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
Classify each of the following as a strong acid or a weak acid and indicate how each should be written in aqueous solution. Classify ... In solution this acid should be written as: weak 1. hydrocyanic acid H3O CN- _______ 2. hydrobromic acid
Answer:
HCN, weak acid
H⁺, Br⁻, strong acid
Explanation:
Hydrocyanic acid is a weak acid, according to the following equation.
HCN(aq) ⇄ H⁺(aq) + CN⁻(aq)
Thus, it should be written in the undissociated form (HCN).
Hydrobromic acid is a strong acid, according to the following equation.
HBr(aq) ⇒ H⁺(aq) + Br⁻(aq)
Thus, it should be written in the ionic form (H⁺, Br⁻).
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.
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-
Describe how you would prepare your assigned ester from a carboxylic acid and an alcohol. You do not need to include a detailed procedure, but you should include any necessary reagents or catalyst (solvents are not needed).
Answer:
The general preparation of esters( for example ethyl ethanoate) is through a process known as ESTERIFICATION.
Explanation:
The formation of an ester by the reaction between an alkanol and an acid is known as esterification. This reaction is extremely slow and reversible at room temperature, and is catalyzed by a high concentration of hydrogen ions.
In the preparation of one of the simpler esters known as ETHYL ETHANOATE the reactants include ethanol(an alcohol) and glacial ethanoic acid(a carboxylic acid) in the presence of concentrated tetraoxosulphate VI acid as a CATALYST. Note that, a catalyst is any substance that is able to increase the rate of a chemical reaction.
The mixture is warmed in a water bath( hot but not boiling) for about 25 minutes. The mixture is poured into a beaker partially filled with a sodium or calcium chloride to remove interacted ethanol. The ethyl ETHANOATE floats on the mixture as oily globules.
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 gbxbcnjsnc'ljda'v'jfsvbfs;kv f;k a'kvb'SNDklv'nSDF"LKnvjkfsk
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]
How many moles of hydrogen are in the sample?
Round your answer to 4 significant digits.
Answer:
1.56 mol H₂
Explanation:
Mg₃(Si₂O₅)₂(OH)₂
There are 4 Si moles per Mg₃(Si₂O₅)₂(OH)₂ mol. With that in mind we can calculate how many Mg₃(Si₂O₅)₂(OH)₂ moles are there in the sample, using the given number of silicon moles:
3.120 mol Si * [tex]\frac{1molMg_3(Si_2O_5)_2(OH)_2}{4molSi}[/tex] = 0.78 mol Mg₃(Si₂O₅)₂(OH)₂Then we can convert Mg₃(Si₂O₅)₂(OH)₂ moles into hydrogen moles, keeping in mind that there are 2 hydrogen moles per Mg₃(Si₂O₅)₂(OH)₂ mol:
0.78 mol Mg₃(Si₂O₅)₂(OH)₂ * 2 = 1.56 mol H₂Rank 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)
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 molesCalculate the moles of H3PO4 that reacted (8). Consult the coefficients in the balance chemical reaction to obtain the mole ratio. Show your calculation here.
Answer:
The number of moles of H₃PO₄ that reacted is 0.000343 moles
Note: Some data is missing. Data from the attachment is used for the calculationsinnthe explanation below.
Explanation:
The reaction is a neutralization reaction between NaOH and H₃PO₄. The equation of the reaction is given as follows:
3 NaOH + H₃PO₄ ---> Na₃PO₄ + 3 H₂O
The molarity of the NaOH solution is 0.238 mol/L.
Average volume of NaOH used during the titration to arrive to endpoint = (4.6 + 3.9 + 4.5) mL / 3 = 4.33 mL
Molarity is defined ratio of the number of moles of solute to the volume of solution. Mathematically, molarity = number of moles/volume in Litres
Number of moles of NaOH reacted = 0.238 mol/L × (4.33mL × 1 L/1000 mL)
Number of moles of NaOH = 0.00103 moles
From the equation of the reaction, 3 moles of NaOH reacts with 1 mole of H₃PO₄
0.00103 moles of NaOH will react with 0.00103 x 1/3 moles of H₃PO₄ = 0.000343 moles of H₃PO₄.
Therefore, number of moles of H₃PO₄ that reacted is 0.000343 moles
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
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.
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
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
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 :)
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].
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
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.
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_____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
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,
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
Indicate type of chemical reactions for 2Mgl2+MN(SO3)2=2MgSO3+Mnl4
Answer:
double decomposition reaction
3. A 3.455-g sample of a mixture was analyzed for barium ion by adding a small excess of sulfuric acid to an aqueous solution of the sample. The resultant reaction produced a precipitate of barium sulfate, which was collected by filtration, washed, dried, and weighed. If 0.2815 g of barium sulfate was obtained, what was the mass percentage of barium in the sample
Answer:
[tex]Ba\ percentage\ in\ Mass=4.8\%[/tex]
Explanation:
From the question we are told that:
Mass of mixture [tex]m=3.455g[/tex]
Mass of Barium [tex]m_b=0.2815g[/tex]
Equation of Reaction is given as
[tex]Ba2+ + H2SO4 => BaSO4 + 2 H+[/tex]
Generally the equation for Moles of Barium is mathematically given by
Since
[tex]Moles of Ba^{2+} = Moles of BaSO_4[/tex]
Therefore
[tex]Moles of Ba^{2+} = \frac{mass}{molar mass of BaSO4}[/tex]
[tex]Moles of Ba^{2+} = \frac{0.2815}{233.39}= 0.0012061 mol[/tex]
Generally the equation for Mass of Barium is mathematically given by
[tex]Mass\ of\ Ba^{2+} = Moles * Molar mass of Ba^{2+}[/tex]
[tex]Mass\ of\ Ba^{2+} = 0.0012061 * 137.33 = 0.1656 g[/tex]
Therefore
[tex]Ba\ percentage\ in\ Mass = mass of Ba^{2+}/mass of sample * 100%[/tex]
[tex]Ba\ percentage\ in\ Mass= \frac{0.1656}{ 3.455 }* 100%[/tex]
[tex]Ba\ percentage\ in\ Mass=4.8\%[/tex]
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.
Do you think a single property is sufficient for conclusively identifying a substance?
Answer:
No
Explanation:A single property is insufficient for identifying a substance
A chemistry grad student measures the performance of the new pump in his lab. The result is:Convert to
Answer:
The correct answer is - 0.0188 mJ*s^-1.
Explanation:
In order to convert the kPA value to PA value--
and then convert PA unit to 1 kgm^-1s^-2
And in finally convert mm to m
The value that come should be convert from kgm^2s^-2 to and then convert J to mJ or kJ by dividing 100 or 100 subsequently.
As no question given this method can be followed for the answer.
(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:.
