Answer:
See the explanation
Explanation:
Our reactions for this question are:
a) Cr+3(aq) + 3e- → Cr(s) ________
b) Ag(s) → 1e- + Ag+1(ag) ________
c) Mn(s) → 5e- + Mn+5(aq) ________
d) Sn+2(aq) + 2e- → Sn(s) ________
Now, we have to remember the definition of oxidation and reduction. In a reduction reaction, the substance will gain electrons, in other words: "The electrons would be placed on the left side, in the side of the reagents. In an oxidation reaction, the substance will lose electrons, in other words: "The electrons would be placed on the right side, in the side of the products".
With this in mind we will have:
a) Cr+3(aq) + 3e- → Cr (s)(Reduction)
b) Ag(s) → 1e- + Ag+1(ag) (Oxidation)
c) Mn(s) → 5e- + Mn+5(aq) (Oxidation)
d) Sn+2(aq) + 2e- → Sn(s) (Reduction)
I hope it helps!
why homoannular diene in uv spectrometry have higher wavlenth than hetroannular diene?
Answer:
Homoanular dienes have a greater base value than heteroanular dienes
Explanation:
Woodward in 1945 gave a set of rules relating the wavelength of maximum absorption to the structure of a compound. These rules were modified by Fieser in 1959. These sets of rules describe the absorption of organic molecules in the UV region of the electromagnetic spectrum.
Each system of diene or triene has a given fixed value at which maximum absorption is expected to occur according to Woodward rules. This given fixed value is called the base or parent value. If the two double bonds are trans to each other, the diene is said to be transoid. If the two double bonds belong to different rings, the system is said to be heteroanular and the base value in each case is 215nm. If the double bonds are cis to each other (cisoid), or the two double bonds are in the same ring (homoanular), then the base value is 253nm.
Since λmax = base value + ∑ substituent contributions + ∑ other contributions, if the other contributions are not very significant, homoanular diene will have a greater λmax because of its larger base value compared to heteroanular diene. This correlates well with the fact that conjugated systems absorb at a longer wavelength.
Consider the three statements below. Which numbered response contains all the statements that are true and no false statements?
I. Hydration is a special case of solvation in which the solvent is water.
II. The oxygen end of water molecules is attracted toward Ca2+ ions.
III. The hydrogen end of water molecules is attracted toward Cl- ions.
a) I, II, and III
b) I and II
c) III
d) I
e) II
Answer:
a) I, II, and III
Explanation:
For the first statement;
Solvation, is the process of attraction and association of molecules of a solvent with molecules or ions of a solute. if the solvent is water, we call this process hydration.
This means the statement is TRUE.
For the second statement;
The negatively-charged side of the water molecules are attracted to positively-charged ions. In the case of water, the oxygen end is the negatively charged side of water. This means the statement is TRUE.
For the third statement;
The positively-charged side of the water molecules are attracted to the negatively-charged chloride ions. In the case of water, the hydrogen end is the positively charged side of water. This means the statement is TRUE.
Going through the options, we can tell that the correct option is option A.
The following skeletal oxidation-reduction reaction occurs under basic conditions. Write the balanced OXIDATION half reaction.
Bi(OH)3 + NO2 → Bi + NO3-
Answer:
[tex]N^{4+}O_2+2OH^-\rightarrow (N^{5+}O_3)^-+1e^-+H_2O[/tex]
Explanation:
Hello,
In this case, for the given reaction, we first start by the writing of the oxidation states of all the involved elements:
[tex]Bi^{3+}(OH)^-+N^{4+}O^{2-}_2\rightarrow Bi^0+(N^{5+}O^{2-}_3)^-[/tex]
In such a way, we are noticing nitrogen is undergoing an increase in its oxidation state, therefore it is being the oxidized species, for which the oxidation half reaction, should be (considering basic conditions):
[tex]N^{4+}O_2+H_2O+2OH^-\rightarrow (N^{5+}O_3)^-+1e^-+2H^++2OH^-\\\\N^{4+}O_2+H_2O+2OH^-\rightarrow (N^{5+}O_3)^-+1e^-+2H_2O\\\\N^{4+}O_2+2OH^-\rightarrow (N^{5+}O_3)^-+1e^-+H_2O[/tex]
Best regards.
