Answer:
See explanation and image attached
Explanation:
The image attached shows the entire scheme of reactions mentioned in the question.
The first reaction is an addition reaction which yields a tertiary alkyl halide as shown in accordance with Markovnikov rule.
The second reaction is a dehydrohalogenation in which the base abstracts a proton from the alkyl halide followed by loss of a bromide ion to yield the corresponding alkene.
This alkene is an isomer of the starting material.
Calculate the molality of each of the following solutions:
(a) 14.3 g of sucrose (C12H22O11) in 685 g of water,
(b) 7.15 moles of ethylene glycol (C2H6O2) in 3505 g of water.
Answer:
(a) The molality of this solution is 0.0613[tex]\frac{moles}{kg}[/tex]
(b)The molality of this solution is 2.04[tex]\frac{moles}{kg}[/tex]
Explanation:
The molality (m) of a solution is defined as the number of moles of solute present per kg of solvent.
The Molality of a solution is determined by the expression:
[tex]Molality=\frac{number of moles of solute}{kilograms of solvent}[/tex]
Molality is expressed in units [tex]\frac{moles}{kg}[/tex]
(a) You have 14.3 g of sucrose (C₁₂H₂₂O₁₁), the solute. With the molar mass of sucrose being 342 [tex]\frac{g}{mole}[/tex], then 14.3 grams of the compound represents the following number of moles:
[tex]14.3 grams*\frac{1 mole}{342 grams} =[/tex] 0.042 moles
Having 685 g= 0.685 kg (being 1000 g= 1 kg) of water, the solvent, molality can be calculated as:
[tex]molality=\frac{0.042 moles}{0.685 kg}[/tex]
Solving:
molality= 0.0613[tex]\frac{moles}{kg}[/tex]
The molality of this solution is 0.0613[tex]\frac{moles}{kg}[/tex]
(b) In this case you have 7.15 moles of ethylene glycol (C₂H₆O₂), the solute, in 3505 g (equal to 3.505 kg) of water, the solvent, molality can be calculated as:
[tex]molality=\frac{7.15 moles}{3.505 kg}[/tex]
Solving:
molality= 2.04[tex]\frac{moles}{kg}[/tex]
The molality of this solution is 2.04[tex]\frac{moles}{kg}[/tex]
1.2.1. Name and explain the purpose of one law /legislation that protects citizens against cyberbullying ?
Answer:
Federal law is a law that may protect people from digital attacks such as cyberbullying.
Explanation:
Although it is mostly based on the situation, federal law does protect those who suffer from cyberbullying. The attackers may be charged with defamation, since cyberbullying consists of spreading false information about someone on the internet, that's when the law comes in.
Gold is a precious metal that is chemically unreactive. It is used mainly in jewelry, dentistry, and electronic devices. A piece of gold ingot with a mass of 301 g has a volume of 15.6 cm3. Calculate the density of gold.
Answer:
Density = 19.3 g/cm³
Explanation:
In order to answer this question we need to keep in mind the following definition of density:
Density = Mass / VolumeAs both the mass and the volume are given by the problem, we can proceed to calculate the density of gold:
Density = 301 g / 15.6 cm³Density = 19.3 g/cm³The information below describes a redox reaction.
Ag+ (aq) + Al(s) — Ag(s) + Al3+ (aq)
Ag+ (aq) + -> Ag(s)
Al(s)->A3+ (aq) + 3e-
What is the coefficient of silver in the final, balanced equation for this reaction?
Answer:
Al°(s) + 3Ag⁺(aq) => Al⁺³(aq) + 3Ag(s)
Explanation:
Oxidation: Al°(s) => Al⁺³(aq) + 3e⁻
Reduction: 3Ag⁺(aq) + 3e⁻ => 3Ag°(s)
_________________________________________
Net Rxn: Al°(s) + 3Ag⁺(aq) => Al⁺³(aq) + 3Ag(s)
One mole of neutral aluminum atoms (Al°(s)) undergo oxidation delivering 3 moles of electrons to 3 moles silver ions (3Ag⁺³(aq)) that are reduced to 3 moles of neutral silver atoms (3Ag°(s)) in basic standard state 25°C; 1atm.
A balanced equation obeys the law of conservation of mass. According to the law of conservation of mass, mass can neither be created nor be destroyed. The coefficient of silver is 3.
What is a balanced equation?A balanced chemical equation can be defined as the chemical equation in which the number of reactants and products on both sides of the equation are equal. The amount of reactants and products on both sides of the equation will be equal in a balanced chemical equation.
