MnO4 - is a stronger oxidizing agent than ReO4 - . Both ions have charge-transfer (LMCT) bands; however, the charge-transfer band for ReO4 - is in the ultraviolet, whereas the corresponding band for MnO4 - is responsible for its intensely purple color. Are the relative positions of the charge-transfer absorptions consistent with the oxidizing abilities of these ions? Explain
Answer:
[tex]$MnO^-_4$[/tex] is a strong oxidizing agent.
Explanation:
The 5d orbitals of Re are higher in energy than 3 d orbitals of Mn. So an LMCT ligand to metal charge transfer excitation requires more energy of [tex]$ReO^-_4$[/tex].
Also, since the molecular orbitals are derived primarily from 3d orbitals of [tex]$MnO^-_4$[/tex] are lower in energy than the corresponding MO's of [tex]$ReO^-_4$[/tex], [tex]$MnO^-_4$[/tex] is better able to accept the electrons.
So it is a better oxidizing agent.
The ligand to metal charge transfer band in ReO4- occurs in the near UV region hence ReO4 - appears colorless.
The electron configuration of Re is Xe 4f14 5d5 6s2 and the electron configuration of Mn is [Ar] 3d5 4s2. We can see that Mn^7+ and Re^7+ have empty d orbitals.
The color of MnO4 - must result from ligand to metal charge transfer hence the purple color of MnO4 -. In the case of ReO4 -, the ligand to metal charge transfer occurs at a much higher energy owing to the fact that 5d orbitals are involved. This transition occurs in the near UV region hence ReO4 - appears colorless. The ligand to metal charge transfer in MnO4- involves lower energy 3d orbitals hence it occurs in the visible region of the spectrum.
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Oxygen and hydrogen are both elements that are found as gases at room temperature. When oxygen combines with hydrogen, they produce the compound water according to the chemical equation below.
O2 + 2 H2 2 H2O
Water is a liquid at room temperature. This example shows that in a chemical equation, the substance that is produced
A.
has properties that are different from the original substances.
B.
can only contain a single type of element.
C.
contains fewer types of elements than the original substances.
D.
always has the same properties as the original substances.
Answer:
A
Explanation:
has properties that are different from the original substances.
How does the conflict in this passage develop a theme?
Mrs. Linde creates a conflict by promising something she cannot give to Krogstad, which develops the theme that empty promises can destroy relationships.
Krogstad continues the conflict by choosing clearing his name over having a life with Mrs. Linde, which develops the theme that you must often choose yourself over others.
Krogstad creates a conflict by deceiving Mrs. Linde about his intentions, which develops the theme that, in love, actions speak louder than words.
Mrs. Linde resolves the conflict by committing to a new life with Krogstad, which develops the theme that new beginnings are always possible.
Answer:
The answer is option D
Explanation:
This is because Mrs. Linde resolves the conflict by committing to a new life with Krogstad, which develops the theme that new beginnings are always possible.
The conflict in this passage develops a theme, as Mrs. Linde resolves the conflict by committing to a new life with Krogstad, which develops the theme that new beginnings are always possible. Option D.
What is conflict?A conflict is defined as a battle or disagreement. A dispute over parenting techniques is an example of conflict, the two parties not seeing eye to eye.
In conclusion, The passages conflict develops the theme that new beginnings are always possible.
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For the following reaction, 33.7 grams of bromine are allowed to react with 13.0 grams of chlorine gas.
bromine (g) + chlorine (g)>bromine monochloride (g)
What is the maximum amount of bromine monochloride that can be formed? __________grams
What is the FORMULA for the limiting reagent?
