In the formation of 1 formula unit of ScBr₃, 3 electrons are transferred.
Let's consider the following balanced redox reaction.
2 Sc + 3 Br₂ ⟶ 2 ScBr₃
We can identify both half-reactions.
Oxidation: 2 Sc ⟶ 2 Sc⁺³ + 6 e⁻
Reduction: 6 e⁻ + 3 Br₂ ⟶ 6 Br⁻
As we can see, 6 electrons are involved in the formation of 2 formula units of ScBr₃. Thus, 3 electrons are involved in the formation of 1 formula unit of ScBr₃.
In the formation of 1 formula unit of ScBr₃, 3 electrons are transferred.
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Why we use two different methods for detection of cogulase enzyme ? Or what other reason or what basic different between them?
A beaker containing a green liquid is left uncovered in a laboratory for one week., After the liquid evaporates, the beaker contains a dry green solid. Was the original liquid in the beaker an element, a compound, or a mixture?
The original liquid is regarded as a mixture.
A mixture is regarded as a material which comprises of two or more
substances which are combined physically. An example is the mixture of
dye and water.
A compound on the other hand is combined chemically and not physically
which when exposed to the same type of situation either remains in the
beaker or nothing at all is found. Since the water evaporates and a green
solid was present then it means the material was a solution in which
water(solvent) and the green solid(solute) was physically combined and not
chemically combined together.
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How many mL of 0.150 M HF solution are required to produce 0.0370 moles of HF
Answer:
247 ml
Explanation:
How many mL of 0.150 M HF solution are required to produce 0.0370 moles of HF 0.150 moles/ liter = 0.150/1000 moles/ml =0.000150 moles/ml
0.000150 x ? = 0.0370 moles HF
? = 0.0370/0.000150 = 247 ml
check
247 ml = 247/1000 L = 0.247
(0.247) x (0.150) =0.370 check
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
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/13043707Thin-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.
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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3.00 L of a gas is collected at 35.0 C and 0.93 atm. What is the volume at STP
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
Tristan wraps some gifts and then brings them to the post office where they are delivered to people in different parts of the country. Which organelle is Tristan most like?
Answer:
Tristan wraps some gifts and then brings them to the post office where they are delivered to people in different parts of the country. Which organelle is Tristan most like?
Answer: Tristan is most like the Golgi Body
Explanation:
What identifies the number of protons in the nucleus of an atom?
Answer: Atomic number
Explanation:
I hope this helps you!
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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2 nitrogen atoms and five chlorine atoms what compound does that make
Answer:
dinitrogen pentachloride
How could you tell if a substance has undergone a physical change or a chemical one?
Answer: Chemical changes occur when a substance combines with another to form a new substance, called chemical synthesis or, alternatively, chemical decomposition into two or more different substances. These processes are called chemical reactions and, in general, are not reversible except by further chemical reactions.
A physical change is are changes affecting the form of a chemical substance, but not its chemical composition. Physical changes are used to separate mixtures into their component compounds, but can not usually be used to separate compounds into chemical elements or simpler compounds.
Suppose that in an equilibrium mixture of HCl, Cl2, and H2, the concentration of H2 is 1.0 x 10-11 mol-L-1and that of Cl2 is 2.0 x 10-10 mol-L-1. What is the equilibrium molar concentration of HCl at 500 K, given Kc = 4.0 x 1018 for H2(g) +Cl2(g) ⇆ 2HCl(g).
Considering the definition of Kc, the equilibrium molar concentration of HCl at 500 K is 0.0894 [tex]\frac{mol}{L}[/tex].
The balanced reaction is:
H₂(g) +Cl₂(g) ⇆ 2 HCl(g)
Equilibrium is a state of a reactant system in which no changes are observed as time passes, despite the fact that the substances present continue to react with each other. In other words, reactants become products and products become reactants and they do so at the same rate.
In other words, chemical equilibrium is established when there are two opposite reactions that take place simultaneously at the same speed.
The concentration of reactants and products at equilibrium is related by the equilibrium constant Kc. Its value in a chemical reaction depends on the temperature and the expression of a generic reaction aA + bB ⇄ cC is
[tex]K_{c} =\frac{[C]^{c} x[D]^{d} }{[A]^{a} x[B]^{b} }[/tex]
That is, the constant Kc is equal to the multiplication of the concentrations of the products raised to their stoichiometric coefficients by the multiplication of the concentrations of the reactants also raised to their stoichiometric coefficients.
In this case, the constant Kc can be expressed as:
[tex]K_{c} =\frac{[HCl]^{2} }{[H_{2} ]x[Cl_{2} ] }[/tex]
You know that in an equilibrium mixture of HCl, Cl₂, and H₂:
the concentration of H₂ is 1.0×10⁻¹¹ [tex]\frac{mol}{L}[/tex]the concentration of Cl₂ is 2.0×10⁻¹⁰ [tex]\frac{mol}{L}[/tex]Kc=4×10¹⁸Replacing in the expression for Kc:
[tex]4x10^{18} =\frac{[HCl]^{2} }{[1x10^{-11} ]x[2x10^{-10} ] }[/tex]
Solving:
[tex]4x10^{18} =\frac{[HCl]^{2} }{2x10^{-21} }[/tex]
[tex]4x10^{18} x 2x10^{-21}=[HCl]^{2}[/tex]
[tex]8x10^{-3} =[HCl]^{2}[/tex]
[tex]\sqrt[2]{8x10^{-3}} =[HCl][/tex]
0.0894 [tex]\frac{mol}{L}[/tex]= [HCl]
Finally, the equilibrium molar concentration of HCl at 500 K is 0.0894 [tex]\frac{mol}{L}[/tex].
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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.
an expression of Avogradros law
Answer:
The formula for Avagadro's law is V1/n1 = V2/n2, where V = volume and n = amount of gas (in moles).
Explanation:
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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A chemical property is a change in _____.
density
physical state
hardness
composition of matter
Answer:
composition of an element
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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Learning Task 2: Read each statement or question below carefully and fill in the blank(s) with the best answer by choosing the words inside the box. Write your answers in a separate sheet of paper. cation 1 -ide -ine nonmetals O ion ionic compound anion metals root name 1. Any atom or molecule with a net charge, either positive or negative, is known as en 2. An atom that gains one extra electron forms an with a 1- charge. 3. A positive ion, called a is produced when one or more electrons are lost from a neutral atom. 4. Unlike a cation, which is named for the parent atom, an anion is named by taking the of the atom and changing the ending. 5. The name of each anions is obtained by adding the suffix to the root of the atom name. 6. The always form positive ions. 7. on the other hand, form negative ions by gaining electrons. 8. It is very important to remember that a chemical compound must have a net charge of
Al2(SO3)3
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
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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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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4. A system contracts from an initial volume of 15.0 L to a final volume of 10.0 L under a constant external pressure of 0.800 atm. The value of w, in J, is?
The value of the work done in joule is 405.3 J.
We know that the work done is obtained using the relation;
w = PΔV
Where;
w = work done
P = pressure
V = volume
Now, substituting values,
w = 0.800( 15.0 - 10.0)
w = 4 atm L
Since;
1 L atm = 101.325 J
4 atm L = 4 atm L × 101.325 J/1 L atm
= 405.3 J
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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????????????
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:
took the test
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.
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.
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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