Consider the equilibrium reaction. 4A+B↽−−⇀3C After multiplying the reaction by a factor of 2, what is the new equilibrium equation?
Aqueous calcium chloride reacts with aqueous potassium carbonate in a double-displacement reaction. Write a balanced equation to describe this reaction. Include states of matter in your answer. Click in the answer box to open the symbol palett
Answer: [tex]CaCl_2(aq)+K_2CO_3(aq)\rightarrow 2KCl(aq)+CaCO_3(s)[/tex]
Explanation:
A double displacement reaction is one in which exchange of ions take place. The salts which are soluble in water are designated by symbol (aq) and those which are insoluble in water and remain in solid form are represented by (s) after their chemical formulas.
The balanced reaction between aqueous calcium chloride reacts with aqueous potassium carbonate is shown as:
[tex]CaCl_2(aq)+K_2CO_3(aq)\rightarrow 2KCl(aq)+CaCO_3(s)[/tex]
Calculate the theoretical percentage of water for the following hydrates.
(a) manganese(II) monohydrate, MnSO4 H2O
(b) manganese(II) tetrahydrate, MnSO4 4H2O
Answer:
(a) [tex]\% H_2O=10.65\%[/tex]
(b) [tex]\% H_2O=32.2[/tex]
Explanation:
Hello.
For this questions we must consider the ratio of the molar mass of water to hydrated compound molar mass as shown below:
(a) In this case, we can consider that inside the manganese (II) sulfate monohydrate, whose molar mass is 169.02 g/mol, there is one water molecule that has a molar mass of 18 g/mol, for which the theoretical percentage of water is:
[tex]\% H_2O=\frac{18g/mol}{169.0g/mol} *100\%\\\\\% H_2O=10.65\%[/tex]
(b) In this case, we can consider that inside the manganese (II) sulfate tetrahydrate, whose molar mass is 223.1 g/mol, there are four water molecules that have a molar mass of 4*18 g/mol, for which the theoretical percentage of water is:
[tex]\% H_2O=\frac{4*18g/mol}{223.1g/mol} *100\%\\\\\% H_2O=32.27\%[/tex]
Best regards.
how many moles of a solute is present in 4.00L of an 8.30M solution
Answer:
The number of moles of solute present in 4.00 L of an 8.30 M solution is 33.2
Explanation:
The Molarity (M) or Molar Concentration is the number of moles of solute per liter of solution; in other words it is the number of moles of solute that are dissolved in a given volume.
The molarity of a solution is calculated by dividing the moles of the solute by the volume of the solution:
[tex]Molarity=\frac{number of moles of solute}{volume}[/tex]
Molarity is expressed in units ([tex]\frac{moles}{liter}[/tex]) or M.
In this case:
molarity= 8.30 Mnumber of moles of solute= ?volume= 4.00 LReplacing:
[tex]8.30 M=\frac{number of moles of solute}{4 L}[/tex]
Solving:
number of moles of solute= 8.30 M* 4 L= 8.30 [tex]\frac{moles}{liter}[/tex] * 4 L
number of moles of solute =33.2
The number of moles of solute present in 4.00 L of an 8.30 M solution is 33.2
Answer:
33.2 is the answer
Explanation:
did the test already :)
. A compound X, containing C, H, and O was found to have a relative molar mass of 6o
whilst 20.0g of X contained 8.0g of C and 1.33g of H. Calculate the empirical formula
of compound X, and hence determine its molecule formula. show calculation
Answer:
empirical formula = CH2O
molecular formula = C2H4O2
Define dew point in complete sentences
half-reaction for the reduction of liquid water to gaseous hydrogen in basic aqueous solution
Answer:
2H₂O (liq) + 2e⁻⇒ H₂ (g) + 2OH⁻ (aq)
Explanation:
In reduction-oxidation reaction two reactions take place, one is oxidation and the other is reduction reaction. In an oxidation reaction, there is the loss of an electron whereas in the reduction reaction there is gain of electron occus.
Reduction reaction occurs on the cathode, in a reduction of water there is gain of 2 electrons to gaseous hydrogen in basic aqueous solution. half-reaction for the reduction of liquid water to gaseous hydrogen in basic aqueous solution-
2H₂O (liq) + 2e⁻⇒ H₂ (g) + 2OH⁻ (aq)
There are five constitutional isomers with the molecular formula C6H14. When treated with chlorine at 300°C, isomer A gives a mixture of two monochlorination products. Under the same conditions, isomer B gives a mixture of five monochlorination products, isomer C gives four monochlorination products, and isomer D gives a mixture of three monochlorination products. From this information, draw the structural formula of isomer D.
Answer:
Pentane or 2,2-dimethylbutane
Explanation:
I've numbered the isomeric hexanes from 1 to 5 and labelled the sets of equivalent hydrogens.
The results are
Isomer 1— three sets of equivalent hydrogens
Isomer 2— five sets of equivalent hydrogens
Isomer 3— four sets of equivalent hydrogens
Isomer 4— two sets of equivalent hydrogens
Isomer 5— three sets of equivalent hydrogens
Each set will give one monochloro substitution product.
4 = A. Two monochloro isomers.
2 = B. Five monochloro isomers.
3 = C. Four monochloro isomers.
Isomers 1 and 5 each give three monochloro isomers.
Thus, we cannot assign Structure D definitively.
D is either pentane or 2,2-dimethylbutane.
How are sodium (Na) and potassium (K) similar?
O
A. They rarely react.
B. They are hard metals.
O O
C. They are highly reactive.
D. They are soft nonmetals.
Answer:
They are also relatively soft metals: sodium and potassium can be cut with a butter knife.
Answer:A
Explanation:
warming oceans temperatures directly lead to all of the following except A sea level rising B coral bleaching C ocean deoxigenation D pollution
Answer:
D. pollution
Explanation:
Increase in ocean temperature is one of the major consequence of global warming which directly leads to rise in sea level, coral bleaching and ocean deoxygenation.
Warming ocean temperatures do not leads to pollution directly whereas pollution leads to warming ocean temperatures. So, in the case of pollution, the effect is opposite.
Hence, the correct option is D.
Identify the type of solid for AgCl. Identify the type of solid for AgCl. metallic atomic solid nonbonding atomic solid molecular solid ionic solid networking atomic solid
Silver chloride (AgCl) is a crystalline solid substance that is composed of silver and chloride ions. AgCl is an ionic solid. Thus, option D is correct.
What are ionic solids?Ionic solids are substances that show the properties of solid matter and have ionic, positive, and negative charges in them. They are linked together by the attraction of the opposite charges.
The silver metal in the molecule has a positive charge and the chloride ions are negative in charge making them establish an ionic bond.
The solid molecule is held together by ionic bonds and not the covalent or other metallic bonds. The cations and anions of AgCl are linked together by the electrostatic forces that make their structure appear strong and brittle.
Therefore, AgCl has been known as an ionic solid.
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Suppose you have a solid that looks like gold but you believe it to be fool’s gold. The mass of the solid is 23.5 grams. When the solid is lowered into a graduated cylinder the water level rises from 47.5 to 52.2 mL. Is the simple fool’s gold
Answer:
The sample is fool's gold
Explanation:
Density is defined as the ratio between mass in grams and volume in mililiters.
A sample of pure gold has a density of 19.3g/mL.
Using Archimedes' principle, the volume of the sample is:
52.2mL - 47.5mL = 4.7mL
As the mass of the sample is 23.5g, the density is:
23.5g / 4.7mL = 5g/mL
The denisty of the sample is very different to density of pure gold, that means:
the sample is fool's gold
How many liters of a 1 M NaOH stock solution would you need to make 785 mL of a 215 mM NaOH dilution? (m.w. = 40.00 g/mol) Do not include units in your answer; report your answer in the requested units.
Answer:
0.1688L of the 1M NaOH stock solution
Explanation:
A 215mM = 0.215M solution of NaOH contains 0.215 moles per liter. As you want to prepare 785mL = 0.785L of the 0.215M you will need:
0.785L × ( 0.215mol / L) = 0.1688 moles of NaOH.
These moles of NaOH comes from the 1M stock solution, that means the volume of 1M NaOH solution you need is:
0.1688 moles NaOH × (1L / 1mol) =
0.1688L of the 1M NaOH stock solutionThe volume of the stock solution of 1 M NaOH stock solution needed to make 785 mL of a 215 mM NaOH dilution is 0.169 L
From the question given above, the following data were obtained:
Molarity of stock (M₁) = 1 M
Volume of diluted solution (V₂) = 785 mL = 785 / 1000 = 0.785 L
Molarity of diluted solution (M₂) = 215 mM = 215 / 1000 = 0.215 M
Volume of stock solution needed (V₁) =?
The volume of the stock solution needed can be obtained as follow:
M₁V₁ = M₂V₂1 × V₁ = 0.215 × 0.785
V₁ = 0.169 LTherefore, the volume of the stock solution needed is 0.169 L
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MnBr2 molecular or ionic:
name:
molecular or ionic
Answer: Manganese(II)bromide
Explanation:
Because it is a chemical compound composed of manganese and bromine
A buffer solution is all of the following EXCEPT: a solution that contains both a weak acid and its conjugate base. a solution that resists a change in pH when a base is added. a solution that regulates pH because it is such a strong acid or base. a solution that resists a change in pH when an acid is added. All of the above are true.
Answer:
A SOLUTION THAT REGULATES pH BECAUSE IT IS SUCH A STRONG ACID OR BASE
Explanation:
A buffer solution is an aqueous solution consisting of a weak acid and its conjugate base. It is an aqueous solution used to keep the pH of solution at a nearly constant value in various chemical processes. It resists change in pH when either a strong acid or a strong base is added. So it is very essential in various chemical applications and even in the human body as the blood pH is kept in nearly constant value by the bicarbonate buffer system in conjunction with the kidneys. The buffer solution is able to keep this nearly constant range of values because of the equilibrium between the weak acid and its conjugate base. So therefore, the incorrect statement in the options is that buffer solution is a solution that regulates pH because it is such a strong acid or base. The other options are correct.
Which of the following best identifies where long-range order would be found?
ОООО
in amorphous solids
in crystalline solids
in thermal plasmas
in nonthermal plasmas
Answer:
in crystalline solids
Hope this answers your question, good luck
The crystalline solids represent the best identification where the long-range order should be found.
What are crystalline solids?Crystalline solids refer to the solid where the atoms, molecules should be make the arrangement. The smallest & repeated pattern of this solid should be called as the unit cell. The unit cell should be treated as the bricks in the wall which means it should be homogenous in the nature and repeated
Therefore, The crystalline solids represent the best identification where the long-range order should be found.
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Arrange the following set of atoms in order of decreasing atomic size: Sn, I, Sr
Which atom has the largest atomic size?
a) Sn
b) I
c) Sr
Which atom has the smallest atomic size?
a) Sn
b) I
c) Sr
What is the electron configuration for N (nitrogen)?
A. 1521p5
B. 1s22s23s 3p
O cly 1s22s22p3
O D. 1s 2s22p 3s
SUBMIT
Answer:
[tex]1s^22s^22p^3[/tex]
Explanation:
Nitrogen has the atomic number = 7
So, No. of electrons = 7
Electronic Configuration:
[tex]1s^22s^22p^3[/tex]
Remember that:
s sub shell holds upto 2 electrons while p sub shell upto 6
A student has a sample of CaSO4 hydrate and it weighs 0.4813 grams. He heats it strongly to drive off the water of hydration, and after subsequent heatings, the student finds the anhydrous compound has a constant mass of 0.3750 grams. Find the formula of the hydrate.(2 points)
Answer:
CaSO4•3H2O.
Explanation:
Let the compound be CaSO4.xH2O.
The following data were obtained from the question:
Mass of hydrated compound (CaSO4.xH2O) = 0.4813g
Mass of anhydrous compound (CaSO4) = 0.3750g
Next we shall determine the mass of the water is the hydrated compound.
This is illustrated below:
Mass of water = mass of hydrated – mass of anhydrous.
Mass of water = 0.4813 – 0.3750
Mass of water = 0.1063g
Next, we shall determine the number of mole of the anhydrous compound and the number of mole of the water present in the compound. This is illustrated below:
Molar mass of anhydrous CaSO4 = 63.5 + 32 + (16x4) = 159.5g/mol
Mass of anhydrous CaSO4 = 0.3750g
Mole of anhydrous CaSO4 =...?
Mole = mass /Molar mass
Mole of anhydrous CaSO4 = 0.3750/159.5 = 2.35×10¯³ mole
Molar mass of H2O = (2x1) + 16 = 18g/mol
Mass of H2O = 0.1063g
Number of mole of H2O =.?
Mole = mass /Molar mass
Mole of H2O = 0.1063/18 = 5.91×10¯³ mole
Next we shall determine the ratio of number of mole of anhydrous CaSO4 to that of H2O. This is illustrated below:
Mole of anhydrous CaSO4 = 2.35×10¯³ mole
Mole of H2O = 5.91×10¯³ mole
Ratio of anhydrous CaSO4 to H2O =>
CaSO4 : H2O => 2.35×10¯³ /5.91×10¯³
CaSO4 : H2O => 1 : 3
Therefore, for 1 mole of the anhydrous CaSO4, there are 3 moles of H2O.
Therefore, the formula for the hydrate compound CaSO4.xH2O => CaSO4•3H2O.
What causes the electrons to flow through a wire in a lead-acid battery?
Answer:
The potential difference between the cathode and anode are set up from the chemical reaction. Inside the battery electrons are pushed by the chemical reaction toward the positive end creating a potential difference. It is this potential difference that drives the electrons through the wire.
What is T2, if T1= 500 k, v1=10L, V2=8L,P1=600 torr,P2=200 torr?
Answer:
T2 = 133.333°K
Explanation:
Using Combined Gas Laws:
(600 torr)(10L)/500°K = (200 torr)(8L)/x°K
[tex]\frac{600 torr(10L)}{500K} =\frac{200 torr(8L)}{xK}[/tex]
Cross multiply:
x°K (600 torr)(10L) = 500°K(200 torr)(8L)
Divide:
x°K = (500°K(200 torr)(8L))/(600 torr)(10L)
[tex]xK = \frac{500K(200 torr)(8L)}{600 torr(10L)}[/tex]
x = 400/3°K or 133.333°K
A piece of bismuth with a mass of 22.5 g is submerged in 46.3 cm3 of water in a graduated cylinder. The water level increases to 48.6 cm3. What is the density of indium to the proper number of significant figures
Answer:
[tex]\rho = 9.78\frac{g}{cm^3}[/tex]
Explanation:
Hello,
In this case, in order to compute the density of bismuth we need to apply the formula:
[tex]\rho =\frac{m_{Bi}}{V_{Bi}}[/tex]
Nonetheless, the volume is computed by the difference:
[tex]V_{Bi}=48.6-46.3=2.30cm^3[/tex]
Therefore:
[tex]\rho = \frac{22.5g}{2.30cm^300}\\\\\rho = 9.78\frac{g}{cm^3}[/tex]
Regards.
What is science?
O A. Attempting to solve ethical problems through research
O B. Learning about the physical world through observation
O C. Determining a body of knowledge that never changes
O D. Using rules and patterns to predict what humans will do
SUBMIT
Answer:
The answer is B
Explanation:
Answer:
B. Learning about the physical world through observation
Explanation:
A p e x 2021 :) Trust me!
Need help with chemistry question
Answer:
See explanation
Explanation:
In this case, we have to check two variables:
1) The leaving group
2) The carbon bonded to the leaving group.
Let's check one by one:
2-chloro-3-methylbutane
In this molecule, the leaving group is "Cl", the carbon bonded to the leaving group has two neighbors. Therefore, we have a secondary substrate.
1-phenylpropan-1-ol
In this molecule, the leaving group is "OH", the carbon bonded to the hydroxyl group has two neighbors also. So, we have a secondary substrate.
(E)-pent-3-en-2-yl 4-methylbenzenesulfonate
In this case, the leaving group is "OTs" (Tosylate), the carbon bonded to the tosylate group has as a neighbor a double bond. Therefore, we have an allylic substrate.
3a-bromooctahydro-1H-indene
In this molecule, the leaving group is "Br", the carbon bonded to the bromine has three neighbors. So, we have a tertiary substrate.
1-iodo-3-methylbutane
In this molecule, the leaving group is "I", the carbon bonded to the iodide has only one neighbor. So, we have a primary substrate.
See figure 1
I hope it helps!
A sample of gold weighs 1.2 oz. The sample is pounded into a thin rectangular sheet with an area of 400. sq. ft. The density of gold is 19.3 grams per cm3. What is the thickness of the foil in centimeters?Given: 28.35 g = 1 oz, 1 ft = 12 inches, 1 inch = 2.54 cm
Answer:
thickness of the gold sheet = 4.74 * 10⁻⁶ cm
Explanation:
mass of gold sample = 1.2 oz,; area of rectangular gold sheet = 400 sq. ft
Converting mass of gold from oz. to g
1 oz. = 28.35 g
mass of gold sample in grams = 1.2 * 28.35 g = 34.02 g
Converting from feet to cm;
1 feet = 12 * 2.54 cm = 30.48 cm
1 sq. ft = (30.48)² = 929.0304 cm²
area of gold in cm² = 400 * 929.0304 cm² = 371612.16 cm²
Since the density of a solid is constant
Density = mass/volume
Volume = mass/density
where volume =area * thickness
therefore, area * thickness =mass/density
thickness = mass/(density * area)
substituting the value; thickness = 34.02 g/(19.3 gcm⁻³ *371612.16 cm²)
thickness of the gold sheet = 4.74 * 10⁻⁶ cm
A compound has a mass percentage of 53.46% C, 6.98% H, and 39.56% O. What is the empirical formula for this compound
Answer:
The empirical formula is: C₂H₃O
Explanation:
The empirical formula, also known as the “minimum formula”, is the simplest expression to represent a chemical compound and indicates the elements that are present and the minimum integer ratio between its atoms.
The percentage composition is the percentage by mass of each of the elements present in a compound.
Having 100 g of the compound as a base, it is possible to express the percentages in grams. That is, assuming you have 100 g of the compound, you have 53.46 g of C , 6.98 g of H , and 39.56 g of O.
Taking into account the molecular mass of each substance, the number of relative atoms of each chemical element is calculated:
C: [tex]53.46 g *\frac{1 mol}{12.01 g } = 4.45 moles[/tex]
H:[tex]6.98 g *\frac{1 mol}{1.01 g } = 6.91 moles[/tex]
O:[tex]39.56 g *\frac{1 mol}{16g } = 2.47 moles[/tex]
Now you divide each value obtained by the least of them:
C: [tex]\frac{4.45 moles}{2.47 moles}= 1.80[/tex]
H:[tex]\frac{6.91 moles}{2.47 moles}= 2.8[/tex]
O:[tex]\frac{2.47 moles}{2.47 moles}=1[/tex]
Decimals approach the nearest integer, then:
C: 2
H: 3
O: 1
Therefore the empirical formula is: C₂H₃O
Answer: The other guy is wrong. The correct answer is C9H14O5
Explanation:
et the mass percentages from the percent composition represent grams in a total mass of 100g. Use the grams to calculate the number of moles of each atom present.
%C:53.46%molC=53.46gC=53.46gC×1molC12.011gC=4.451molC
%H:6.98%molH=6.98gH=6.98gH×1molH1.008gH=6.924molH
%O:39.56%molO=39.56gO=39.56gO×1molO15.999gO=2.473molO
Divide by the smallest number of moles.
subscriptC=4.451molC2.473molO≈1.800≈95
subscriptH=6.924molH2.473molO≈2.800≈145
subscriptO=2.473molO2.473molO=1
Now, multiply each subscript by 5 to achieve whole number subscripts. Therefore, the empirical formula is C9H14O5.
Once the chromatographic column has been prepared, why is it important to allow the level of the liquid in the column to drop the level of the alumina before applying the solution of the compound to be separated?
Answer: Chromatography is a technique which is used for the separation of components present in a mixture into sub-components.
Explanation:
The solvent or liquid used for the separation of the chromatographic mixture should lie below the spot point where the mixture is loaded in an alumina column. This is done to prevent the air bubbles formations, which can lead to poor sanitation. Also this is done to prevent the mixing of the mixture to be separated with the solvent instead of running with the solvent via capillary action.