What amount of the excess reagent remains after the reaction is complete? __________grams
Explanation:
The reaction is given as;
Br2(g) + Cl2(g) ----> 2BrCl(g)
From the equation;
1 mol of Br2 reacts with 1 mol of Cl2
Converting the masses given to moles, using the formular;
Number of moles = Mass / Molar mass
Br2;
Number of moles = 33.7 g / 159.808 g/mol = 0.21088 mol
Cl2;
Number of moles = 13.0 g / 70.906 g/mol = 0.18334 mol
From the values;
0.18334 mol of Cl2 would react with 0.18334 mol of Br2 with an excess of 0.02754 mol of Br2
What is the maximum amount of bromine monochloride that can be formed? __________grams
1 mol of Cl2 produces 2 mol of Bromine Monochloride
0.18334 mol of Cl2 would produce x
Solving for x;
x = 0.18334 * 2 = 0.36668 mol
Converting to mass;
Mass = Number of moles * Molar mass = 0.36668 mol * 115.357 g/mol
Mass = 42.299 g
What is the FORMULA for the limiting reagent?
The limiting reagent is Cl2 as it determines the amount of product formed. The moment the reaction uses up Cl2, the reaction stops.
What amount of the excess reagent remains after the reaction is complete? __________grams
The excess reagent is Br2
The number of moles left is;
0.02754 mol of Br2
Converting to mass;
Mass = Number of moles * Molar mass = 0.02754 mol * 159.808 g/mol
Mass = 4.401 g
7. A solution containing 90grams of KNO3 per 100. grams of H2O at
50.°C is considered to be
(1) dilute and unsaturated
(2) dilute and supersaturated
(3) concentrated and unsaturated
(4) concentrated and supersaturated
Answer:
(4) concentrated and supersaturated
Explanation:
At 50.°C, 90g of KNO3 lies above the solubility curve [on the Regents Reference Table G]. This indicates that the solution is supersaturated, meaning it contains more solute than will naturally dissolve, and was formed when a saturated solution cooled. Furthermore, the percent concentration of this solution is 90% KNO3 making this solution concentrated. This can be calculated using the formula for mass percent concentration.
Percent Mass = Mass of Solute (g) x 100
Mass of Solution (g)
Given solution is considered as concentrated and supersaturated.
What is supersaturated solution?Supersaturated solutions are those solutions in which no. of dissolved solute is more and added to the saturated solution of that solute.
In the question given that,
Weight of solute or KNO₃ = 90 g
Weight of solvent or water = 100 g
Temperature = 50°C
According to the solubility curve, given solution is supersaturated because it contains excess amount of solute and it is concentrated because solute in dissolve in the saturated solution at nearly high temperature and after cooling we get the consistent solution.
Hence, option (4) is correct i.e. given solution is concentrated and supersaturated.
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Which of the following would NOT be a suitable solvent for organolithium and organomagnesium reagents?
a. CH3OCH2CH2OCH3
b. CH3CH2OH
c. THF (tetrahydrofuran)
d. CH3CH2OCH2CH3
Answer:
b
Explanation:
it could also be a, c, or d
Consider the reaction 2Al + 6HBr → 2AlBr3 + 3H2. If 12 moles of Al react with 12 moles of HBr, what is the limiting reactant?
Answer:
the limiting reactant is HBr
Explanation:
if you tried to make the products using 12 mol Al and 12 mol HBr, the HBr will run out first
What is the half life of the graphed material?
Answer:
3 hours
Explanation:
To know the the correct answer to the question given above, it is important we know the definition of half-life.
The half-life of a substance is simply defined as the time taken for half the substance to decay.
Considering the diagram given above, the initial mass of the substance is 100 g.
Half of the initial mass = 100 / 2 = 50 g
Now, we shall determine the time from the graph taken to get to 50 g.
Considering the diagram given above, the time taken to get to 50 g is 3 hours.
Therefore, the half-life of the material is 3 hours.
Hydroxylamine hydrochloride is a powerful reducing agent which is used as a polymerization catalyst. It contains 5.80 mass % H, 20.16 mass % N, 23.02 mass % O, and 51.02 mass % Cl. What is its empirical formula? Determine the molecular formula of the compound with molar mass of 278 g.
