Work = force x distance
Power = work/time
Work is said to be done if the force applied to a body causes the body to move through a distance.
Student B used more work and power
Workdone = Force * Distance
For student A:
Force = 3N
distance = 5meters
Time taken = 10 secs
Workdone by Student A = 3 * 5
Workdone by student A = 15Nm
Power used up = workdone/time
Power used up = 15/10
Power used up = 1.5 Watts
For student B:
Force = 6N
distance = 5meters
Time taken = 5 secs
Workdone by Student A = 6 * 5
Workdone by student A = 30Nm
Power used up = workdone/time
Power used up = 30/5
Power used up = 6 Watts
This shows that student B used more work and power
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In experiment 9, in one operation, we heat up the alcohol with acid and do a concurrent distillation. What was the purpose of doing this
Answer:
we heat up because the component with lower boiling evaporates first,
leaving the other behind
Cesium-137 is part of the nuclear waste produced by uranium-235 fission. The half-life of cesium-137 is 30.2 years. How much time is required for the activity of a sample of cesium-137 to fall to 8.32 percent of its original value
Answer:
la primera va con la última
To what pH should you adjust a standard hydrogen electrode to get an electrode potential of -0.128 V ? (Assume that the partial pressure of hydrogen gas remains at 1 atm.) Express your answer using two decimal places.\
The pH of the standard hydrogen electrode that has electrode potential of -0.128 V is 4.3.
The equation of the hydrogen electrode is;
2H^+(aq) + 2e ⇄ H2(g)
The standard electrode potential of hydrogen is 0.00 V
Using the Nernst equation;
Ecell = E°cell - 0.0592/n log Q
Now;
E°cell = 0.00 V
n = 2
Q = 1/[H^+]
-0.128 = 0.00 - 0.0592/2 log 1/[H^+]
-0.128 = 0.00 - 0.0296 log 1/[H^+]
-0.128 = - 0.0296 log 1/[H^+]
-0.128/ - 0.0296 = log 1/[H^+]
1/[H^+] = Antilog (4.32)
[H^+] = 4.79 × 10^-5
Now;
pH = -log[H^+]
pH = -log (4.79 × 10^-5)
pH = 4.3
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Calculate the Ka of your acetic acid solution. Discuss this calculation. Based on the value of Ka, is acetic acid a strong acid or a weak acid
Based on our knowledge of strong and weak acids, we can confirm that the Ka value for acetic acid will be relatively low since it is a weak acid.
Acids can be strong or weak. This is determined by its tendency to break apart into ions or stay together to form molecules. Although somewhat counter-intuitive, strong acids are those that are most likely to break apart and therefore contain a high number of ions within their solutions.
Weak acids, on the other hand, are those that tend to stay together in the form of molecules and therefore possess very low ion counts in their solutions. The acid dissociation constant, Kₐ, is used to measure whether an acid is weak or strong and how much so. In the case of Acetic acid, the ka measurement will offer a low value, indicating a weak acid.
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Convert 1.36x10 to standard form
Answer:
13.6 is the correct answer written in standard form.
Explanation:
1.36, move the decimal once to the right to get 13.6
Answer:
13.6
Explanation:
The standard form is 13.6
Using the Kf value of 1.2×109 calculate the concentration of Ni2+(aq) and Ni(NH3)62+ that are present at equilibrium after dissolving 1.47 gNiCl2 in 100.0 mL of NH3(aq) solution such that the equilibrium concentration of NH3 is equal to 0.20 M.
Express your answers in moles per liter to two significant figures separated by a comma.
This problem is describing the equilibrium whereby the formartion of a complex is attained when 1.47 g of nickel(II) chloride is dissolved in 100.0 mL of ammonia so that the latter's equilibium concentration is 0.20 M. Thus, it is asked to calculate the equilibrium concentrations of both nickel(II) ions and that of the complex.
