Answer:
34.1g of C₅H₅NHCl the student need to dissolve to the solution.
Explanation:
Full question is:
A chemistry graduate student is given 500.mL of a 0.20M pyridine C5H5N solution. Pyridine is a weak base with =Kb×1.7x10−9 . What mass of C5H5NHCl should the student dissolve in the C5H5N solution to turn it into a buffer with pH =4.76 ?
Using H-H expression for weak bases, it is possible to find pH of a buffer thus:
pOH = pKb + log [BH⁺] / [B]
Where pKb is -log Kb = 8.77, [BH⁺] concentration of C₅H₅NHCl and [B] concentration of C₅H₅N (It is possible to take the moles of both compounds and not its concentration.
As pH the student wants is 4.76, pOH is:
pOH = 14 - pH = 14 - 4.76 = 9.24
Replacing:
9.24 = 8.77 + log [C₅H₅NHCl] / [C₅H₅N]
Moles of C₅H₅N are:
0.500L × (0.20mol / L) = 0.10mol C₅H₅N
Replacing again:
9.24 = 8.77 + log [C₅H₅NHCl] / [0.10mol]
2.9512 = [C₅H₅NHCl] / [0.10mol]
0.29512 moles = [C₅H₅NHCl].
As molar mass of C₅H₅NHCl is 115.56g/mol, mass of 0.29512 moles are:
0.29512 moles C₅H₅NHCl × (115.56g / mol) =
34.1g of C₅H₅NHCl the student need to dissolve to the solution.4Ga + 3S2 → 2Ga2S3
1. How many grams of Gallium burned if 200.0 grams of Gallium(III)Sulfide formed?
Answer:
118.4 g
Explanation:
4 Ga + 3 S₂ → 2 Ga₂S₃
According to the equation, for every 4 moles of gallium burned, 2 moles of gallium(III) sulfide.
First, convert grams of Ga₂S₃ to moles. The molar mass is 235.641 g/mol.
(200.0 g)/(235.641 g/mol) = 0.8487 mol
Use the relationship above to convert moles of Ga₂S₃ to moles of Ga.
(0.8487 mol Ga₂S₃) × (4 mol Ga)/(2 mol Ga₂S₃) = 1.697 mol Ga
Convert moles of Ga to grams. The molar mass is 69.723 g/mol.
(1.697 mol Ga) × (69.723 g/mol) = 118.4 g
Which of the following describes green design?
A. Materials that will fail in a predictable and safe way
B. The use of new engineering technologies in building construction
c. Building structures made of composite materials to make them
safer
D. Materials and design techniques that reduce the negative
environmental impact of a structure
Answer: materials and design Techniques that reduce the negative environmental impact of a structure
Explanation:
Where would you find the following symbol on a diagram of a water molecule: δ+ ?
Answer:
On the 2 hydrogen atoms.
Explanation:
δ+ indicates the atom has a lower electronegativity than the other atom it is bonded with. This only exist in polar covalent bonds, where the 2 atoms have different electronegativity values. When they have different electronegativity values, the one with higher electronegativity has a higher tendency to "pull" the shared electrons towards itself, they have a δ- symbol.
Back to H2O, since the electronegativity of elements increases from left to right horizontally and upwards vertically in the periodic table (except for noble gases, they are unreactive. Note that fluorine has the highest electronegativity), O atom has a higher electronegativity than hydrogen (hydrogen sits at the centre top of the table). hence, we can find δ+ on the hydrogen atoms.
Given the specific heat for aluminum is 0.900 J/g.°C, how much heat is released when a 3.8 g sample
of Al cools from 450.0°C to 25°C?
A. 1.5 kJ
B. 54 J
C. 60J
D. 1.7 kJ
E. 86 J
Answer:
Q = 1.5 kJ
Explanation:
It is given that,
The specific heat for aluminum is 0.900 J/g°C
Mass of sample, m = 3.8 g
Initial temperature, [tex]T_i=450^{\circ} C[/tex]
Final temperature, [tex]T_f=25^{\circ} C[/tex]
We need to find the heat released. The amount of heat released is given by the formula:
[tex]Q=mc\Delta T\\\\Q=mc(T_f-T_i)\\\\Q=3.8\times 0.9\times (25-450)\\\\Q=1453.5\ J\\\\Q=1.45\ kJ[/tex]
or
[tex]Q=1.5\ kJ[/tex]
So, the correct option is (A) i.e. 1.5 kJ.