Answer: The molecular formula will be [tex]H_{16}NOCl[/tex]
Explanation:
If percentage are given then we are taking total mass is 100 grams.
So, the mass of each element is equal to the percentage given.
Mass of H = 5.80 g
Mass of N = 20.16 g
Mass of O = 23.02 g
Mass of Cl = 51.02 g
Step 1 : convert given masses into moles.
Moles of H =[tex]\frac{\text{ given mass of H}}{\text{ molar mass of H}}= \frac{5.80g}{1g/mole}=5.80moles[/tex]
Moles of N =[tex]\frac{\text{ given mass of N}}{\text{ molar mass of N}}= \frac{20.16g}{14g/mole}=1.44moles[/tex]
Moles of O =[tex]\frac{\text{ given mass of O}}{\text{ molar mass of O}}= \frac{23.02g}{16g/mole}=1.44moles[/tex]
Moles of Cl =[tex]\frac{\text{ given mass of O}}{\text{ molar mass of O}}= \frac{51.02g}{35.5g/mole}=1.44moles[/tex]
Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.
For H = [tex]\frac{5.80}{1.44}=4[/tex]
For N = [tex]\frac{1.44}{1.44}=1[/tex]
For O = [tex]\frac{1.44}{1.44}=1[/tex]
For Cl = [tex]\frac{1.44}{1.44}=1[/tex]
The ratio of H: N: O: Cl= 4: 1: 1: 1
Hence the empirical formula is [tex]H_4NOCl[/tex]
The empirical weight of [tex]H_4NOCl[/tex] = 4(1)+1(14)+ 1(16) + 1(35.5)= 69.5 g.
The molecular weight = 278 g/mole
Now we have to calculate the molecular formula.
[tex]n=\frac{\text{Molecular weight }}{\text{Equivalent weight}}=\frac{278}{69.5}=4[/tex]
The molecular formula will be=[tex]4\times H_4NOCl=H_{16}NOCl[/tex]
WHEN YOU SEE A BLUE CAR WHAT COLER IS BEING REFLECTED
Answer:
violet
Explanation:
just violet
oh and you spelled "COLER" wrong, its color or colour if you live somewhere else
What might an organism do if there is a change to it's ecosystem?
A. Move
B. Die
C. Survive
D. All of these
Answer:
I would say C) Survive
An organism will survive, move or die if there is a change to its ecosystem. Therefore, option (D) is correct.
What is an ecosystem?An ecosystem contains all the organisms and the physical environment with which they interact. The biotic and abiotic components are connected together through energy flows and nutrient cycles. Energy injects into the system through photosynthesis into plant tissue.
Animals play an important role by feeding the plants in the movement of matter and energy through the system. Decomposers release carbon back into the atmosphere and facilitate nutrient cycling by breaking down dead organic matter.
External and internal factors control the ecosystem. External factors such as climate control the overall structure of an ecosystem but are not influenced by the ecosystem. Internal factors are controlled such as decomposition, shading, root competition, disturbance, succession, and the types of species present.
The change in the ecosystem makes the organism to adapt the environment in order to survive, move to a new place or die.
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Gas law practice problems
LE
gas
A 3.0L sample of carbon dioxide gas at 155 kPa is injected into a 5.0 L vessel. What is the new pressure of the
gas
Answer:
93 kPa
Explanation:
Step 1: Given data
Initial volume of carbon dioxide gas (V₁): 3.0 LInitial pressure of carbon dioxide gas (P₁): 155 kPaFinal volume of carbon dioxide gas (V₂): 5.0 LFinal pressure of carbon dioxide gas (P₂): ?Step 2: Calculate the final pressure of carbon dioxide gas
If we assume carbon dioxide behaves as an ideal gas, we can calculate the final pressure of carbon dioxide gas using Boyle's law.
P₁ × V₁ = P₂ × V₂
P₂ = P₁ × V₁/V₂
P₂ = 155 kPa × 3.0 L/5.0 L = 93 kPa
What is the molar mass of Ammonium Carbonate?