The numbers which are used to balance the chemical equation are called the coefficients. The coefficients are the numbers which are added in front of the formula.
The balanced chemical equation for the given redox reaction is given as:
Al (s) + 3 Ag⁺ (aq) → Al³⁺ (aq) + 3Ag (s)
Thus the coefficient of silver is 3.
To know more about balanced equation, visit;
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All bases dissociate
True or false
Answer:
verdadero
Explanation:
porque esoo [tex]\lim_{n \to \infty} a_n x_{123} \frac{x}{y} \sqrt[n]{x} x^{2} \sqrt{x} \pi \neq \lim_{n \to \infty} a_n \left[\begin{array}{ccc}1&2&3\\4&5&6\\7&8&9\end{array}\right] \int\limits^a_b {x} \, dx \left \{ {{y=2} \atop {x=2}} \right.[/tex]
An unknown organic compound composed of carbon, hydrogen and oxygen was analyzed and found to be 50.84% C, 8.53% H and the rest being oxygen. Which of the following represents the correct empirical formula for the compound?
(a) CH2O
(b) C3H602
(c) C4H803
(d) C2H40
(e) C5H1003
Answer:
e) C5H10O3
Explanation:
According to the information given in this question;
50.84% represents C
8.53% represents H
[100% - (50.84 + 8.53)]
= 100 - 59.37
= 40.63% of Oxygen (O)
This percentage means that;
Carbon = 50.84g
Hydrogen = 8.53g
Oxygen = 40.63g
By dividing by their respective atomic masses, we convert to moles:
C = 50.84g ÷ 12 = 4.24mol
H = 8.53g ÷ 1 = 8.53mol
O = 40.63g ÷ 16 = 2.54mol
Next we divide by the smallest no of the values (2.54mol)
C = 4.24 ÷ 2.54 = 1.669
H = 8.53 ÷ 2.54 = 3.358
O = 2.54 ÷ 2.54 = 1
To get a simple whole number ratio, we multiply the results by 3
C = 1.669 × 3 = 5.007
H = 3.358 × 3 = 10.074
O = 1 × 3 = 3
Simple whole number ratio of carbon, hydrogen and oxygen is 5:10:3. Hence, the empirical formula is C5H10O3.
QUESTIONS :
1.
Many of the flavours and smells of fruits are esters. A learner prepared an ester with a sme
Ilke banana in the school laboratory using pentanol and ethanoic acid. She set up the
apparatus as shown in the diagram below.
PAPER TOWEL DIPPED
-WATER BATH
IN COLD WATER
PEITANOL ETHANOIC
ACID+ 4 DROPS OF
SULPHURIC ACID
1.1 Which property of sulphuric acid makes it suitable to use as a catalyst for the
preparation of esters?
1.2 Why do we heat the test tube in a water bath and not directly over a flame?
1.3 With reference to the characteristic smells of esters, name TWO examples where
esters are used in different industries.
1.4 State ONE function of the wet paper towel in the opening of the test tube.
1.5 Write down the IUPAC name of an ester fomed.
Answer:
See explanation
Explanation:
Esterification is a reaction that involves the combination of an alkanoic acid and an alkanol. The product is always a sweet smelling substance.
Sulphuric acid acts as a catalyst in this reaction because it is a dehydrating agent thereby pushing the equilibrium position towards the right by the removal of water molecules.
The test tubs is heated in a water bath and not directly moved the flame because the alcohol is flammable. Also heating in a water bath helps to separate the reaction mixture from the newly formed ester.
Esters are used in industries that produces soaps and perfumes. There is a great need for the use of fragrances which are ester compounds in these industries.
The wet paper towel in the opening of the test tube cools the top of the test tube. It usually serves as a kind of condenser preventing an excess loss of vapour from the reaction mixture.
The reaction of pentanol and ethanoic acid yields pentyl ethanoate according to IUPAC nomenclature.
g Select the statement that best answers the following question What effect does the cation of an ionic compound have on the appearance of the solution? The cation affects the intensity of the color more than the color of the solution. The cation affects the color of the solution more than the intensity of the color. The cation does not affect the color or color intensity of the solution. The cation only affects the intensity of the color in a solution.
Answer:
The cation affects the intensity of the color more than the color of the solution.
Explanation:
According to Beer Lambert law, the intensity of the colour of the solution depends on the concentration of the specie responsible for the colour in the solution.
Let us recall that transition metal compounds are coloured in solution due to electronic transitions.
Therefore, the cation affects the intensity of the color more than the color of the solution.
anuvia, the trade name for sitagliptin, was introduced in 2006 for the treatment of type 2 diabetes. In what type of orbital does the lone pair on each N atom reside.
Answer: hello your question is poorly written below is the complete question
answer:
For N1 : sp³ orbital
For N2: p orbital
For N3 : p orbital
For N4 : sp² orbital
For N5 : sp² orbital
Explanation:
Determining the type of orbital in which the lone pair on each N atom will reside.
From the configuration attached below we can determine the type of orbital and they are ;
For N1 : sp³ orbital
For N2: p orbital
For N3 : p orbital
For N4 : sp² orbital
For N5 : sp² orbital
Which is an example of poor safety practices when working outdoors
Answer:
C
Explanation:
u can't touch a chemical with bare skin
Answer:
touching a chemical with his or her bare skin.
How many grams of KCl solid do I need to make 500 ml of 8% KCl solution
Answer:
40 grams of KCI are required to make 500 ml of 8% solution.
You combine 0.75 moles formate and 0.85 moles formic acid to make a buffer solution. The Ka of formic acid is 1.8x10-4 what is the pH of the solution
Answer:
pH = 3.68
Explanation:
We can solve this problem by using Henderson-Hasselbach's equation:
pH = pKa + log[tex]\frac{[Formate]}{[Formic Acid]}[/tex]Where pKa = -log(Ka)pKa = -log(1.8x10⁻⁴) = 3.74Assuming we have 1 L of the buffer solution then the molar concentrations of formate and formic acid would be:
[Formate] = 0.75 mol / 1 L = 0.75 M[Formic Acid] = 0.85 mol / 1 L = 0.85 MWe now have all required data to calculate the pH:
pH = 3.74 + log[tex]\frac{0.75}{0.85}[/tex]pH = 3.68What causes an ice cube to melt when removed from a freezer?
Answer:
the melting process begins right away because the air temperature around the ice cubes is warmer than the temperature in the freezer
Guys! I need a long inforation about...
"Colloidal solutions
Types of colloids
Formation of colloids"
Please! And thank you
A colloid is a mixture in which one substance of microscopically dispersed insoluble particles are suspended throughout another substance. However, some definitions specify that the particles must be dispersed in a liquid, and others extend the definition to include substances like aerosols and gels.
The types of colloids includes sol, emulsion, foam, and aerosol.
Sol is a colloidal suspension with solid particles in a liquid.Emulsion is between two liquids.Foam is formed when many gas particles are trapped in a liquid or solid.Aerosol contains small particles of liquid or solid dispersed in a gas.Explanation:
everything can be found in the picture
an alkane group has a formula of CxH6, determine the value of x
Answer:
[tex]x = 2[/tex]
Explanation:
General formula for alkanes
[tex]C _{n}H _{2n + 2}[/tex]
since H = 6
[tex]2n + 2 = 6 \\ 2n = 4 \\ n = 2 \\ since \: x = n\\ \therefore \: x = 2[/tex]
Answer:
x=2
Explanation:
Because alkanes have a general formula of CnH(2n+2).
We known that 2n+2=6, so we solve for n.
2n+2=6
2n=4
n=2
Thus, there are 2 carbon atoms.
I think I put my in it when I asked Jane about her ex-husband
root
a. Giải thích vì sao tính bazơ tăng từ LiOH đến CsOH?
I WILL GIVE BRAINLIEST IF YOU ANSWER CORRECTLY
How many grams in 3.75 x 1024 atoms of F?
Answer:
96 grams
Explanation:
I NEED THIS NOW NO LINKS OR ILL REPORT PLZZZ
What is an example of a chemical property?
Odensity
O reactivity
Omalleability
O solubility
Three acid samples are prepared for titration by 0.01 M NaOH:
1. Sample 1 is prepared by dissolving 0.01 mol of HCl in 50 mL of water.
2. Sample 2 is prepared by dissolving 0.01 mol of HCl in 60 mL of water.
3. Sample 3 is prepared by dissolving 0.01 mol of HCl in 70 mL of water.
a. Without performing a formal calculation, compare the concentrations of the three acid samples (rank them from highest to lowest).
b. When the titration is performed, which sample, if any, will require the largest volume of the 0.01 M NaOH for neutralization?
A sample of oxygen gas occupies a volume of 2.,0cm3 at pressure of 700K pa. what will be pressure of the same sample occupies a volume of 150cm, assume temperature remains constant
Answer:
The pressure will be 933.33 Kpa
Explanation:
Given that:
Volume V₁ = 200 cm³ (note, there is a mistake in the volume. It is supposed to be 200 cm³)
Pressure P₁ = 700 Kpa
Pressure P₂ = ??? (unknown)
Volume V₂ = 150 cm³
Temperature = constant
Using Boyle's law:
PV = constant
i.e.
P₁V₁ = P₂V₂
700 Kpa × 200 cm³ = P₂ × 150 cm³
P₂ = (700 Kpa × 200 cm³)/150 cm³
P₂ = 933.33 Kpa
A) The average molecular speed in a sample of Ar gas at a certain temperature is 391 m/s. The average molecular speed in a sample of Ne gas is ______ m/s at the same temperature.
B) The rate of effusion of Xe gas through a porous barrier is observed to be 7.03×10-4 mol / h. Under the same conditions, the rate of effusion of SO2 gas would be ______ mol / h
Answer:
For A: The average molecular speed of Ne gas is 553 m/s at the same temperature.
For B: The rate of effusion of [tex]SO_2[/tex] gas is [tex]1.006\times 10^{-3}mol/hr[/tex]
Explanation:
For A:
The average molecular speed of the gas is calculated by using the formula:
[tex]V_{gas}=\sqrt{\frac{8RT}{\pi M}}[/tex]
OR
[tex]V_{gas}\propto \sqrt{\frac{1}{M}}[/tex]
where, M is the molar mass of gas
Forming an equation for the two gases:
[tex]\frac{V_{Ar}}{V_{Ne}}=\sqrt{\frac{M_{Ne}}{M_{Ar}}}[/tex] .....(1)
Given values:
[tex]V_{Ar}=391m/s\\M_{Ar}=40g/mol\\M_{Ne}=20g/mol[/tex]
Plugging values in equation 1:
[tex]\frac{391m/s}{V_{Ne}}=\sqrt{\frac{20}{40}}\\\\V_{Ne}=391\times \sqrt{2}=553m/s[/tex]
Hence, the average molecular speed of Ne gas is 553 m/s at the same temperature.
For B:
Graham's law states that the rate of diffusion of a gas is inversely proportional to the square root of the molar mass of the gas. The equation for this follows:
[tex]Rate\propto \frac{1}{\sqrt{M}}[/tex]
Where, M is the molar mass of the gas
Forming an equation for the two gases:
[tex]\frac{Rate_{SO_2}}{Rate_{Xe}}=\sqrt{\frac{M_{Xe}}{M_{SO_2}}}[/tex] .....(2)
Given values:
[tex]Rate_{Xe}=7.03\times 10^{-4}mol/hr\\M_{Xe}=131g/mol\\M_{SO_2}=64g/mol[/tex]
Plugging values in equation 2:
[tex]\frac{Rate_{SO_2}}{7.03\times 10^{-4}}=\sqrt{\frac{131}{64}}\\\\Rate_{SO_2}=7.03\times 10^{-4}\times \sqrt{\frac{131}{64}}\\\\Rate_{SO_2}=1.006\times 10^{-3}mol/hr[/tex]
Hence, the rate of effusion of [tex]SO_2[/tex] gas is [tex]1.006\times 10^{-3}mol/hr[/tex]
TIME REMAINING
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Scientists use different types of microscopes to observe objects that are not visible to the naked eye. A scientist is
reviewing various samples of pond water to determine what species of microorganisms live in the pond. The scientist
wishes to make drawings of the structure of each microorganism and study each one's method of movement. Which of
the following microscopes would be best for the scientist to use?
transmission electron microscope
b. scanning electron microscope
c. compound light microscope
d. dissecting microscope
a.
Please select the best answer from the choices provided
ОА
ОВ
D
Nox
Submit
Save and Exit
Mark this and retum
Sono
Answer:
compound light microscope
A sample of titrations of 0.200 M sodium hydroxide (NaOH) being added to 30.0 mL of hydrochloric acid (HCl) of unknown concentration. Measure the volume of sodium hydroxide needed to neutralize the HCl, and calculate the concentration of the HCl.
Answer:
Assuming 20.0 mL of 0.2 M NaOH is used to neutralize 30 mL of the HCl of unknown concentration, the concentration of the hydrochloric acid is 0.133 M.
Note: The volume of NaOH used isnot given and issp to be determined by titration.
Explanation:
The question is a lab activity. Therefore, the volume of 0.2 M NaOH required for the neutralization of 30.0 mL of hydrochloric acid of unknown concentration is to be determined from titration. However, assuming a certain volume of the base is used for the complete neutralization of the acid, the concentration of the acid can be calculated.
Assuming 20.0 mL of 0.2 M NaOH is used to neutralize 30 mL of the HCl of unknown concentration, the concentration of the hydrochloric acid can be calculated thus:
Step 1: Balanced chemical equation of reaction:
HCl + NaOH ---> NaCl + H₂O
From the equation of the reaction, 1 mole of NaOH reacts with 1 mole of HCl to produce 1 mole of NaCl and water.
Step 2: Determine the number of moles of NaOH in 20.0 mL of 0.200 M solution
Molarity = number of moles/volume in Litres
Number of moles = molarity × volume in litres
Volume of NaOH in litres = 20 ml × 1 litre/1000ml = 0.02 litres
Number of moles NaOH = 0.02 L × 0.2 M = 0.004 moles
Therefore, 0.00 M of NaOH reacted with 0.004 moles of HCl
Step 3: Determine concentration of HCl
Molarity or concentration = number of moles / volume in litres
Number of moles of HCl = 0.004 moles
Volume of HCl in litres = 30.0 mL × 1 L/1000 mL = 0.03 L
Concentration of HCl = 0.004 moles / 0.03 L = 0.133 M
Therefore, assuming 20.0 mL of 0.2 M NaOH is used to neutralize 30 mL of the HCl of unknown concentration, the concentration of the hydrochloric acid is 0.133 M.
during the process of photosynthesis, green plants produce...
Answer: photosynthesis
Explanation:
Can someone help me with this one
Answer:
Easy my dude let me help you out
A.In
B.27
C.73
D.49
E.56
F.56
G.114
H.180
Also with protons and electrons they equal the same atomic number
If we have 1.23 mol of NaOH in solution and 0.85 mol of Cl2 gas is available to react, which one is the limiting reactant? Give your reason.
Answer:
NaOH is the limiting reactant.
Explanation:
Hello there!
In this case, since the reaction taking place between sodium hydroxide and chlorine has is:
[tex]NaOH+Cl_2\rightarrow NaCl+NaClO+H_2O[/tex]
Which must be balanced according to the law of conservation of mass:
[tex]2NaOH+Cl_2\rightarrow NaCl+NaClO+H_2O[/tex]
Whereas there is a 2:1 mole ratio of NaOH to Cl2, which means that the moles of the former that are consumed by 0.85 moles of the latter are:
[tex]n_{NaOH}=0.85molCl_2*\frac{2molNaOH}{1molCl_2}\\\\n_{ NaOH}=1.7molNaOH[/tex]
Therefore, since we just have 1.23 moles out of 1.70 moles of NaOH, we infer this is the limiting reactant.
Regards!
Chrysanthenone is an unsaturated ketone. If Chrysanthenone has M+ = 150 and contains 2 double bond(s) and 2 ring(s); what is its molecular formula? Enter the formula in the form CH first, then all other atoms in alphabetical order; do not use subscripts. The formula is case-sensitive.
Answer:
the Molecular formula will be; C10H14O
Explanation:
Given the data in the question;
Chrysanthenone is an unsaturated ketone,
it has M+ = 150 and contains 2 double bond(s) and 2 ring(s).
molecular formula = ?
we know that ketone contain 1 oxygen and mass of oxygen is 16
so mass of the C and H remaining will be;
⇒ 150 - 16 = 134
Now we determine the number of C atoms;
⇒ 134 / 13 = 10
hydrocarbon with 10 hydrogen atom have CnH2n+2 means
⇒ ( 10 × 2 ) +2 = 22 hydrogens
But then we have 3 unsaturation meaning 6 hydrogens less and also we have ring meaning 2 more hydrogens
⇒ 22 - 6 - 2 = 14
Hence the Molecular formula will be; C10H14O
what's the ph of 0.0000067 m hcl solution
Answer:
[tex]pH = - log[H {}^{ + } ] \\ = - log(0.0000067) \\ pH = 5.17[/tex]
Guanidine is a neutral compound but is an extremely powerful base. In fact, it is almost as strong a base as a hydroxide ion. Identify which nitrogen atom in guanidine is so basic, and explain why guanidine is a much stronger base than most other amines.
Answer:
See explanation
Explanation:
The molecule called guanidine is shown in the image attached to this answer. It contains three nitrogen atoms. Two among them are sp3 hybridized while one of them is sp2 hybridized.
Guanidine is more basic than other amines because its protonanation leads to the formation of three equivalent resonance structures thereby making its protonated form quite stable. This effect is not observed in other amines.
Also, the sp2 hybridized nitrogen atom is more basic and more easily protonated because when it is protonated, three equivalent resonance structures are obtained.