Firstly, we can write out the chemical equation to be considered:
[tex]Ni^{2+}+6NH_3\rightleftharpoons Ni(NH_3)_6^{2+}[/tex]
Next, we can calculate the initial concentration of nickel(II) ions by using the concept of molarity:
[tex][Ni^{2+}]=\frac{1.47gNiCl_2*\frac{1molNiCl_2}{129.6g}*\frac{1molNi^{2+}}{1molNiCl_2} }{0.1000L}=0.113M[/tex]
Afterwards, we set up an equilibrium expression for this chemical reaction:
[tex]Kf=\frac{[Ni(NH_3)_6^{2+}]}{[Ni^{2+}][NH_3]^6}[/tex]
Which can be written in terms of the reaction extent, [tex]x[/tex]:
[tex]Kf=\frac{x}{(0.113-x)(0.2)^6}[/tex]
Now, for the calculation of [tex]x[/tex], we plug in Kf, and solve for it:
[tex]1.2x10^9=\frac{x}{(0.113-x)(0.2)^6}\\\\1.2x10^9=\frac{x}{(0.113-x)(6.4x10^{-5})}\\\\7.68x10^4(0.113-x)=x\\\\x=0.112999 M[/tex]
Which is about the same to the initial concentration of nickel(II) ions because the Kf is too large.
Thus, the required concentrations at equilibrium are about:
[tex][Ni(NH_3)_6^{2+}]=0.113M[/tex]
[tex][Ni^{2+}]=0M[/tex]
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https://brainly.com/question/13043707The boiling point of a substance is tested. After 10 tests, the result is given as 37+/−3°C. Which conclusion can be drawn from this result? (1 point)
The scientists do not need to collect more data because they have narrowed down the range of the results.
The scientists should not report these results until they have the exact number.
The actual boiling point is either 34°C or 40°C.
The actual boiling point is probably between 34°C and 40°C.
Answer:
Explanation:
The actual boiling point is probably between 34C and 40C.
Temperature measures the average kinetic energy of particles of the substances. Therefore, the correct option is option D that is the actual boiling point is probably between 34°C and 40°C.
What is temperature?Temperature is used to measure degree or intensity of heat of a particular substance. Temperature is measured by an instrument called thermometer.
Temperature can be measured in degree Celsius °c, Kelvin k or in Fahrenheit. Temperature is a physical quantity. Heat always flow from higher temperature source to lower temperature source.
We can convert these units of temperature into one another. The relationship between degree Celsius and Fahrenheit can be expressed as:
°C={5(°F-32)}÷9
The actual boiling point is probably between 34°C and 40°C.
Therefore, the correct option is option D.
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HELP!! what are the usual products of combustion reactions?
Explanation:
Carbon dioxide and water
I hope it helps
Answer:
The usual products of combustion reactions are carbon dioxide and water.
Explanation:
Combustion reaction is when a substance reacts with oxygen gas, resulting in a release of energy in the form of light and heat. Combustion reactions must have oxygen (O2) as one of the reactants.
A chemical property is a change in _____.
density
physical state
hardness
composition of matter
Answer:
composition of an element
2 nitrogen atoms and five chlorine atoms what compound does that make
Answer:
dinitrogen pentachloride
Describe your observations when the paper clip was gently crushed with the pestle
Answer:
We can do an experiment with vinegar to see what happens to the calcium in an eggshell when it is exposed to an acid.
analyze why is the result of the stage of meiosis shown below an advantage for organisms that reproduce sexually.
Answer:
Meiosis produces haploid cells (gametes), which contain single chromosomes, or on-half the number of chromosomes in diploid cells. When a sperm and an egg join, the single chromosomes pair up, which results in genetic diversity in the offspring
How many of sodium (Na) are needed to make 4.5 liters of a 1.5mol/L of Na solution?
Answer:
Explanation:
First you will find the mole from the molarity and then the desired mass from the mole.
the solubility of nitrogen gas is 1.90 mL/dL of blood at 1.00 atm. what is the solubility of nitrogen gas in a deepsea divers blood at a depth of 200 feet and pressure of 7.00 atm
The solubility of nitrogen gas in water is 1.90 mL/dL at 1.00 atm and 13.3 mL/dL at 7.00 atm.
We want to relate the solubility of a gas with its partial pressure.
We can do so using Henry's law.
What does Henry's law state?Henry's law states that the amount of dissolved gas in a liquid is proportional to its partial pressure above the liquid.
C = k × P
where,
C is the concentration of a dissolved gas. k is the Henry's Law constant. P partial pressure of the gas.The solubility of nitrogen gas is 1.90 mL/dL of blood at 1.00 atm.
Since the solvent is basically water, we can understand that the concentration of nitrogen gas is 1.90 mL/dL at 1.00 atm.
We can use this information to calculate Henry's Law constant.
k = C/P = (1.90 mL/dL)/1.00 atm = 1.90 mL/dL.atm
We want to calculate the solubility of nitrogen gas at a pressure of 7.00 atm.
We will use Henry's law.
C = k × P = (1.90 mL/dL.atm) × 7.00 atm = 13.3 mL/dL
The solubility of nitrogen gas in water is 1.90 mL/dL at 1.00 atm and 13.3 mL/dL at 7.00 atm.
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Radio waves bounce off of _____________ before returning to Earth
Answer: ionosphere
Explanation: First it bounces off a top layer of the atmosphere called the ionosphere, then it bounces back to the Earth (this is reflection. It then bounces up again to the ionosphere, and continues bouncing back again until it reaches the radio receiver. This is called a skywave, which works around 3 to 30 MHz.
Natural gas is almost entirely methane. A container with a volume of 2.65L holds 0.120mol of methane. What will the volume be if an additional 0.182mol of methane is added to the container under constant temperature and pressure? Give your answer in three significant figures.
The final volume of the methane gas in the container is 6.67 L.
The given parameters;
initial volume of gas in the container, V₁ = 2.65 Linitial number of moles of gas, n₁ = 0.12 moladditional concentration, n = 0.182 molThe total number of moles of gas in the container is calculated as follows;
[tex]n_t = 0.12 + 0.182 = 0.302 \ mol[/tex]
The final volume of gas in the container is calculated as follows;
[tex]PV = nRT\\\\\frac{V}{n} = \frac{RT}{P} \\\\\frac{V_1}{n_1} = \frac{V_2}{n_2} \\\\V_2 = \frac{V_1 n_2}{n_1} \\\\V_2 = \frac{2.65 \times 0.302}{0.12} \\\\V_2 = 6.67 \ L[/tex]
Thus, the final volume of the methane gas in the container is 6.67 L.
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At 298 K, the reaction 2 HF (g) ⇌ H2 (g) + F2 (g) has an equilibrium constant Kc of 8.70x10-3. If the equlibrium concentrations of H2 and F2 gas are both 1.33x10-3 M, determine the initial concentration of HF gas assuming you only started with HF gas and no products initially.
This problem is describing the equilibrium whereby hydrofluoric acid decomposes to hydrogen and fluorine gases at 298 K whose equilibrium constant is 8.70x10⁻³, the equilibrium concentrations of all the reactants are both 1.33x10⁻³ M and asks for the initial concentration of hydrofluoric acid which turns out to be 2.86x10⁻³ M.
Then, we can write the following equilibrium expression for hydrofluoric acid once the change, [tex]x[/tex], has taken place:
[tex][HF]=[HF]_0-2x[/tex]
Now, since both products are 1.33x10⁻³ M we infer the reaction extent is also 1.33x10⁻³ M, and thus, we can calculate the equilibrium concentration of HF via the law of mass action (equilibrium expression):
[tex]8.70x10^{-3}=\frac{(1.33x10^{-3} M)^2}{[HF]} }[/tex]
[tex][HF]=\frac{(1.33x10^{-3} M)^2}{8.70x10^{-3}} }=2.03x10^{-4}M[/tex]
Finally, the initial concentration of HF is calculated as follows:
[tex][HF]_0=[HF]+2x=2.033x10^{-4}+2*(1.33x10^{-3})=2.86x10^{-3}M[/tex]
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a. Count the number of Sulfur atom
b. How many total atoms are given in the compound
Please helppp
Answer:
from the words below underline six example of rhetorical patterns
A mixture of solids containing a ketone, a carboxylic acid, and an amine, are dissolved in DCM. What is the best way to begin an extraction to separate the amine from the mixture
There are different ways of extraction. The best way to begin an extraction to separate the amine from the mixture is to extract with dilute NaOH.
An acid-base extraction is often used in the extraction of carboxylic acids from the organic layer and thereafter into the aqueous layer.NaOH is known to be the most common compound that is used to convert a carboxylic acid into its more water-soluble ionic carboxylate form.
But if the mixture has a compound that you want, and that can react with NaOH, another milder base such as sodium bicarbonate is preferably used.
See full question below
A mixture of solids containing a ketone, a carboxylic acid, and an amine, are dissolved in DCM. What is the best way to begin an extraction in order to separate the carboxylic acid from the mixture?
A) Extract with dilute NaOH
B) Extract with dilute HCl
C) Extract with dichloromethane
D) Extract with water
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he hybridization of carbon in diamond is _________. Enter your answer with no superscripts or subscripts, i.e., ab3.
Diamond is composed of hexagonal rings in which sp3 hybridized carbon atoms are linked together.
Hybridization refers to the mixing of atomic orbitals to yield hybrid orbitals that are suitable for bonding. The energy of orbitals that combine to form hybrid orbitals must be close enough for such combination to take place.
Diamond is composed of hexagonal rings in which sp3 hbridized carbon atoms are linked together. Each carbon atom in diamond is tetrahedral.
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The mixing of the two different orbitals to form a compound is called hybridization. For example mixing of s and p orbits.
The correct answer is sp3.
The arrangement of the elements in a different manner to form a new compound is called allotropes. For example, diamond and graphite are the allotropes of carbon.
The valence electrons are in p orbitals and p orbit mixed after the s orbitals.
Therefore, the correct answer is sp3
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Pleeeeasee someone who’s good at chemistry?! 10 grade
ASAP
I’ll give points, just help please
Answer:
where is the question????????????
The sum of the number of proteins and neutrons in an atoms nucleus is its __________ ___________.
Answer:
Mass Number
Explanation:
In nuclear physics, the sum of the numbers of protons and neutrons present in the nucleus of an atom.
Please vote brainliest!
Use the diagram below to answer 3 questions
Answer:
The element X is sulfur.
Sulfur will gain 2 electrons to become an ion.
The charge of sulfide ion is (2-).
Explanation:
Doing Labs at home
I’m a junior and I’m staying home for this semester and I have to take chemistry and a lot of my work is Labs but I don’t know how to do them since I don’t have the materials at home to do the labs. Someone please help!!!
Answer:
go get the stuff.
Explanation:
Thin-layer chromatography explain ?????
Answer:
Explanation:
Thin-layer chromatography (TLC) is a chromatography technique used to separate non-volatile mixtures. ... After the sample has been applied on the plate, a solvent or solvent mixture (known as the mobile phase) is drawn up the plate via capillary action.
If 38.6 grams of iron react with an excess of bromine gas, what mass of FeBr2 can form?
Answer:
›› FeBr2 molecular weight. Molar mass of FeBr2 = 215.653 g/mol. This compound is also known as Iron(II) Bromide. Convert grams FeBr2 to moles or moles FeBr2 to grams. Molecular weight calculation: 55.845 + 79.904*2 ›› Percent composition by element
Explanation:
If 38.6 grams of iron react with an excess of bromine gas, the mass of FeBr2 can form is 149 grams.
What is mass?Mass is defined as a way to gauge how much matter there is in a substance or thing. The kilogram (kg) is the fundamental SI unit of mass, while lower masses can also be measured in grams (g). Atoms make up everyday matter. A majority of an atom's mass is contained in its nucleus.
Given Fe = 38.6 g.
Fe has a molar mass = 55.845 g/mol.
Given mass/molar mass equals 38.6g/55.845gmol-1, or 0.6912 moles of iron.
The reaction is described as Fe + Br2 FeBr2.
One mole Fe yields 1 mole of FeBr2.
FeBr2 would be produced from 0.6912 moles of Fe.
FeBr2 has a molar mass of 215.65 g/mol.
Moles of FeBr2 x Molar mass of FeBr2
= 215.65 g/mole x 0.6912 mole
= 149.06 g FeBr2 produced is the formula.
Thus, if 38.6 grams of iron react with an excess of bromine gas, the mass of FeBr2 can form is 149 grams.
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Oil does not dissolve in water because
Explanation:
Oils are made up of long hydrocarbon chains which are non polar. Since water is a polar solvent oil doesn't dissolve in water well.( Polar solutes dissolve in polar solvents and non polar solutes dissolve in non polar solvents) The hydrocarbon chains are hydrophobic.
A pan containing 40 grams of water was allowed to cool from a temperature of 91.0 °C. If the amount of heat released is 1,300 joules, what is the approximate final temperature of the water? 74 °C 78 °C 81 °C 83 °C
Answer:
d. 83 °c
Explanation:
Answer:
D. 83 c
Explanation:
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