Explanation:
Molar mass
96.09 g/mol
glad to help....
which one of the following are pure substances : water,salt water,copper,brass,air,oxygen
Answer:
Copper, Water, and Oxygen
Explanation:
Because they are compounds and compounds are pure substances <3
Answer: copper
Explanation: because all of the atoms that make up copper are copper i.e its not a compound or mixture. not made od molecules
Using the van der Waals equation, determine the pressure exerted by 4.30 mol Ar in 3.6 L at 325K.
Answer:
37.7 atm
Explanation:
Using the relation;
(P + an^2/V^2) (V - nb) = nRT
(P + an^2/V^2) = nRT/(V - nb)
a = 0.0341 atm dm^2 Mol^2
b = 0.0237 dm/mol
P = nRT/(V - nb) - an^2/V^2
P = [4.3 * 0.082 * 325 / (3.6 - (4.3 * 0.0237))] - (0.0341 * (4.3^2))/(3.6^2)
P = 114.595/(3.498) - 0.0487
P = 37.7 atm
Enter the electron configuration for I+ using noble gas shorthand notation.
In the first box enter the noble gas (notice the brackets). In the following boxes enter the number that goes in front of the orbital followed by the superscript.
For example, the electron configuration for sulfur is: [Ne]3s2 3p4
so the first box would have Ne in it followed by 3, then 2, then 3 then 4.
If you do not need an orbital, just enter 0 (zero) in the boxes for the coefficient and superscript.
[ ] s f d p
Find the element in the periodic table and count over
to the right the number of negative charges on your anion.
This element has the same electron configuration as your anion.
Which noble gas precedes the element? Knowing the s, p, d, and f blocks
of elements in the periodic table, deduce the the electron configuration
of the element from the preceding noble gas. Remember that the
1p 1d, 2d, 1f, 2f, and 3f orbitals are forbidden energy levels (they do not exist).
Answer:
[Kr] 4d10 5s2 5p4
Explanation:
The Symbol I represents Iodine. It has atomic number of 53. The full electronic configuration is given as;
1s2 2s2 2p6 3s2 3p6 3d10 4s2 4p6 4d10 5s2 5p5
However the question requested for the configuration of I+.
I+ is a cation and it simply refers to an iodine atom that has lost a single electron. The electronic configuration of I+ is given as;
1s2 2s2 2p6 3s2 3p6 3d10 4s2 4p6 4d10 5s2 5p4
Using Noble gas shorthand representation, we have;
[Kr] 4d10 5s2 5p4
a log burns to ashes in a fireplace what kind of change is this
Answer:
A log burning to ashes is a chemical change.
Explanation:
Answer:
Chemical Change
Explanation:
Burning of wood is a chemical change as new substances which cannot be changed back [e.g. carbon dioxide] are formed. For example, if wood is burned in a fireplace, there is not wood anymore but ash.
Consider the reaction: S(s) O2(g)SO2(g) Write the equilibrium constant for this reaction in terms of the equilibrium constants, Ka and Kb, for reactions a and b below: a.) 2 S(s) 3 O2(g) 2 SO3(g) Ka b.) SO2(g) 1/2 O2(g) SO3(g) Kb
Answer:
[tex]Ka=\frac{[SO_3]^2}{[O_2]^3} \\\\Kb=\frac{[SO_3]^3}{[SO_2][O_2]^{1/2}}[/tex]
Explanation:
Hello!
In this case, according to the reactions:
a.) 2 S(s) 3 O2(g) ⇔ 2 SO3(g) Ka
b.) SO2(g) + 1/2 O2(g) ⇔ SO3(g) Kb
Thus, according to the law of mass action, we can write Ka and Kb as follows:
[tex]Ka=\frac{[SO_3]^2}{[O_2]^3} \\\\Kb=\frac{[SO_3]^3}{[SO_2][O_2]^{1/2}}[/tex]
Whereas solid carbon is not inserted in the equilibrium expression.