A 5.024 mg sample of an unknown organic molecule containing carbon, hydrogen, and nitrogen only was burned and yielded 13.90 mg of CO2 and 6.048 mg of H2O. What is the empirical formula
Answer:
C8H17N
Explanation:
Mass of the unknown compound = 5.024 mg
Mass of CO2 = 13.90 mg
Mass of H2O = 6.048 mg
Next, we shall determine the mass of carbon, hydrogen and nitrogen present in the compound. This is illustrated below:
For carbon, C:
Molar mass of CO2 = 12 + (2x16) = 44g/mol
Mass of C = 12/44 x 13.90 = 3.791 mg
For hydrogen, H:
Molar mass of H2O = (2x1) + 16 = 18g/mol
Mass of H = 2/18 x 6.048 = 0.672 mg
For nitrogen, N:
Mass N = mass of unknown – (mass of C + mass of H)
Mass of N = 5.024 – (3.791 + 0.672)
Mass of N = 0.561 mg
Now, we can obtain the empirical formula for the compound as follow:
C = 3.791 mg
H = 0.672 mg
N = 0.561 mg
Divide each by their molar mass
C = 3.791 / 12 = 0.316
H = 0.672 / 1 = 0.672
N = 0.561 / 14 = 0.040
Divide by the smallest
C = 0.316 / 0.04 = 8
H = 0.672 / 0.04 = 17
N = 0.040 / 0.04 = 1
Therefore, the empirical formula for the compound is C8H17N
Calculate the pH of a 0.0255 M solution of ammonium bromide (NH4Br). The Kb of ammonia is 1.76 x 10-5
Answer:
5.42
Explanation:
Step 1: Consider the dissociation of NH₄Br
NH₄Br(aq) ⇒ NH₄⁺(aq) + Br⁻(aq)
Br⁻ is the conjugate base of HBr, a strong acid, so it doesn´t react with water. NH₄⁺ is the conjugate acid of NH₃, so it does react with water.
Step 2: Consider the acid reaction of NH₄⁺
NH₄⁺(aq) + H₂O(l) ⇄ NH₃(aq) + H₃O⁺(aq)
Step 3: calculate the acid dissociation constant for NH₄⁺
We will use the following expression.
[tex]K_a \times K_b = K_w\\K_a = \frac{K_w}{K_b} = \frac{1.00 \times 10^{-14} }{1.76 \times 10^{-5}} = 5.68 \times 10^{-10}[/tex]
Step 4: Calculate the concentration of H₃O⁺
We will use the following expression.
[tex][H_3O^{+} ]= \sqrt{K_a \times C_a } = \sqrt{5.68 \times 10^{-10} \times 0.0255 } = 3.81 \times 10^{-6}M[/tex]
Step 5: Calculate the pH
We will use the following expression.
[tex]pH = -log [H_3O^{+} ] = -log (3.81 \times 10^{-6}) = 5.42[/tex]
The pH of 0.0255 M solution should be 5.42.
Calculation of the pH of 0.0255 M solution:Since we know that
ka * kb = kw
So,
ka = kw/kb
= 1.00*10^-14 / 1.76*10^-5
= 5.68*10^-10
Now the concentration of H3O should be
= √ka * Ca
= √5.68*10^-10 * 0.0255
= 3.81*10^-6M
Now the pH value should be
= -log(H3O+)
= -log(3.81*10^-6)
= 5.42
hence, The pH of 0.0255 M solution should be 5.42.
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The element potassium forms a _______ with the charge . The symbol for this ion is , and the name is . The number of electrons in this ion is
Answer:
The element potassium forms a cation with the charge +1 . The symbol for this ion is K⁺, and the name is potassium ion. The number of electrons in this ion is 18.
