The reduction of the species defines the gain of electrons. The iron is most likely to be reduced when reacts with zinc. Thus option A is correct.
What is oxidizing agent?Oxidizing agents are the species that gain electrons and get reduced, their oxidation number gets reduced when the metal reacts.
In the reactivity, series zinc is placed before iron and hence is a reducing agent that gets oxidized. Down the series, the reducing ability decreases while the oxidizing increases.
Therefore, option A. iron will be reduced when reacts with zinc.
Learn more about the reactivity series here:
https://brainly.com/question/2288636
#SPJ1
Br NaOCH2CH3 + CH3CH-OH + NaBr CH3 CH3 a. Identify the mechanism of the reaction. b. Suggest steps for the mechanism of this reaction. Use curved arrows to show the electron motions. c. How would the rate be affected if the concentration of sodium ethoxide, NaOCH CH3 is increased? Justify your answer.
Answer:
a) The mechanism of the reaction is the Elimination Bimolecular or E2.
b) Steps for the mechanism of this reaction is given as follows,
c) Reaction rate = K[Organic compound][[tex]NaoCH_{2} CH_{3}[/tex]].
Explanation:
a) The mechanism of the reaction is the Elimination Bimolecular or E2.
c) This is an E2 reaction, so it depends on the concentration of both substrate and reactant. If we increase the concentration of [tex]NaoCH_{2} CH_{3}[/tex], the reaction rate will be increased.
Reaction rate = K[Organic compound][[tex]NaoCH_{2} CH_{3}[/tex]].
b) Steps for the mechanism of this reaction is given as follows,
Suppose, in an experiment to determine the amount of sodium hypochlorite in bleach, you titrated a 23.92 mL sample of 0.0100 M K I O 3 with a solution of N a 2 S 2 O 3 of unknown concentration. The endpoint was observed to occur at 13.80 mL . How many moles of K I O 3 were titrated
Answer:
2.39x10⁻⁴ moles
Explanation:
As the problem asks us the number of moles of KIO₃ that were titrated, all that is required of us is to calculate how many moles of KIO₃ are there in 23.92 mL of a 0.0100 M solution (All moles in the samples are titrated).
We can do so by using the definition of molarity:
Molarity = moles / litersconverting 23.92 mL ⇒ 23.92 / 1000 = 0.02392 L
moles = 0.0100 M * 0.02392 Lmoles = 2.39x10⁻⁴ molesLiquid A is poured into Liquid B and a single, clear layer results. More Liquid A is added and two distinct layers form. The solution is now_____and there is a_____equilibria occurring.
Answer:
Explanation:
Liquid A is poured into Liquid B and a single, clear layer results. More Liquid A is added and two distinct layers form. The solution is now unsaturated and there is a static equilibria occurring
For which of the following transitions would a hydrogen atom absorb a photon with the longest wavelength?
a. n = 1 to n = 2
b. n = 3 to n = 2
c. n = 5 to n = 6
d. n = 7 to n = 6
Answer:
Hence among the options a and c, option d is that the correct answer because it has rock bottom energy ( as n value increases, energy decreases as energy levels come closer).
Explanation:
The relation between energy and wavelength is:
[tex]\Lambda = hC/E[/tex]
From this equation, it's clear that wavelength and energy are inversely proportional to every other. The Lower the energy of a specific transition, the longest will the wavelength be of that specific transition.
Among the given options, options b and d are often ruled out, since those transitions produce to release of a photon because it is coming down from an excited state.
The position of the equilibrium for a system where K = 6.4 × 10 9 can be described as being favoring ________________
Answer:
to the right (products side)
Explanation:
The equilibrium constant K describes the ratio between the concentration of products and reactants at equilibrium. For a general reaction:
a A + b B → c C + d D
The equilibrium constant expression is:
[tex]K = \frac{[C]^{c} [D]^{d} }{[A]^{a} [B]^{b} }[/tex]
A low value of K indicates that the concentration of products (C and D) is low in relation with the concentration of reactants (A and B).
Conversely, a high value of K indicated that the concentration of products is high compared with the concentration of reactants.
Since K = 6.4 × 10⁹ is a high value, the concentration of products is higher than the concentration of reactants at equilibrium. Thus, the position of the equilibrium is favored to the right.
Organic foods do not contain chemicals.
True
Or
False
Answer:
True
Explanation:
The word Organic refers to the methods used to cultivate and process farm agricultural products. Organic foods are edible and nutritious substances consumed (both plants and animals) that are free from the use of synthetics and chemicals. In plants, the include the use of organic manure that serves as fertilizers and carrying out the weeding process by hand weeding. In animals, diseases can be prevented by maintaining a clean house or rotational grazing.
The benefit of organic foods are to produce food substances with no chemical substances.
An ionic compound contains an unknown ion X and has the formula X3N2. Ion X contains 10 electrons. Write down the chemical symbol of X?
Answer:
Mg3N2
Explanation:
it would be magnesium as it would loss to electron so it would have 10 electron. you can see in the picture above .
hope this helps :)
In an endothermic reaction, reactants are __ products.
equal to
less stable than
more stable than
equally stable than
Which factors would increase the rate of a reaction?
1. Lowering the temperature
II. Increasing the concentration of readiants
JUL. Adding a catalyst
I and
Oland
O only
I only
Tony
Next
Adding a catalyst would increase the rate of a reaction
when a polar bond is formed between 2 atoms which atom receives a partial positive charge
Answer:
The more electronegative atom in a covalent bond
Why the catalytic and optical properties of nanomaterial are different from bulk material
Answer:
The material properties of nanostructures are different from the bulk due to the high surface area over volume ratio and possible appearance of quantum effects at the nanoscale. ... Yu; they found that the structural distortions on the quantum dots depend both on the kind of dopant and on the size of the dots.
Explanation:
hope it helps
When 1-methylcyclopentene undergoes acid catalyzed hydration, the major product formed is 1-methylcyclopentanol. What change(s) in the reactants would result in 2-methylcyclopentanol being the major product
Answer:
Conversion of 1-methyl cyclopentene to 2-methyl cyclopentanol.
Explanation:
When alkenes react with diborane followed by the reaction with hydrogen peroxide and NaOH then the least substituted carbon in the double bond will be substituted.
The formation of 2-methyl cyclopentanol from 1-methyl cyclopentene is shown below:
bxbcnjsnc'ljda'v'jfsvbfs;kv f;k a'kvb'SNDklv'nSDF"LKnvjkfsk
Draw structures corresponding to the following IUPAC names:(a) (Z)-2-Ethyl-2-buten-1-ol (b) 3-Cyclohexen-1-ol(c) trans-3-Chlorocycloheptanol (d) 1,4-Pentanediol(e) 2,6-Dimethylphenol (f ) o-(2-Hydroxyethyl)phenol
Answer:
Draw structures corresponding to the following IUPAC names:(a) (Z)-2-Ethyl-2-buten-1-ol (b) 3-Cyclohexen-1-ol(c) trans-3-Chlorocycloheptanol (d) 1,4-Pentanediol(e) 2,6-Dimethylphenol (f ) o-(2-Hydroxyethyl)phenol
Explanation:
According to IUPAC rules, the name of a compound is:
Prefix+root word+suffix
1) Select the longest carbon chain and it gives the root word.
2) The substituents give the prefix.
3) The functional group gives the secondary suffix and the type of carbon chain gives the primary suffix.
The structure of the given compounds are shown below:
(S)-Pentan-2-ol was treated sequentially with methanesulfonyl chloride (CH3SO2Cl) and then potassium iodide. What is the final product that forms
Answer:
(S)-Pentan-2-ol was treated sequentially with methanesulfonyl chloride (CH3SO2Cl) and then potassium iodide. What is the final product that forms
Explanation:
Alcohols are poor leaving groups.
To make -OH group a better-leaving group, it should be treated with sulfonyl chlorides.
Then, methane sulfonyl group makes will be substituted on the -OH group and forms sulfonyl esters and makes it a better leaving group.
After that treating with KI proceeds through nucleophilic bimolecular substitution and the final product formed is shown below:.
write any two things that should be remembered while writing chemical equation
Answer:
the product and the reactant must be balanced
if u are required to give the mechanism if the reaction it must be written
Indicate type of chemical reactions for 2Mgl2+MN(SO3)2=2MgSO3+Mnl4
Answer:
double decomposition reaction
Arrange these compounds by their expected boiling point. Highest boiling point Lowest boiling point CH3OH, CH3CI CH4.
Answer: The given compounds are arranged according to decreasing boiling point as [tex]CH_{3}OH > CH_{3}Cl > CH_{4}[/tex].
Explanation:
The temperature at which vapor pressure of a substance becomes equal to the atmospheric pressure is called boiling point.
Stronger is the intermolecular forces present the atoms of a molecule more heat will be required by it to break the bond between its atoms. Hence, more will the boiling point of the molecule.
In [tex]CH_{3}OH[/tex] (methanol), there is hydrogen bonding present which is a stronger force. So, it will have highest boiling point as compared to [tex]CH_{3}Cl[/tex] and [tex]CH_{4}[/tex].
In [tex]CH_{3}Cl[/tex] (chloroform), there is more electronegative atom attached (Cl) is attached to less electronegative atom (C and H). So, electrons are more pulled towards the chlorine atom. So, boiling point of [tex]CH_{3}Cl[/tex] is more than methane [tex](CH_{4})[/tex].
Thus, we can conclude that given compounds are arranged according to decreasing boiling point as [tex]CH_{3}OH > CH_{3}Cl > CH_{4}[/tex].
define the following terms atom
Di- n- pentyl ether can be converted to 1- bromopentane by treatment with HBr through essentially a(n) ________ mechanism.
Answer:
SN1 mechanism
Explanation:
The mechanism of this reaction is shown in the image attached.
The Di- n- pentyl ether is first protonated. The CH3(CH2)4OH is now a good leaving group as shown.
The attack of the bromide ion on the cation formed completes the mechanism to yield 1- bromopentane as shown in the mechanism.
A student conducted an experiment 4 times. His results were very close to each other each time he ran the experiment and
were very close to the true or actual value. His results showed
A. None of these answers are correct
B. poor accuracy and poor precision
C. good accuracy and good precision
D. poor accuracy and good precision
E. good accuracy and poor precision
Answer:
d is the answer have a good one
define saturated and unsaturated fats
Explanation:
Saturated fats are defined as the fat where fatty acid chains contain only single bonds.
For example, stearic acid, palmitic acid etc.
Unsaturated fats are defined as the fat where fatty acids contain one or more number of double bonds on the carbon atoms.
For example, oleic acid, palmitoleic acid etc.
If 11g of a gas occupies 5.6dm'3at s.t.p., calculate it's vapour density (1.0mol of a gas occupies 22.4dm'3at s.t.p.).
Answer:
[tex]{ \boxed{ \bf{vapour \: density = 2 \times molecular \: mass}}} \\{ \tt{ PV= (\frac{m}{ m_{r}}) RT}} \\ { \tt{3 \times 5.6 = \frac{11}{m _{r}} \times 0.0831 \times 273}} \\ { \tt{m _{r} = 14.85 \: g}} \\ \\ { \bf{vapour \: density = 2 \times m _{r}}} \\ = 2 \times 14.85 \\ = 29.7 \: { \tt{g {dm}^{ - 3} }}[/tex]
Rank the following substances/solutions in order of lowest boiling point to highest boiling point where 1 has the lowest boiling point and 5 has the highest boiling point. pure water; 1.0 m NaCl; 0.5 m KBr, 0.75 m CaCl2; 1.5 m glucose (C6H12O6)
Answer:
1) pure water
2) 0.75 m CaCl2
3) 1.0 m NaCl
4) 0.5 m KBr
5) 1.5 m glucose (C6H12O6)
Explanation:
Boiling point elevation is a colligative property. Coligative properties are properties that depend on the amount of solute present in the system. The boiling point of solvents increase due to the presence of solutes.
The boiling point elevation depends on the number of particles the solute forms in solution and the molality of the solute. The more the number of particles formed by the solute and the greater the molality of the solute, the greater the magnitude of boiling point elevation.
The order of decreasing hoping point elevation is;
1) 0.75 m CaCl2
2) 1.0 m NaCl
3) 0.5 m KBr
4) 1.5 m glucose (C6H12O6)
For each of the following changes at equilibrium, indicate whether the equilibrium shifts in the direction of products, reactants, or does not change: CaCO3(s)+heat⇌CaO(s)+CO2(g)
1) increasing the temperature
shifts equilibrium in the direction of the reactants
does not change
shifts equilibrium in the direction of the products
2) decreasing the volume of the container
shifts equilibrium in the direction of the reactants
shifts equilibrium in the direction of the products
does not change
3) adding a catalyst
shifts equilibrium in the direction of the reactants
shifts equilibrium in the direction of the products
does not change
4) adding more CaO(s)
does not change
shifts equilibrium in the direction of the reactants
shifts equilibrium in the direction of the products
Answer:
shifts equilibrium in the direction of the products
shifts equilibrium in the direction of the reactants
does not change
shifts equilibrium in the direction of the reactants
Explanation:
When a constraint such as a change in pressure, concentration or temperature is imposed on a reaction system in equilibrium, the equilibrium position will shift in such a way as to annul the constraint.
The reaction is endothermic as written. Hence, increase in temperature increases the rate of forward reaction thereby shifting the equilibrium position towards the products.
When the volume of a reaction is decreased, the equilibrium position shifts in the direction which produces the least total volume. In this case, decrease in volume shifts the equilibrium position towards the reactants.
A catalyst has no effect on the equilibrium position. However, a catalyst may cause equilibrium to be achieved faster or at a lower temperature.
When more CaO is added, the equilibrium position shifts towards the reactants side and more CaCO3 is produced.
find out the equivalent weight of Ca(OH)2
Answer:
The equivalent weight of calcium hydroxide is 1/2 he mass of a mol of calcium hydroxide. 1 mol Ca(OH)2 = 74 grams Ca(OH)2 ; 1 equivalent Ca(OH)2 = 37 grams Ca(OH)2......
Explanation:
HOPE IT HELPS YOU
Part A of the lab involved adding 4 mL increments of distilled water to 5.00 mL of antimony trichloride solution. The antimony trichloride solution contains 0.10 M SbCl3 in 4.5 M HCl. Calculate the concentrations of SbCl3 and H /Cl- in the test tube after 12.0 mL of distilled water has been added. Assume dilution only.
Answer:
0.0238M SbCl3, 1.07M H+, 1.14M Cl-
Explanation:
The total volume of the solution is:
4mL + 5.00mL + 12.0mL = 21mL
As the volume of the SbCl3 is 5.00mL, the dilution factor is:
21mL / 5.00mL = 4.2 times
The concentration of SbCl3 is:
0.10M SbCl3 / 4.2 times = 0.0238M SbCl3
The concentration of H+ = [HCl]:
4.5M / 4.2 times = 1.07M H+
The initial concentration of Cl- is:
3 times SbCl3 + HCl = 0.10M*3 + 4.5M =
3 times SbCl3 because 1 mole of SbCl3 contains 3 moles of Cl-
4.8M Cl- / 4.2 times = 1.14M Cl-
Which diagram correctly depicts the trend in electronegativity?
a.
b.
c.
d.
The electronegativity increases across the period and decreases down the group. Thus, option B is correct.
Electronegativity can be defined as the tendency of an atom to gain or attract an electron. The electronegativity has been dependent on the size of the atom, as well as the atomic number and valence electrons.
The atom with the requirement of a less number of atoms to complete its octet can easily gain the electron and thereby have high electronegativity. The atomic size also plays a role in the electronegativity of the atom.
The atom with a bigger size has the lesser force of attraction from the nucleus and thus has difficulty attracting the electron, however, the smaller size atom can easily attract the electron with the attraction force from the nucleus.
Thus, the elements with smaller sizes and a high number of valence electrons are more electronegative. In the periodic table, on moving from left to right the valence electrons increase, thus the electronegativity increases.
On moving down the group, the element size increase, thus the electronegativity decreases down the group.
The electronegativity increases across the period and decreases down the group. Thus, option B is correct.
For more information about electronegativity, refer to the link:
https://brainly.com/question/2060520
Choose the substance with the higher entropy in each pair. Assume constant temperature, except in part (5)
(i) 1 mol of SO2(g) or 1 mol of SO3(g)
(ii) 1 mol of CO2(s) or 1 mol of CO2(g)
(iii) 3 mol of O2(g) or 2 mol of O3(g)
(iv) 1 mol of KBr(s) or 1 mol of KBr(aq)
(v) Seawater at 2°C or at 23°C
(vi) 1 mol of CF4(g) or 1 mol of CCl4(g)
Answer:
I) 1 mol of SO3(g)
2) 1 mol of CO2(g)
3) 3 mol of O2(g)
4) 1 mol of KBr(aq)
5) Seawater at 23°C
6) 1 mol of CCl4(g)
Explanation:
In molecules having greater numbers of atoms, there is an increase the number of ways by which the molecule vibrates thereby leading to a higher number of possible microstates and overall increase in entropy of the system. Hence, 1 mol of SO3(g) has a higher entropy than 1 mol of SO2.
Gases have a higher entropy than liquids and liquids have a higher entropy than gases.
Also, the greater the molecular weight of a molecule, the higher the entropy. Higher number of moles of a gas as well as the increase in temperature of a substance are also factors that lead to higher entropy.
Gaseous methane will react with gaseous oxygen to produce gaseous carbon dioxide and gaseous water . Suppose 9.6 g of methane is mixed with 64.9 g of oxygen. Calculate the maximum mass of water that could be produced by the chemical reaction. Round your answer to significant digits.
Answer:
21.6 g
Explanation:
The reaction that takes place is:
CH₄ + 2O₂ → CO₂ + 2H₂OFirst we convert the given masses of both reactants into moles, using their respective molar masses:
9.6 g CH₄ ÷ 16 g/mol = 0.6 mol CH₄64.9 g O₂ ÷ 32 g/mol = 2.03 mol O₂0.6 moles of CH₄ would react completely with (2 * 0.6) 1.2 moles of O₂. As there are more O₂ moles than required, O₂ is the reactant in excess and CH₄ is the limiting reactant.
Now we calculate how many moles of water are produced, using the number of moles of the limiting reactant:
0.6 mol CH₄ * [tex]\frac{2molH_2O}{1molCH_4}[/tex] = 1.2 mol H₂OFinally we convert 1.2 moles of water into grams, using its molar mass:
1.2 mol * 18 g/mol = 21.6 g