Best regards!
Methyl isocyanate, shown as resonance structure 1, can also be represented by other resonance structures. Draw the next most important resonance contributor. Then add curved arrows to each structure to show delocalization of electron pairs to form the other structure.
Include lone pairs of electrons, formal charges, and hydrogen atoms. You can add condensed hydrogens using the More menu, selecting +H and clicking on the carbon as many times as needed.
Solution :
Structure I
The formal charge on both Carbon (C) atom is = 4 valance [tex]$e^-$[/tex] - bonds = 0
Formal charge (O) = 6 V.E - 2 bonds - 4 non bonding electrons = 0
Formal charge on (N) = 5 V.E - 3 bonds - 2 non bonding electrons = 0
F.C. on H = 1 V.E. - 1 bond = 0
Overall charge on the molecule = 0 charge
Structure II
Formal charge on both C atom = 4 valence [tex]$e^-$[/tex] - 4 bonds = 0
Formal charge (O) = 6 V.E. - 1 bonds - 6 non bonding electrons = -1 charge
Formal charge on (N) = 5 V.E. - 4bonds - 0 non bonding electrons = +1 charge
F.C on H = 1 V.E. - 1 bond = 0
Overall charge on the molecule = +1 -1
= 0 charge
If a solution containing 24.0 g of a substance reacts by first-order kinetics, how many grams remain after three half-lives?
Answer:
3.0g remain
Explanation:
The half-life is defined as the time required for a reactant to decrease its concentration in exactly the half of the initial amount of reactant. Having this in mind:
In one half-life, the mass will be:
24.0g / 2 = 12.0g
In two half-lifes:
12.0g / 2 = 6.0g
And in three half-lifes, the mass that remain is:
6.0g / 2 =
3.0g remainA chemistry student is given 650. mL of a clear aqueous solution at 33.°C. He is told an unknown amount of a certain compound X is dissolved in the solution. The student allows the solution to cool to 17.9 0C. At that point, the student sees that a precipitate has formed. He pours off the remaining liquid solution, throws away the precipitates, and evaporates the water from the remaining liquid solution under vacuum. More precipitate forms. The student washes, dries and weighs the additional precipitate. It weighs 0.150 kg.
1. Using only the information above, can you calculate the solubility of X in water at 17 0C.
2. If you said yes, calculate it.
Answer:
No, you can not calculate the solubility of X in water at 17 0C.
Explanation:
Solubility refers to the amount of a substance that dissolves in 1000 L of water.
To calculate the solubility of a solute in water, all the water is evaporated and the solid is carefully collected, washed, dried and weighed. The mass of solid obtained can now be used to calculate the solubility of the solute in water as long as there was no loss in mass of solid during the experiment.
In this case, the student threw away part of the solid that precipitated. As a result of this, the mass of solid obtained at the end of the experiment is not exactly the total mass of solute that dissolved in the solvent. Hence, the solubility of X in water at 17 0C can not be accurately calculated.
Solid nickel is added to aqueous iron nitrate. Using the metal activity series, which chemical equation reflects this reaction and its outcome? Question 5 options: A) 3Ni(NO3)2 (aq) + 2Fe (s) → NR B) Ni (s) + Fe(NO3)3 (aq) → NR C) 3Ni(NO3)2 (aq) + 2Fe (s) → 3Ni (s) + 2Fe(NO3)3 (aq) D) 3Ni (s) + 2Fe(NO3)3 (aq) → 3Ni(NO3)2 (aq) + 2Fe (s)
Answer:
3 Ni + Fe(NO₃)₃ --> 3 Ni(NO₃) + Fe
Explanation:
In this question, the reactants are;
Solid Nickel and Iron nitrate. This is represented as;
Ni and Fe(NO3)3
The equation of the reaction is given as;
3 Ni + Fe(NO₃)₃ --> 3 Ni(NO₃) + Fe
The option that closely matches this is;
Option D. (Although the reaction is not balanced)
& Oregonians eat about 9503 metric tons of food each day. What is this consumption rate in grams
per second?
