Answers
Answer:
B. s, p, d, f
Explanation:
These things are often referred to as suborbitals and you normally have s,p,d,f.
S has 1 two orbitals
P has 3 orbitals
D has 5 orbitals
F has 7 orbitals
and each orbital can house 2 electrons
Related Questions
find the mass of h2 produced Binary compounds of alkali metals and hydrogen react with water to produce H2(g). The H2H2 from the reaction of a sample of NaH with an excess of water fills a volume of 0.505 L above the water. The temperature of the gas is 35 ∘C∘C and the total pressure is 755 mmHg
Answers
Answer: Mass of hydrogen produced is 0.0376 g.
Explanation:
The reaction equation will be as follows.
[tex]NaH(aq) + H_{2}O(l) \rightarrow H_{2}(g) + NaOH(aq)[/tex]
Now, formula for total pressure will be as follows.
[tex]P_{total} = P_{H_{2}} + P_{H_{2}O}[/tex]
Hence, [tex]P_{H_{2}} = P_{total} - P_{H_{2}O}[/tex]
= 755 mm Hg - 42.23 mm Hg
= 712.77 mm Hg
[tex]P_{H_{2}} = \frac{712.77 \times 1 atm}{760 mm Hg}[/tex]
= 0.937 atm
Now, we will calculate the moles of [tex]H_{2}[/tex] as follows.
[tex]P_{H_{2}}V = nRT[/tex]
[tex]0.937 atm \times 0.505 L = n \times 0.0821 \times 308.15 K[/tex]
n = [tex]\frac{0.473}{25.29}[/tex] mol
= 0.0187 mol
Therefore, mass of [tex]H_{2}[/tex] will be calculated as follows.
[tex]m_{H_{2}} = \frac{0.0187 mol \times 2.0158 g}{1 mol}[/tex]
= 0.0376 g
Thus, we can conclude that mass of hydrogen produced is 0.0376 g.
Design your own experiment: Factors that affect the rate of a reaction
Did anybody do the lab?
Answers
Answer:
Reactant concentration, the physical states of the reactants, surface area, temperature and the presence of catalyst.
The quantum theory of energy levels within atoms was aided by:
study of the sun's light spectrum
emission line spectra of various elements
alpha particles
gamma rays
Answers
Answer:
it was based on the studies by emission line spectra of various elements
Explanation:
Answer:
The answer is: emission line spectra of various elements
Explanation:
Which of the following would be more reactive than magnesium (Mg)?
A. Calcium (Ca)
B. Potassium (K)
C. Argon (Ar)
D. Beryllium (Be)
Answers
Answer:potassium is more reactive than Mg because both lie in the same group and the element potassium has more electropositivity than magnesium
Explanation:
I hope it will help you
Answer: B. Potassium(K)
Explanation:
PdPd has an anomalous electron configuration. Write the observed electron configuration of PdPd. Express your answer in complete form in order of orbital filling. For example, 1s22s21s22s2 should be entered as 1s^22s^2. View Available Hint(s)
Answers
Answer:
1s²,2s²,2p⁶,3s²,3p⁶,4s²,3d¹⁰,4p⁶,5s⁰,4d¹⁰.
Explanation:
Palladium is a chemical element with the symbol Pd and atomic number 46.
The electronic configuration is;
[Kr] 4d¹⁰
The full electronic configuration observed for palladium is given as;
1s²,2s²,2p⁶,3s²,3p⁶,4s²,3d¹⁰,4p⁶,5s⁰,4d¹⁰.
The reason for for the anomlaous electron configuration is beacuse;
1. Full d orbitals are more stable than partially filled ones.
2. At higher energy levels, the levels are said to be degenerated which means that they have very close energies and then electrons can jump from one orbital to another easily.
Copper sulfate is a blue solid that is used to control algae growth. Solutions of copper sulfate that come in contact with the surface of galvanized ( Zinc-plated) steel pails undergo the following reaction that forms copper metal on the zinc surface. How many grams of Zinc would react with 454g (1lb) of copper sulfate (160g/mol)?
CuSO4(aq)+ Zn(s)>>>>Cu(s) + ZnSO4(aq)
Answers
Answer:
185.49 grams of Zinc would react with 454g (1lb) of copper sulfate
Explanation:
Yo know the following balanced reaction:
CuSO₄(aq)+ Zn(s) →Cu(s) + ZnSO₄(aq)
You can see that by stoichiometry of the reaction (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of reagents and products are part of the reaction:
CuSO₄: 1 moleZn: 1 moleCu: 1 moleZnSO₄: 1 moleBeing:
Cu: 63.54 g/moleS: 32 g/moleO: 16 g/moleZn: 65.37 g/molethe molar mass of the compounds participating in the reaction is:
CuSO₄:63.54 g/mole + 32 g/mole + 4*16 g/mole= 159.54 g/mole ≅ 160 g/moleZn: 65.37 g/moleCu: 63.54 g/moleZnSO₄: 65.37 g/mole + 32 g/mole + 4*16 g/mole= 161.37 g/moleThen, by stoichiometry of the reaction, the following amounts of mass of reagent and product participate in the reaction:
CuSO₄: 1 moles* 160 g/mole= 160 gZn: 1 mole* 65.37 g/mole= 65.37 gCu: 1 mole* 63.54 g/mole= 63.54 gZnSO₄: 1 mole* 161.37 g/mole= 161.37 gNow you can apply the following rule of three: if 160 grams of CuSO₄ react with 65.37 grams of Zn by this reaction stoichiometry, 454 grams of CuSO₄ with how much mass of Zn will it react?
[tex]mass of Zn=\frac{454 grams of CuSO_{4} *65.37 grams of Zn}{160 grams of CuSO_{4}}[/tex]
mass of Zn= 185.49 grams
185.49 grams of Zinc would react with 454g (1lb) of copper sulfate
25.00 mL of a H2SO4 solution with an unknown concentration was titrated to a phenolphthalein endpoint with 28.11 mL of a 0.1311 M NaOH solution. What is the concentration of the H2SO4 solution
Answers
Answer:
Concentration of the H₂SO₄ solution is 0.0737 M
Explanation:
Equation of the neutralization reaction between the acid, H₂SO₄, and the base, NaOH, is given below:
H₂SO₄ + 2NaOH -----> Na₂SO₄ + 2H₂O
From the above equation, one mole of acid requires 2 moles of base for complete neutralization which occurs at phenolphthalein endpoint.
mole ratio of acid to base, nA/nB = 1:2
Concentration of the base, Cb = 0.1311 M
Volume of base, Vb, = 28.11 mL
Concentration of acid, Ca = ?
Volume of acid, Va + 25.0 mL
Using the formula, CaVa/CbVb = nA/nB
making Ca subject of the formula, Ca = Cb*Vb*nA/Va*nB
substituting the values into the equation
Ca = (0.1311 * 28.11 * 1) / 25.0 * 2 = 0.0737 M
Therefore, concentration of the H₂SO₄ solution is 0.0737 M
A certain reaction with an activation energy of 155 kJ/mol was run at 495 K and again at 515 K . What is the ratio of f at the higher temperature to f at the lower temperature
Answers
Answer:
4.32 is the ratio of f at the higher temperature to f at the lower temperature
Explanation:
Using the sum of Arrhenius equation you can obtain:
ln (f₂/f₁) = Eₐ / R ₓ (1/T₁ - 1/T₂)
Where f represents the rate constant of the reaction at T₁ and T₂ temperatures. Eₐ is the energy activation (155kJ / mol = 155000J/mol) and R is gas constant (8.314J/molK)
Replacing:
ln (f₂/f₁) = 155000J/mol / 8.314J/molK ₓ (1/495K - 1/515)
Where 2 represents the state with the higher temperature and 1 the lower temperature.
ln (f₂/f₁) = 155000J/mol / 8.314J/molK ₓ (1/495K - 1/515)
ln (f₂/f₁) = 1.4626
f₂/f₁ = 4.32
4.32 is the ratio of f at the higher temperature to f at the lower temperature
Refer to the example about diatomic gases A and B in the text to do problems 20-28.
It was determined that 1 mole of B2 is needed to react with 3 moles of A2.
How many grams in one mole of B2?
__g
Answers
Answer:
28g.
Explanation:
Hello,
In this case, considering the statement, we can infer that the monoatomic atomic mass of B is 14 g in one one mole. In such a way, since it is diatomic, we can notice that one mole of B2, is having 28 g of B2, as monoatomic atomic mass is considered twice.
Regards.
Suppose you are performing an extraction procedure with a carboxylic acid, like a benzoic or toluic acid Which layer should contain the carboxylic acid in a two-layer mixture of water and an organic solvent, like diethyl ether? A. The organic layer B. The aqueous layer C. Neither layer
Answers
Answer:
The correct answer is option A, that is, the organic layer.
Explanation:
One knows that non-polar solvent dissolves non-polar substances and the polar solvent dissolves polar substances, that is, like dissolved like. The toluic acid or benzoic acid refers to the carboxylic acids, which are non-polar or are very less polar in characteristics.
Therefore, they possess the tendency to remain in the organic solvent, which is non-polar like diethyl ether. The toluic acid's or benzoic acid's non-polar characteristic is because of the existence of huge hydrophobic rings of benzene. Hence, the correct answer is option A, that is, the organic layer.
Decide which element probably forms a compound with chlorine that has a chemical formula most and least similar to the chemical formula of the compound formed by chlorine and strontium.
Answers
Answer:
a. Beryllium,
b. Nitrogen
Explanation:
Most similar to strontium is beryllium, they both are in the 2d group.
Least similar to strontium is nitrogen. Strontium is a metal in the 2d group, Nitrogen is non-metal in the 15th group
The most common isotopic forms of hydrogen are ordinary hydrogen (1H) and deuterium (2H), which have percent compositions of 99.98% and 0.0115%, respectively. Convert the percent isotopic composition value of 2H to decimal form.
Answers
Answer:
0. 000115
Explanation:
A percentage is defined as a ratio with a basis of 100 as total substance. Convert a percentage to decimal implies to divide the percentage in 100 because decimal form has as basis 1.
For the isotopic forms:
1H: 99.98% → As percent.
99.98% / 100 = 0.9998 → As decimal form.
2H: 0.0115% → As percent.
0.0115% / 100 = 0. 000115→ As decimal form.
The percent should be 0. 000115
The calculation is as follows:For the isotopic forms:
1H: 99.98% → As percent.
Now
[tex]99.98\% \div 100[/tex]= 0.9998 → As decimal form.
Now
2H: 0.0115% → As percent.
And,
[tex]0.0115\% \div 100[/tex]= 0. 000115→ As decimal form.
Learn more: https://brainly.com/question/6789603?referrer=searchResults
Consider the reaction:
2H2O(l)2H2(g) + O2(g)
Using standard thermodynamic data at 298K, calculate the entropy change for the surroundings when 1.73 moles of H2O(l) react at standard conditions.
Answers
Answer:
The correct answer is -1659.17 J/K.
Explanation:
The reaction given is:
2H₂O (l) ⇔ 2H₂ (g) + O₂ (g)
In the given case, first there is a need to find ΔHreaction, which is equivalent to ΔHf (products) - ΔHf(reactants)
Based on the standard thermodynamic table, the ΔHf(H₂O) is -285.8 KJ/mol, the ΔHf(H₂) is 0 KJ/mol, and the ΔHf(O₂) is 0 KJ/mol.
On putting the values, the ΔHreaction will be,
ΔHreaction = 2 × ΔHf(H₂) + ΔHf(O₂) - 2 × ΔHf(H₂O)
= 2 × 0 + 0 - 2 × (-285.8 KJ/mol) = 571.6 KJ
The calculated value of ΔHreaction is for the two moles of H₂O, now for 1.73 moles of H₂O it will be,
ΔHreaction = +571.6 KJ / 2 mol × 1.73 mol = 494.434 KJ
The temperature given in the question is 298 K, now ΔSsurrounding will be,
ΔSsurrounding = -ΔHreaction/T = -494434 J/298 K = -1659.17 J/K.
a) What substances are present in an aqueous buffer composed of HC2H3O2 and C2H3O2 - ?b) What happens when LiOH is added to a buffer composed of HC2H3O2 and C2H3O2 - ? Write a chemical equation for that reaction.c) What happens when HBr is added to this buffer? Write a chemical equation for that reaction.
Answers
Answer:
a) HC₂H₃O₂, C₂H₃O₂⁻, H₃O⁺, H₂O, OH⁻
b) HC₂H₃O₂ + LiOH ⇄ H₂O + LiC₂H₃O₂
c) C₂H₃O₂⁻ + HBr ⇄ HC₂H₃O₂ + Br⁻
Explanation:
a) In a HC₂H₃O₂/C₂H₃O₂⁻ buffer system, the following reactions take place:
HC₂H₃O₂ + H₂O ⇄ C₂H₃O₂⁻ + H₃O⁺
C₂H₃O₂⁻ + H₂O ⇄ HC₂H₃O₂ + OH⁻
Thus, the species present are: HC₂H₃O₂, C₂H₃O₂⁻, H₃O⁺, H₂O, OH⁻.
b) When LiOH is added to the buffer system, it is partially neutralized according to the following equation.
HC₂H₃O₂ + LiOH ⇄ H₂O + LiC₂H₃O₂
c) When HBr is added to the buffer system, it is partially neutralized according to the following equation.
C₂H₃O₂⁻ + HBr ⇄ HC₂H₃O₂ + Br⁻
How unwanted radiation in AES, AFS, AES rejected
Answers
Answer:
Reduce the intensity of radiations
Explanation:
Atomic Emission Spectroscopy is the study of electromagnetic radiations that are emitted by matter when it is excited. It is used to analyze intensity of light emitted. To end the radiations of AFS and AES their intensity is reduced and they are gradually rejected. It is not possible to completely eliminate the radiations immediately.
Naturally occurring sulfur consists of four isotopes: 32S (31.97207 u, 95.0%); 33S (32.97146 u, 0.76%); 34S (33.96786 u, 4.22%); and 36S (35.96709 u, 0.014%). Calculate the average atomic mass of sulfur in atomic mass units.
Answers
Answer:
32.062
Explanation:
The following data were obtained from the question:
Mass of isotope A (32S) = 31.97207 u
Abundance of isotope A (A%) = 95.0%
Mass of isotope B (33S) = 32.97146 u Abundance of isotope B (B%) = 0.76%
Mass of isotope C (34S) = 33.96786 u
Abundance of isotope C (C%) = 4.22%
Mass of isotope D (36S) = 35.96709 u Abundance of isotope D (D%) = 0.014%
Average atomic mass of S =..?
The average atomic mass of sulphur, S can be obtained as follow:
Average atomic mass = [(Mass of A x A%)/100] + [(Mass of B x B%)/100] + [(Mass of C x C%)/100] + [(Mass of D x D%)/100]
Average atomic mass of sulphur =
[(31.97207 x 95)/100] + [(32.97146 x 0.76)/100] + [(33.96786 x 4.22)/100] + [(35.96709 x 0.014)/100]
= 30.373 + 0.251 + 1.433 + 0.005
= 32.062
Therefore, the average atomic mass of sulphur is 32.062
Thermal decomposition of 5.00 metric tons of limestone to lime and carbon dioxide requires 9.00 × 106 kJ of heat. Convert this energy to (a) joules; (b) calories; (c) British thermal units. Give your answers in scientific notation.
Answers
Answer:
Take a look at the attachment below
Explanation:
Hope that helps!
what are the differences between strong and weak acids?
Answers
Strong acids are completely ionised and weak acids are partly ionised
Answer:
Como forman los iones en soluciión
Explanation:
Los ácidos fuertes y las bases fuertes se refieren a especies que se disocian completamente para formar los iones en solución. Por el contrario, los ácidos y bases débiles se ionizan solo parcialmente y la reacción de ionización es reversible.
The solvent was propanone. Which of the three basic colours is most soluble in propanone?
Answers
Answer:
Red dye
Explanation:
In the given question, the complete question has not been provided but the propanone is used as a solvent in paper chromatography. The paper chromatography was performed for the black ink in which the black ink got separated in the red, blue and yellow colour.
From these three colours that are red, blue and yellow, the dye which is most soluble in propanone was red as red colour moved the most in the given chromatogram and the dye which travelled the most is most soluble in propanone.
Thus, red dye is the correct answer.
The value of ΔG°′ΔG°′ for the conversion of glucose-6-phosphate to fructose-6-phosphate (F6P) is +1.67 kJ/mol+1.67 kJ/mol . If the concentration of glucose-6-phosphate at equilibrium is 2.65 mM2.65 mM , what is the concentration of fructose-6-phosphate? Assume a temperature of 25.0°C25.0°C .
Answers
Answer:
The concentration of fructose-6-phosphate F6P ≅ 1.35 mM
Explanation:
Given that:
ΔG°′ is the conversion of glucose-6-phosphate to fructose-6-phosphate (F6P) = +1.67 kJ/mol = 1670 J/mol
concentration of glucose-6-phosphate at equilibrium = 2.65 mM
Assuming temperature = 25.0°C
=( 25 + 273)K
= 298 K
We are to find the concentration of fructose-6-phosphate
Using the relation;
ΔG' = -RT In K_c
where;
R = 8.314 J/K/mol
1670 = - (8.314 × 298 ) In K_c
1670 = -2477.572 × In K_c
1670/ 2477.572 = In K_c
0.67 = In K_c
[tex]K_c = e^{-0.67}[/tex]
[tex]K_c =[/tex] 0.511
Now using the equilibrium constant [tex]K_c[/tex]
[tex]K_c = \dfrac{[F6P]}{[G6P]}[/tex]
[tex]0.511 = \dfrac{[F6P]}{[2.65]}[/tex]
F6P = 0.511 × 2.65
F6P = 1.35415
F6P ≅ 1.35 mM
The simplest carboxylic acid is called *
O Formaldehye
O formic acid
acetic acid
O
acetone
Answers
Consider the insoluble compound silver bromide , AgBr . The silver ion also forms a complex with ammonia . Write a balanced net ionic equation to show why the solubility of AgBr (s) increases in the presence of ammonia and calculate the equilibrium constant for this reaction. For Ag(NH3)2+ , Kf = 1.6×107 . Use the pull-down boxes to specify states such as (aq) or (s).
Answers
Answer:
- [tex]AgBr(s)+2NH_3(aq)\rightleftharpoons Ag(NH_3)_2^+(aq)+Br^-(aq)[/tex]
- [tex]K=1.2x10^{-5}[/tex]
Explanation:
Hello,
In this case, by considering the dissolution of silver bromide:
[tex]AgBr(s)\rightleftharpoons Ag^+(aq)+Br^-(aq) \ \ \ Ksp=[Ag^+][Br^-]=7.7x10^{-13}[/tex]
And the formation of the complex:
[tex]Ag^+(aq)+2NH_3(aq)\rightleftharpoons Ag(NH_3)_2^+(aq)\ \ \ Kf=\frac{[Ag(NH_3)_2^+]}{[Ag^+][NH_3]^2}=1.6x10^7[/tex]
We obtain the balanced net ionic equation by adding the aforementioned equations:
[tex]AgBr(s)+Ag^+(aq)+2NH_3(aq)\rightleftharpoons Ag(NH_3)_2^+(aq)+Br^-(aq)+Ag^+(aq)\\\\AgBr(s)+2NH_3(aq)\rightleftharpoons Ag(NH_3)_2^+(aq)+Br^-(aq)[/tex]
Now, the equilibrium constant is obtained by writing the law of mass action for the non-simplified net ionic equation:
[tex]AgBr(s)+Ag^+(aq)+2NH_3(aq)\rightleftharpoons Ag(NH_3)_2^+(aq)+Br^-+Ag^+\\\\K=[Ag^+][Br^-]*\frac{[Ag(NH_3)_2^+]}{[Ag^+][NH_3]^2}[/tex]
So we notice that the equilibrium constant contains the solubility constant and formation constant for the initial reactions:
[tex]K=Ksp*Kf=7.7x10^{-13}*1.6x10^{7}\\\\K=1.2x10^{-5}[/tex]
Best regards.
Vanadium (V) and oxygen (O) form a series of compounds with the following compositions: Mass % V 76.10 67.98 61.42 56.02 Mass % O 23.90 32.02 38.58 43.98 Compound Mass % N 1 33.28 2 39.94 Mass % Si 66.72 60.06 10. What are the relative numbers of atoms of oxygen in the compounds for a given mass of vanadium
Answers
Answer:
For every given mass of Vanadium, the relative number of oxygen atoms present or the mole ratio of Oxygen to Vanadium is:
A. 1:1
B. 3:2
C. 2:1
D. 5:2
Note: The question is stated more clearly below:
Vanadium (V) and oxygen (O) form a series of compounds with the following compositions: Mass % V 76.10 67.98 61.42 56.02 Mass % O 23.90 32.02 38.58 43.98 Compound Mass % N 1 33.28 2 39.94.
What are the relative numbers of atoms of oxygen in the compounds for a given mass of vanadium?
Explanation:
Number of moles in 100 g mass = % mass / molar mass
Molar mass of Vanadium, V = 51 g/mol
Molar mass of oxygen atom, O = 16 g/mol
1. Percentage mass of V and O is 76.10% and 23.90% respectively.
Number of moles of each atom;
V = 76.10/51.0 = 1.5 moles
O = 23.9/16 = 1.5 moles
Mole ratio of oxygen to vanadium = 1.5/1.5 = 1 : 1
2. Percentage mass of V and O is 67.98% and 32.02% respectively
Number of moles of each atom:
V = 67.98/51 = 1.33
O = 32.02/16 = 2
Mole ratio of oxygen to vanadium = 2/1.33 = 1.5 : 1 = 3 : 2
3. Percentage mass of V and O is 61.42% and 38.58% respectively
Number of moles of each atom:
V = 61.42/51 = 1.2
O = 38.58/16 = 2.4
Mole ratio of oxygen to vanadium = 2.4/1.2 = 2 : 1
4. Percentage mass of V and O is 56.02% and 43.98% respectively
Number of moles of each atom:
V = 56.02/51 = 1.10
O = 43.98/16 = 2.75
Mole ratio of oxygen to vanadium = 2.75/1.10 = 2.5 : 1 = 5 : 2
Mass of the Vanadium, number of O2 atoms present, or the mole ratio of 1:1 , 3:2 , 2:1 , 5:2 . As Vanadium (V) and oxygen (O) form a series of compounds is given with masses of 76.10 67.98, 23.90 32.02, 33.28 2 39.94, etc.
As per No of moles in 100 g mass = % mass / molar mass Mass of Vanadium, V = 51 g/ mol e, Mass of oxygen atom, O = 16 g/mole O = 23.9/16 = 1.5 moles for oxygen to vanadium = 1.5/1.5 = 1 : 1 2. Percentage mass of V and O is 67.98% and 32.02%. Mole ratio of oxygen to vanadium = 2/1.33 = 1.5 : 1 = 3 : 2 3. Percentage mass of V and O is 61.42% and 38.58% Mole ratio of oxygen to vanadium = 2.4/1.2 = 2 : 1 4. Percentage mass of V and O is 56.02% and 43.98%. Mole ratio of oxygen to vanadium = 2.75/1.10 = 2.5 : 1 = 5 : 2Learn more about the Vanadium (V) and oxygen (O).
brainly.com/question/2145642.
key A 0.150 M sodium chloride solution is referred to as a physiological saline solution because it has the same concentration of salts as normal human blood. Calculate the mass of solute needed to prepare 275.0 mL of a physiological saline solution.
Answers
Answer:
2.41065 grams
Explanation:
Here we have to apply molarity, particularly in reference to the equation molarity = moles of solute / volume. I would like to rewrite this formula, but with respect to the units - grams = moles / Liters,
We can use molarity to determine the number of moles. After doing so, we can determine the mass of the solute with respect to the formula moles = mass / molar mass. The molar mass of NaCl is 58.44 grams.
_______________________________________________________
275 mL = 0.275 L,
Number of Moles of NaCl = 0.150 * 0.275 = 0.04125 moles,
Mass = 0.04125 * 58.44 = 2.41065 grams,
Solution - Mass of NaCl = 2.41065 grams
Hope that helps!
Answer:
2.41g
Explanation:
Data obtained from the question include the following:
Molarity of NaCl = 0.150M
Volume = 275mL
Mass of NaCl =..?
Next, we shall determine the number of mole of NaCl in the solution. This is illustrated below:
Molarity of a solution is simply defined as the mole of solute per unit litre of the solution. Mathematically, the molarity is expressed as:
Molarity = mole /Volume
Molarity = 0.150M
Volume = 275mL = 275/1000 = 0.275L
Mole =..?
Molarity = mole /Volume
0.150 = mole /0.275
Cross multiply
Mole = 0.15 x 0.275
Mole = 0.04125 mole
Therefore, the number of mole of solute, NaCl in the solution is 0.04125 mole.
Finally, we shall convert 0.04125 mole of NaCl to farms. This is illustrated below:
Molar mass of NaCl =23 + 35.5 = 58.5g/mol
Mole of NaCl = 0.04125 mole
Mass of NaCl =..?
Mole = mass/molar mass
0.04125 = Mass /58.5
Cross multiply
Mass = 0.04125 x 58.5
Mass = 2.41g
Therefore, the mass of the solute, NaCl needed to prepare the solution is 2.41g
Spell out the full name of the compound.
Answers
Answer:
4–octene.
Explanation:
To name the compound given in the question, we must determine the following:
1. Determine the functional group of the compound and locate its position by giving it the lowest possible count.
2. Locate the longest continuous carbon. This gives the parent name of the compound.
3. Combine the above to obtain the name of the compound.
Now, let us determine the name of the compound bearing in mind the information given above. This is illustrated below:
1. The functional group of the compound is double bond i.e alkene and it located at carbon 4.
2. The longest continuous carbon chain is 8. Since the compound is an alkene, the name becomes octene.
3. Therefore, the name of the compound is:
4–octene.
Consider the three isomeric alkanes n-hexane,2,3-dimethylbutane, and 2-methylpentane. Which of the following correctly lists these compounds in order of increasing boiling point
a. 2,3-dimethylbutane < 2-methylpentane < n-hexane
b. 2-methylpentane
c. 2-methylpentane < 2,3-dimethylbutane
d. n-hexane < 2-methylpentane < 2,3-dimethylbutane
e. n-hexane < 2,3-dimethylbutane < 2-methylpentane
Answers
Answer:
a. 2,3-dimethylbutane < 2-methylpentane < n-hexane
Explanation:
The boiling point of alkanes is highly affected by the degree of branching in the molecule. Branched alkanes generally have a lower boiling point than unbranched alkanes.
The reason for the higher boiling point of unbranched alkanes is because they have greater vanderwaals forces acting between their molecules due to their larger surface area. Recall that branched alkanes have a lesser surface area compared to unbranched alkanes.
n-hexane is an unbranched alkane hence it will have the highest boiling point followed by 2-methyl pentane and lastly 2,3-dimethyl butane. The boiling point continues to decrease as the extent of branching increases.
A student mixes baking soda and vinegar in a glass. Are there any new substances created from this mixture?
Answers
Answer:
Explanation:
1. A student mixes baking soda and vinegar in a glass. The results are shown at left. ... Yes I do belive that new substances are being formed because there is a chemical reaction between the baking soda and vinegar turning it into a bubbly substances instead of a powder and liquid.
Yes, there are new substances created from this mixture.
1-hexanol was prepared by reacting an alkene with either hydroboration-oxidation or oxymercuration-reduction. Draw the structure of the alkene that was used to prepare the alcohol in highest yield. You do not have to consider stereochemistry. Indicate the method of preparation by drawing either BH3 (for hydroboration-oxidation), or Hg (for oxymercuration-reduction), in a separate sketcher. If there is more than one alkene that can be used for a given method, draw all of them. If either hydroboration-oxidation or oxymercuration-reduction can be used, just give the structures for one method. Separate structures with signs from the drop-down menu.
Answers
Answer:
Alkene form hexan-1-ol with oxidation in presence of NaOH with highest yield
Explanation:
The following reaction: NO2 (g) --> NO (g) 1/2 O2 (g) is second-order in the reactant. The rate constant for this reaction is 3.40 L/mol*min. Determine the time needed for the concentration of NO2 to decrease from 2.00 M to 1.50 M.
Answers
Answer:
t = 0.049 mins or 2.94 secs
Explanation:
For a simple second order reaction, the integrated law which describes the concentration of reactants at a given time t, is as follows: 1/[A] = 1/[A]o + Kt;
Where [A] is concentration of reactant at time, t, [A]o is initial concentration of A; K is rate constant; t is time at a given instant.
Using the integrated rate law:
I/[NO2]t - 1/[NO2]o = Kt
Where K = 3.40 L/mol/min
[NO2]t = 1.5 mol/L
[N02]o = 2.0 mol/L
t = ?
Making t subject of formula;
t = (1/[NO2]t - 1/[NO2]o) / K
t = (1/1.5 - 1/2.0)/3.40
t = 0.049 mins or 2.94 secs
Prepare a solution that is 0.1 M acetic acid and 0.1 M sodium acetate by measuring out 5.0 mL of the 1.0 M acetic acid solution and 5.0 mL of the 1.0 M sodium acetate solution in a 100 mL graduated cylinder, diluting the 10.0 mL to a final volume of 50.0 mL with deionized water, and then stirring. Pour this solution into a clean, dry 100 mL breaker. By knowing that the Ka for acetic acid is 1.8 x -5 10 , calculate the theoretical pH of the solution.
Answers
Answer:
4.74
Explanation:
It is possible to find pH of a buffer (The mixture of a weak acid: Acetic acid, with its conjugate base: Sodium acetate) using Henderson-Hasselbalch equation:
pH = pKa + log₁₀ [A⁻] / [HA]
Where pKa is -log Ka of the weak acid, [A⁻] concentration of the conjugate base and [HA] concentration of the weak acid
pKa of acetic acid is -log 1.8x10⁻⁵ = 4.74
The concentration of both, acetic acid and sodium acetate is 0.1M. Replacing in H-H equation:
pH = pKa + log₁₀ [A⁻] / [HA]
pH = 4.74 + log₁₀ [0.1] / [0.1]
pH = 4.74 + log₁₀ 1
pH = 4.74
Theoretical pH is 4.74