Explanation:
Potassium is a metal. It belongs to the group 1 elements. Metals form cations by losing electrons. Since potassium is a group element, it forms a cation by losing one electron. The charge it has is +1 due to the excess of the protons compared t the electrons by 1.
Potassium has 19 electrons. Potassium io on the other hand has 19-1 = 18 electrons.
1. If a carbohydrate, like xylulose, has five carbon atoms and a carbonyl group on the second carbon, it is called a(n):_______.
2. Glyceraldehyde is an example of a(n):_____, because it has three carbon atoms.
3. A monosaccharide is a(n):_______ if the carbonyl group is on the end of the carbon chain.
4. Any carbohydrate with the carbonyl group on the second carbon is a(n):_______.
5. The most common carbohydrate, , has six carbon atoms.
6. With the carbonyl group on the end of a six-carbon chain, the carbohydrate would be classified as a(n):_________.
Answer:
Following are the answer to this question:
Explanation:
The answer are:
1) ketopentose
2) Triose
3) Aldose
4) Ketose
5) Glucose
6) Aldohexose
The pentose has 2-position contain a personal ketone group. The triose is a monosaccharide or simple sugar that contains three atoms of carbon. The Aldose and ketose are simple carbohydrates, both also called monosaccharides. In aldose, it has a functional group of aldehydes within its structure. The ketose sugars have workable ketone groups. Stereoisomerism has been found in aldose sugars that contain more than three carbon atoms. Glucose is also one of the main molecules which function as plant and animal energy sources. It's also derived from plant sap and seems to be present in the bloodstream of humans, that's why it is called "blood sugar." The aldohexose is a hexose is a group of aldehydes on one end, it has a total of 16 possible aldohexose stereoisomers in four chiral centers.Discuss the contrary aspect of proton NMR and C-13 NMR by elucidating the structure of 2 chloro pentanal?
Answer:
See explanation
Explanation:
1H NMR
In the 2-chloro-pentanal we have 4 different types of hydrogens. Therefore, we will have 4 different signals. (See figure 1)
Red hydrogen
For the red hydrogens we have only 1 neighbor. So, if we follow the n+1 rule we can calculate the multiplicity of this hydrogen. In this case a doublet.
Blue hydrogens
In this case, we have 3 neighbors (one in the right, two in the left). Therefore we will have a quartet.
Purple hydrogens
For these hydrogens, we have also will have a quartet, because we have 3 neighbors (one in the right, two in the left).
Green hydrogens
In the green hydrogen,s we have 5 neighbors (2 in the right 3 in the left). Therefore a sextet would be produced.
Orange hydrogens
Finally, in these hydrogens, we have 2 neighbors. Therefore a triplet is expected.
13C NMR
For the 13C NMR, we have again 4 different kinds of carbons. Therefore we will have 4 signals. The most deshielded carbon, in this case, is the red one (see figure 2), so this carbon would be on the left side (around 190). Then the next deshield carbon is the blue one, due to the "Cl" atom placed on this carbon.
I hope it helps!
The thermochemical equation is for the reaction of hydrogen bromide gas to form hydrogen gas and bromine liquid. 2HBr(g) = H 2 (g)+ Br 2 (l) 72.6 kJ How many grams of HBr (g) would be made to react if 11.4 energy were provided?
Answer:
the mass of HBr that would react is 25.41 g of HBr
Explanation:
attached is the calculations.
Which example involves a phase change in which heat energy is released by the substance?
Ofreezing ice cream
O cooking a pot of soup
O melting ice under sunlight
O watching frost disappear into air
Answer:
Cooking a pot of soup
Explanation:
id say that because when you freeze ice cream, its already frozen, so no heat is being released. melting ice wouldn't be the answer because, once again, it is already frozen, and no heat is being released.
Answer:
the correct answer is freezing ice cream
Explanation:
i took the test & got this question correct. also, heat energy is released when freezing because there is no heat energy involved.
A transition in the balmer series for hydrogen has an observed wavelength of 434 nm. Use the Rydberg equation below to find the energy level that the transition originated. Transitions in the Balmer series all terminate n=2.
Delta E= -2.178 x10-18J ( 1/n2Final - 1/n2Initial )
The number is 5.
What is the energy of this transition in units of kJ/mole? ( hint: the anser is NOT 4.58x10-22kJ/mole or -4.58x10-22kJ/mole)
Answer:
i. n = 5
ii. ΔE = 7.61 × [tex]10^{-46}[/tex] KJ/mole
Explanation:
1. ΔE = (1/λ) = -2.178 × [tex]10^{-18}[/tex]([tex]\frac{1}{n^{2}_{final} }[/tex] - [tex]\frac{1}{n^{2}_{initial} }[/tex])
(1/434 × [tex]10^{-9}[/tex]) = -2.178 × [tex]10^{-18}[/tex] ([tex]\frac{n^{2}_{initial} - n^{2}_{final} }{n^{2}_{final} n^{2}_{initial} }[/tex])
⇒ 434 × [tex]10^{-9}[/tex] = (1/-2.178 × [tex]10^{-18}[/tex])[tex]\frac{n^{2}_{final} *n^{2}_{initial} }{n^{2}_{initial} - n^{2}_{final} }[/tex]
But, [tex]n_{final}[/tex] = 2
434 × [tex]10^{-9}[/tex] = (1/2.178 × [tex]10^{-18}[/tex])[tex]\frac{2^{2} n^{2}_{initial} }{n^{2}_{initial} - 2^{2} }[/tex]
434 × [tex]10^{-9}[/tex] × 2.178 × [tex]10^{-18}[/tex] = [tex](\frac{4n^{2}_{initial} }{n^{2}_{initial} - 4 })[/tex]
⇒ [tex]n_{initial}[/tex] = 5
Therefore, the initial energy level where transition occurred is from 5.
2. ΔE = hf
= (hc) ÷ λ
= (6.626 × 10−34 × 3.0 × [tex]10^{8}[/tex] ) ÷ (434 × [tex]10^{-9}[/tex])
= (1.9878 × [tex]10^{-25}[/tex]) ÷ (434 × [tex]10^{-9}[/tex])
= 4.58 × [tex]10^{-19}[/tex] J
= 4.58 × [tex]10^{-22}[/tex] KJ
But 1 mole = 6.02×[tex]10^{23}[/tex], then;
energy in KJ/mole = (4.58 × [tex]10^{-22}[/tex] KJ) ÷ (6.02×[tex]10^{23}[/tex])
= 7.61 × [tex]10^{-46}[/tex] KJ/mole
The initial energy level is 5 and the energy of this transition in units of kJ/mole is 7.57 * 10^-43 kJ/mole
We must first calculate ΔE as follows;
ΔE = hc/λ
h = Plank's constant = 6.6 * 10^-34 Js
c = speed of light = 3 * 10^8 m/s
λ = wavelength = 434 * 10^-9
ΔE = 6.6 * 10^-34 * 3 * 10^8/434 * 10^-9
ΔE = 0.0456 * 10^-17 J
ΔE = [tex]ΔE = -2.178 x10^-18 (\frac{1}{n^2final} - \frac{1}{n^2initial}) \\ΔE = -2.178 x10^-18 (\frac{1}{2^2} - \frac{1}{n^2initial} )\\\\4.56 * 10^-19/2.178 x10^-18 = (\frac{1}{2^2} - \frac{1}{n^2initial})\\0.210 = (\frac{1}{2^2} - \frac{1}{n^2initial})\\\frac{1}{n^2initial} = 0.25 - 0.210\\\frac{1}{n^2final} = 0.04\\n = (\sqrt{(0.04)^-1} \\n = 5[/tex]
Energy of this transition in units of kJ/mole = 4.56 * 10^-19/ 6.02 * 10^23
= 7.57 * 10^-43 kJ/mole
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Q3. Which of the following objects are chiral and which are achiral? a
club b. teacup c. football d. corkscrew e. tennis racket f. shoe g. portrai
pencil (8 marks)
Q4. Write a structure for each of the compounds listed. Explain why th
name given is incorrect, and give a correct name in each case (20 mai
a. l-methylbutane
b. 1.1,3-trimethylhexane
c. 5-octyne
d. 2-ethyl-1-propanol
e. 2.2-dimethyl-3-butanol
Submit via aduwiejuah a uds.edu.gh and deadline for submission is
July, 2020 at 11:00 AM.
Answer:
See explanation
Explanation:
In this question, we have to follow the IUPAC rules. Lets analyze each compound:
a. 1-methylbutane
In this compound we have a chain of 5 carbons, so the correct name is Pentane.
b. 1,1,3-trimethylhexane
In this compound, we longest chain is made of 7 carbons, so, we have to use the name "heptane". Carbon one would be the closet one to the methyl group, so the correct name is 2,4-dimethylheptane.
c. 5-octyne
In this case, carbon 1 would be the closet one to the triplet bond. With this in mind, the correct name is oct-3-yne.
d. 2-ethyl-1-propanol
In this compound, we longest chain is made of 4 carbons, so, we have to use the name "butane". Carbon one would be the carbon with the "OH" group, so the correct name is 2-methylbutan-1-ol.
e. 2.2-dimethyl-3-butanol
In this case, carbon 1 would be the closet one to the "OH". With this in mind, the correct name is 3,3-dimethylbutan-2-ol.
See figure 1
I hope it helps!
The percent errors of your experimental values of the specific heats may be quite large. Identify several sources of experimental error.
Answer:
The various sources of such errors are given below.
Explanation:
Sources of uncertainty or error could include necessary splattering of water leading to reduced cold water density as well as elevated temperatures of equilibration.The temperature might not have been reasonably stable when developers evaluated at every phase of the investigation or research.So that the percentage of someone specific produces heat exploratory value systems inaccuracies can be somewhat massive.
1. What form of matter is made from only one type of atom?
A molecule
B compound
C element
6.66%
D material
Answer:
A molecule is the answer.
Calculate the combustion of gaseous dimethyl ether CH 3 OCH 3 (g)+3O 2 (g) 2CO 2 (g)+3H 2 O(l) using standard molar enthalpies of formation Molecule AH H l ^ 0 (k)/mol) CH 3 OCH 3 (g) - 184.1
Answer:
[tex]\Delta _cH=-1328.3kJ/mol[/tex]
Explanation:
Helllo,
In this case, for the given chemical reaction in gaseous state:
[tex]CH_3OCH_3+3O_2\rightarrow 2CO_2+3H_2O[/tex]
We comoute the combustion enthalpy as the reaction enthalpy for one mole of fuel (dimethyl ether) considering the formation enthalpy of each given substance and whether they are reactants (subtracting) or products (adding), therefore we write:
[tex]\Delta _cH=2*\Delta _fH_{CO_2}+3*\Delta _fH_{H_2O}-\Delta _fH_{CH_3OCH_3}-3*\Delta _fH_{O_2}[/tex]
Whereas the formation enthalpies for carbon dioxide, water, dimethyl ether and oxygen are -393.5, -241.8, -184.1 and 0 kJ/mol respectively, thereby, the combustion enthalpy turns out:
[tex]\Delta _cH=2(-393.5)+3*(-241.8)-(-184.1)-3(0)\\\\\Delta _cH=-1328.3kJ/mol[/tex]
Notice that enthalpy of formation of oxygen is zero since forming an element has no chemical sense, it just exists as it has been early demonstrated.
Regards.
4. If 13 percent of the carbon-14 in a sample of cotton cloth remains, what's the approximate age of the cloth? Show your work
The approximate age of the cloth is 17190 years.
We'll begin by calculating the number of half-lives that has elapsed. This can be obtained as follow:
Original amount (N₀) = 100%Amount remaining (N) = 13%Number of half-lives (n) =?2ⁿ = 100 / 13
2ⁿ = 8
2ⁿ = 2³
n = 3
Finally, we shall determine the age of the cloth.
Half-life (t½) = 5730 yearsNumber of half-lives (n) = 3Time (t) =?t = n × t½
t = 3 × 5730
t = 17190 years
Thus, the approximate age of the cloth is 17190 years
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The molecular weight of table salt, NaCl, is 58.5 g/mol. A tablespoon of salt weighs 6.37 grams. Calculate the number of moles of salt in one tablespoon.
Finally, solve (remember significant figures):
Answer:
0.109 mol/tablespoon
Explanation:
6.37 g/ 58.5 mol = 0.10888888 mol (0.109 significantly)
Answer:
A: 0.109
Explanation:
Edge 2020
Click the "draw structure" button to launch the drawing utility. Draw the product of the reaction of propanal with lithium aluminum hydride, followed by water.
Answer:
Product: propan-1-ol
Explanation:
IIn this case, we have to remember that [tex]LiAlH_4[/tex] is a reduction agent. So, this is a reduction reaction. The [tex]LiAlH_4[/tex] has the ability to produce hydride ions [tex]H^-[/tex]. This ion can attack the carbonyl group generating a negative charge in the oxygen. In the next step, the negative charge in the oxygen can attack a water molecule to protonate the molecule and produce propan-1-ol.
See figure 1
I hope it helps!
you mix 45 ml of .20M KOH in calorimeter. The temperature of both reactions before mixing is 21.5 C. The Cp of the calorimeter was 36 J/K. If the final temperature of the mixture is 23.6 C, what is the enthalpy change per mole of water produced?
Answer:
THE ENTHALPY CHANGE PER MOLE OF KOH IS 8400 Joules/ mole OF HEAT.
Explanation:
Heat = heat capacity * change in temperature
Heat capacity = 36 J/K
Temperature of the mixture before mixing = 21.5 C
Temperature of mixtire after mixing = 23.6 C
Calculate the change in temperature:
Change in temperature = 23.6 C - 21.5 C = 2.1 C
Heat = 36 * 2.1
Heat = 75.6 J of heat
In essence, 45 ml of 0.20 M of KOH produces 75.8 J of heat
The enthalpy change per mole of water:
It is important t obtain the number of moles involved in the reaction of 45 mL of 0.20 M of KOH
n = C V
n = 0.20 M * 45 *10^-3
n = 0.009 moles
Since number of moles = mass / molar mass
The mass of 45 ml of 0.20 M of KOH is then:
Molar mass = ( 39 + 16 + 1) g/mol = 56 g/mol
Mass = number of moles * molar mass
Mass = 0.009 * 56
Mass = 0.504 g
So therefore 0.504 g of KOH produces 75.6 J of heat
1 mole of KOH will produce x J of heat
1 mole of KOH = 56 g of KOH
0.504 g = 75.6 J
56 g = x J
x J = 56 * 75.6 / 0.504
x J = 8400 J / mole of KOH
If 5.00 mL of a 0.5 M solution is diluted to a final volume of 100.0 mL, what is the concentration of the final dilute solution?
Answer:
0.025 M
Explanation:
The following data were obtained from the question:
Initial volume (V1) = 5 mL
Initial concentration (C1) = 0.5 M
Final volume (V2) = 100 mL
Final concentration (C2) =..?
Using the dilution formula, we can obtain the final concentration of the diluted solution as follow:
C1V1 = C2V2
0.5 x 5 = C2 x 100
Divide both side by 100
C2 = (0.5 x 5)/100
C2 = 0.025 M
Therefore, the final concentration of the diluted solution is 0.025 M
The concentration of the final diluted solution is 0.025M
The dilution formula is expressed according to the formula:
[tex]C_1V_1=C_2V_2[/tex]
Given the following parameters
[tex]C_1=0.5M\\V_1=5.00mL\\V_2=100.0mL\\C_2=?[/tex]
Substitute the given parameters into the formula:
[tex]C_1V_1=C_2V_2\\0.5(5)=100C_2\\2.5=100C_2\\C_2=\frac{2.5}{100}\\C_2= 0.025M[/tex]
Hence the concentration of the final diluted solution is 0.025M
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Homolysis, or homolytic bond dissociation, produces a very specific type of product under certain reaction conditions. In Part 1, select all the products (in formulae and general chemical terms) that could result from homolysis. In Part 2, select the reaction conditions that are most likely to promote homolysis.
Part 1. Choose all that may occur as possible products of a homolysis reaction.
Choose one or more:_______.
a. hydride ion
b. R3CO
c. Br2
d. H
e. a carbocation
f. H3C
g. H3CO-
h. hydrogen ion
i. a carbon free radical
Part 2. Choose the conditions under which homolysis is likely to occur.
Choose one or more:_______.
a. strong base
b. ultraviolet irradiation
c. high temperature
d. strong acid
e. infrared irradiation
f. low temperature
Answer:
1) R₃CO , H, H₃C, a carbon free radical
2) high temperature, ultraviolet irradiation
Explanation:
1) Homolysis leads to the formation of free radicals (species having a free electron). Thus, answer is :
R₃CO
H
H₃C
a carbon free radical
2) Homolysis require high temperature, ultraviolet irradiation.
Which type of rock is formed from existing rock or organisms?
Answer:
Sedimentary rocks
Explanation:
My explanation is that when an animal decomposes it body returns to the ground eventually being used in the rock cycle and rocks form this through the rock cycle when broken down by weathering and erosion.
Hope this helps you
Answer:
sedimentary rocks
They form from deposits that accumulate on the Earth's surface.
How many valence electrons are in the electron dot structures for the elements in group 3A(13)?
Answer:
here, as we have known the elements of group 3A(13) such as aluminium , boron has three valance electron and in perodic table the elements are kept with similar proterties in same place so, their valance electron is 3.
hope it helps...
The number of valence electrons are in the electron dot structures for the elements in group 3A(13) is three.
What are Groups in the Periodic Table?The periodic table is organized into groups (vertical columns), periods (horizontal rows), and families (groups of elements that are similar). Elements in the same group have the same number of valence electrons.
Groups are the columns of the periodic table, and periods are the rows. There are 18 groups, and there are 7 periods plus the lanthanides and actinides.
There are two different numbering systems that are commonly used to designate groups, and you should be familiar with both.
The traditional system used in the United States involves the use of the letters A and B. The first two groups are 1A and 2A, while the last six groups are 3A through 8A. The middle groups use B in their titles.
Therefore, The number of valence electrons are in the electron dot structures for the elements in group 3A(13) is three.
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When 75.5 grams of phosphorus pentachloride react with an excess of water, as shown in the unbalanced chemical equation below, how many moles of hydrochloric acid will be produced? Please show all your work for the calculations for full credit. PCl5 + H2O --> H3PO4 + HC
Answer:
Explanation: M(PCL5)= 31 + 5(35.5)
=208.5g/mol
M(H20)= 18g/mol
n(PCL5) = 75.5÷208.5
= 0.362mol
n(HCl)/n(PCL5)= 5/1
n(HCl)= 5×0.362
=1.81mol of HCl
Silver crystallizes in a face-centered cubic structure. What is the edge length of the unit cell if the atomic radius of silver is 144 pm?
Answer:
Edge length of the unit cell is 4.07x10⁻¹⁰m
Explanation:
In a face-centered cubic structure, the edge, a, could be obtained using pythagoras theorem knowing the hypotenuse of the unit cell, b, is equal to 4r:
a² + a² = b² = (4r)²
2a² = 16r²
a = √8 r
That means edge lenght is = √8 r
adius
As radius of Silver is 144pm = 144x10⁻¹²m:
a = √8 r
a = √8 ₓ 144x10⁻¹²m
a = 4.07x10⁻¹⁰m
Edge length of the unit cell is 4.07x10⁻¹⁰mSelect the correct answer
What determines the average kinetic energy of the particles in a gas?
ОА
the number of collisions
OB.
the number of particles
OC. the size of the particles
OD. the temperature
Answer:
Temperature
Explanation:
Kinetic energy of gass molecules is directly propotional to the temperature.
help please i have 5 minutes to do this !!!
Answer:
A) One that occurs on its own
When silver nitrate is added to the Fe/SCN equilibrium, why is the colorless intense and a precipitate forms?
Answer:
Here's what I get
Explanation:
You have an equilibrium reaction between Fe³⁺/ SCN⁻ and FeSCN²⁺.
[tex]\underbrace{\hbox{Fe$^{3+}$}}_{\text{pale yellow-green}} +\underbrace{\hbox{SCN$^{-}$}}_{\text{colourless}} \, \rightleftharpoons \, \underbrace{\hbox{Fe(SCN)$^{2+}$}}_{\text{deep blood red}} \\[/tex]
When you add AgNO₃, the Ag⁺ reacts with the SCN⁻. It forms a colourless precipitate of Ag(SCN).
Ag⁺(aq) + SCN⁻(aq) ⟶ AcSCN(s)
According to Le Châtelier's Principle, when we apply a stress to a system at equilibrium, the system will respond in a way that tends to relieve the stress.
If you add Ag⁺ to the equilibrium solution, it removes the SCN⁻ [as an Ag(SCN) precipitate].
The system responds by trying to replace the missing SCN⁻:
The Fe(SCN)²⁺ dissociates to form SCN⁻, so the position of equilibrium shifts to the left,
You now have more Fe³⁺ and SCN⁻ and less of the highly coloured Fe(SCN)²⁺ at the new equilibrium.
The deep red colour becomes less intense.
When silver nitrate is added to the Fe/SCN equilibrium, the colourless intense and precipitate forms because it settles at the bottom.
What is chemical equilibrium?Chemical equilibrium is the condition in the course of a reversible chemical reaction in which no net change in the amounts of reactants and products occurs.
The added silver nitrate, [tex]AgNO_3[/tex] , effectively removes thiocyanate ions, [tex]SCN^{-1}[/tex], from the equilibrium system via a precipitation reaction when the [tex]Ag^{+1}[/tex] combines with [tex]SCN^{-1}[/tex] to produce insoluble silver thiocyanate, AgSCN, which settles to the bottom of the test tube.
Ag⁺(aq) + SCN⁻(aq) ⟶ AcSCN(s)
According to Principle, when we apply stress to a system at equilibrium, the system will respond in a way that tends to relieve the stress.
Adding Ag⁺ to the equilibrium solution, it removes the SCN⁻ [as an Ag(SCN) precipitate].
The system responds by trying to replace the missing SCN⁻:
The Fe(SCN)²⁺ dissociates to form SCN⁻, so the position of equilibrium shifts to the left,
You now have more Fe³⁺ and SCN⁻ and less of the highly coloured Fe(SCN)²⁺ at the new equilibrium.
The deep red colour becomes less intense.
Learn more about chemical equilibrium here:
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Which of the following is NOT one of the types of bonds? A. Ionic B. Metallic C. Covalent D. Valence
Considering the definition of bond and the different type of bonds, valence is not one of the types of bonds.
What is a chemical bondA chemical bond is defined as the force by which the atoms of a compound are held together. These are electromagnetic forces that give rise to different types of chemical bonds.
In other words, a chemical bond is the force that joins atoms to form chemical compounds and confers stability to the resulting compound.
Covalent bondThe covalent bond is the chemical bond between atoms where electrons are shared, forming a molecule. Covalent bonds are established between non-metallic elements, such as hydrogen H, oxygen O and chlorine Cl. These elements have many electrons in their outermost level (valence electrons) and have a tendency to gain electrons to acquire the stability of the electronic structure of noble gas. The shared electron pair is common to the two atoms and holds them together.
Ionic bondAn ionic bond is produced between metallic and non-metallic atoms, where electrons are completely transferred from one atom to another. During this process, one atom loses electrons and another one gains them, forming ions.
Metallic bondMetallic bonds are a type of chemical bond that occurs only between atoms of the same metallic element. In this way, metals achieve extremely compact, solid and resistant molecular structures, since the atoms that share their valence electrons.
SummaryIn summary, valence is not one of the types of bonds. The types of bonds are covalent, ionic and metallic.
Learn more about chemical bonds:
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