Answer:
109988 grams per second
Explanation:
To solve this problem first we convert 9503 metric tons into grams, keeping in mind that:
1 metric ton = 1000 kg1 kg = 1000 gMeaning that:
9503 metric ton * [tex]\frac{1000kg}{1metricTon}*\frac{1000g}{1kg}[/tex] = 9503x10⁶ gThen we calculate how many seconds are there in one day:
1 day * [tex]\frac{24h}{1day} *\frac{60min}{1h} * \frac{60s}{1min}[/tex] = 86400 sFinally we calculate the consumption rate:
9503x10⁶ g / 86400 s = 109988 g/s03
A force of 20 N acts upon a 5 kg block. Calculate the acceleration of the object.
In the laboratory, a general chemistry student measured the pH of a 0.328 M aqueous solution of acetylsalicylic acid (aspirin), HC9H7O4 to be 1.987. Use the information she obtained to determine the Ka for this acid.
Answer: [tex]K_a[/tex] for the acid is [tex]3.34\times 10^{-4}[/tex]
Explanation:
[tex]HC_9H_7O_4\rightarrow H^+C_9H_7O_4^-[/tex]
cM 0 0
[tex]c-c\alpha[/tex] [tex]c\alpha[/tex] [tex]c\alpha[/tex]
Give c = 0.328 M and [tex]pH=1.987[/tex]
[tex]1.987=-log[H^+][/tex]
[tex][H^+]=0.0103[/tex]
[tex][H^+]=c\times \alpha[/tex]
[tex]0.0103=0.328\times \alpha[/tex]
[tex]\alpha=0.0314[/tex]
So dissociation constant will be:
[tex]K_a=\frac{(c\alpha)^{2}}{c-c\alpha}[/tex]
Putting in the values we get:
[tex]K_a=\frac{(0.328\times 0.0314)^2}{(0.328-0.328\times 0.0314)}[/tex]
[tex]K_a=3.34\times 10^{-4}[/tex]
why is plastic a better choice for the slide than aluminum?
Answer:
Plastic is better because it doesn't melt in the microwave but aluminum burns in the microwave
Explanation:
Starting with 5.00 g barium chloride n hydrate yields 4.26 g of anhydrous barium chloride after heating. Determine the integer n.
Answer:
BaCl₂·(H₂O)₂
Explanation:
The reaction that takes place is:
BaCl₂·(H₂O)ₙ → BaCl₂ + nH₂OThe information given by the problem tells us that 5.00 g of reactant were consumed. Of those 5.00 g, 4.26 g were converted into BaCl₂, this means that the other 0.74 g were converted into water.
We convert 0.74 g of water into moles, using its molar mass:
0.74 g H₂O ÷ 18 g/mol = 0.041 mol H₂OThen we convert 4.26 g of BaCl₂ into moles:
4.26 g BaCl₂ ÷ 208.23 g/mol = 0.0204 mol BaCl₂We can write these results as (BaCl₂)₀.₀₂·(H₂O)₀.₀₄. We multiply those coefficients by 50 in order to make them integers, and we're left with:
BaCl₂·(H₂O)₂No pain no gain . which figure of speech is this
Answer:
No pain, no gain is a proverb that means in order to make progress or to be successful, one must suffer. This suffering may be in a physical or mental sense. The phrase no pain, no gain was popularized in the 1980s by the American actress, Jane Fonda.
Consider a 0.70 M solution of HOCl. If the molarity was decreased to 0.3 M, what would happen to the percent dissociation?
Answer:
The percent dissociation would increase
Explanation:
Percent dissociation is the ratio of amount dissociated to the initial concentration.
Decreasing the concentration of the solution leads to an increase in percent dissociation.
What is the concentration of a Kl solution of 20.68 g of solute was dissolved in enough water to form 100. ml of solution?
Answer:
1.25M
Explanation: