A 2.950×10−2 M solution of glycerol (C3H8O3) in water is at 20.0∘C. The sample was created by dissolving a sample of C3H8O3 in water and then bringing the volume up to 1.000 L. It was determined that the volume of water needed to do this was 998.7 mL . The density of water at 20.0∘C is 0.9982 g/mL.Calculate:
a. molality
b. mole fraction of glycerol in this solution
c. the concentration of the glycerol solution in percent by mass
d. the concentration of the glycerol solution in parts per million

Answers

Answer 1

Answer:

a. 2.959x10⁻²m

b. 5.327x10⁻⁴

c. 0.272%

d. 2718 PPM

Explanation:

A solution of 2.950x10⁻²M contains 2.950x10⁻² moles of Glycerol per Liter of solution. As the volume of the solution made was 1.000L, moles of glycerol are 2.950x10⁻².

a. molality: Molality is defined as the ratio between moles of solute (2.950x10⁻²) in kg of solvent. As there are 998.7mL of solvent and density is 0.9982g/mL, kg are:

998.7mL ₓ (0.9982g/mL) ₓ (1kg / 1000g) = 0.9969kg of solvent.

Molality: 2.950x10⁻² moles / 0.9969kg of solvent = 2.959x10⁻²m

b. Mole fraction is the ratio between moles of solute and total moles. Moles of water are:

998.7mL ₓ (0.9982g/mL) ₓ (1mol / 18.01g) = 55.35 moles of water.

Mole fraction glycerol:

2.950x10⁻² moles / (2.950x10⁻²moles + 55.35) = 5.327x10⁻⁴

c. Percent by mass Is the ratio by mass of solute and solution multiplied 100 times.

Mass of glycerol (Molar mass:  92.09g/mol):

2.950x10⁻² moles × (92.09g / mol) = 2.717g of glycerol

Mass of water:

998.7mL ₓ (0.9982g/mL) = 996.9g of water.

Percent by mass:

2.717g of glycerol / (996.9g of water + 2.717g) × 100 = 0.272%

d. Parts per million are defined as the ratio between mg of solute and kg of solution.

mg of 2.717g of glycerol are 2717mg

kg of solution are (996.9g + 2.717g) / 1000 = 0.9996kg

Parts per million:

2717mg / 0.9996kg = 2718 PPM


Related Questions

An empty beaker is weighed and found to weigh 23.1 g. Some potassium chloride is then added to the beaker and weighed again. The second weight is 24.862 g. What is the mass of the potassium chloride

Answers

Answer:Mass of Potassium chloride =1.762g

Explanation:

Mass of empty beaker = 23.100 g

Mass of beaker with Potassium chloride = 24.862g

Mass of Potassium chloride = Final weight - initial weight = Mass of beaker with Potassium chloride  - Mass of empty beaker = 24.862-23.100 = 1.762g

need helpp asapp please

Answers

Answer:

B. None of these

Explanation:

Sulfur has less ionization energy than phosphorus because sulfur has a pair of electron in its 3p subshell that increases electron repulsion in sulfur and sulfur electrons can easily remove from its sub-level.

While, there are no electron pairs in 3p subshell of phosphorus, therefore it requires more energy to remove an electron from 3p subshell.

Hence, the reason is electron repulsion and the correct answer is B.

Which spheres of earth are represented in tropical rainforests?

Answers

Answer:

Biosphere

Explanation:

because of the rain it gets from huge water sources.

I hope that was useful.

In the first 15.0 s of the reaction, 1.7×10−2 mol of O2 is produced in a reaction vessel with a volume of 0.440 L . What is the average rate of the reaction over this time interval?

Answers

Answer:

[tex]Rate=2.57x10^{-3}\frac{M}{s}[/tex]

Explanation:

Hello,

In this case, for the reaction:

[tex]2N_2O(g) \rightarrow 2N_2(g)+O_2(g)[/tex]

We can easily compute the average rate by firstly computing the final concentration of oxygen:

[tex][O_2]=\frac{0.017mol}{0.440L}=0.0386M[/tex]

Then, we compute it by using the given interval of time: from 0 seconds to 15.0 seconds and concentration: from 0 M to 0.0386M as oxygen is being formed:

[tex]Rate=\frac{0.0386M-0M}{15.0s-0s}\\ \\Rate=2.57x10^{-3}\frac{M}{s}[/tex]

Regards.

The average rate of the reaction will be "[tex]2.57\times 10^{-3} \ M/s[/tex]".

According to the question,

Volume = 0.440 LTime = 15.0 sMol of O₂ = 1.7×10⁻²

The reaction will be:

[tex]2 N_2 O (g) \rightarrow 2 N_2 (g) +O_2 (g)[/tex]

Now,

The final concentration of O₂ will be:

→ [tex][O_2] = \frac{0.017}{0.440}[/tex]

          [tex]= 0.0386 \ M[/tex]

hence,

The rate of reaction will be:

= [tex]\frac{0.0386-0}{15.0-0}[/tex]

= [tex]2.57\times 10^{-3} \ M/s[/tex]

Thus the above approach is right.

Learn more about volume here:

https://brainly.com/question/15050688

A weather balloon is inflated to a volume of 27.6 L at a pressure of 755 mmHg and a temperature of 29.9 ∘C. The balloon rises in the atmosphere to an altitude where the pressure is 385 mmHg and the temperature is -14.1 ∘C. Assuming the balloon can freely expand, calculate the volume of the balloon at this altitude.

Answers

Answer: The volume of the balloon at this altitude is 46.3 L

Explanation:

Combined gas law is the combination of Boyle's law, Charles's law and Gay-Lussac's law

The combined gas equation is,

[tex]\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}[/tex]

where,

[tex]P_1[/tex] = initial pressure of gas = 755 mm Hg

[tex]P_2[/tex] = final pressure of gas (at STP) = 385 mm Hg

[tex]V_1[/tex] = initial volume of gas = 27.6 L

[tex]V_2[/tex] = final volume of gas = ?

[tex]T_1[/tex] = initial temperature of gas = [tex]29.9^0C=(29.9+273)K=302.9K[/tex]

[tex]T_2[/tex] = final temperature of gas = [tex]-14.1^0C=((-14.1)+273)K=258.9K[/tex]

Putting all the values we get:

[tex]\frac{755\times 27.6}{302.9}=\frac{385\times V_2}{258.9}[/tex]

[tex]V_2=46.3L[/tex]

Thus the volume of the balloon at this altitude is 46.3 L

Question 1
1 pts
2B+6HCI --
| --> 2BCl3 + 3H2
How many moles of boron chloride will be produced if you start with 8.752 moles of HCI
(hydrochloric acid)? (Round to 3 sig figs. Enter the number only do not include units.)

Answers

Answer:

2.92 mol

Explanation:

Step 1: Write the balanced equation

2 B(s) + 6 HCI(aq) ⇒ 2 BCl₃(aq) + 3 H₂(g)

Step 2: Establish the appropriate molar ratio

The molar ratio of hydrochloric acid to boron chloride is 6:2.

Step 3: Calculate the moles of boron chloride produced from 8.752 moles of hydrochloric acid

[tex]8.752molHCl \times \frac{2molBCl_3}{6molHCl} = 2.92molBCl_3[/tex]

What type of chemical reaction occurs between AgNO3 (sliver nitrate) and Cu (Copper)?

Answers

Answer:

The answer is option c.

I hope this helps you.

calculate how many moles of CaCl2•2H2O are present in 1.50 g of CaCl2•2H2O and then calculate how many moles of pure CaCl2 are present in the 1.50 g of CaCl2•2H2O.

Answers

Answer:

[tex]0.0102~mol~CaCl_2*2H_2O[/tex]

[tex]0.0102~mol~CaCl_2[/tex]

Explanation:

For this question, we have to start with the molar mass calculation of [tex]CaCl_2*2H_2O[/tex]. For this, we have to know the atomic mass of each atom:

O: 16 g/mol

Cl: 35.45 g/mol

H: 1 g/mol

Ca: 40 g/mol

If we take into account the amount of each atom in the formula we will have:

[tex](40*1)+(35.45*2)+(1*4)+(16*2)=~147.01~g/mol[/tex]

So, in 1 mol of [tex]CaCl_2*2H_2O[/tex] we will have 147.01 g. Now we can do the conversion:

[tex]1.50~g~CaCl_2*2H_2O\frac{1~mol~CaCl_2*2H_2O}{147.01~g~CaCl_2*2H_2O}=0.0102~mol~CaCl_2*2H_2O[/tex]

Additionally, in 1 mol of [tex]CaCl_2*2H_2O[/tex] we will have 1 mol of [tex]CaCl_2[/tex]. Therefore, we have a 1:1 mol ratio . With this in mind, we will have the same number of moles for [tex]CaCl_2[/tex]

[tex]0.0102~mol~CaCl_2*2H_2O=0.0102~mol~CaCl_2[/tex]

I hope it helps!

A certain mass of carbon reacts with 9.53 g of oxygen to form carbon monoxide. ________ grams of oxygen would react with that same mass of carbon to form carbon dioxide, according to the law of multiple proportions.

Answers

Answer: 9.53 *2= 19.06

Explanation:

The law of multiple proportions states that if two elements combines to form more than one compound the ratio of masses of the second element which combines to the fixed mass of the first element will always be the ratios of the small whole numbers.

in case of carbon monoxide, mass of carbon will be the same of mass of oxygen.

But in case of carbon dioxide, if carbon is 9.53 units then oxygen will be twice as that of carbon.

CO2, so 9.53*2= 19.06 grams of oxygen will combine with 9.53 grams of carbon to form carbon dioxide.

In electrophilic aromatic substitution reactions, a chlorine substituent is ________. In electrophilic aromatic substitution reactions, a chlorine substituent is ________. an activator and a meta-director a deactivator and an ortho,para-director an activator and an ortho,para-director a deactivator and a meta-director none of the above

Answers

Answer:

d) is an activator and an ortho/para- director

Explanation:

We options for this question are:

a) is a deactivator and an ortho/para- director

b) is a deactivator and a meta-director

c) is an activator and a meta-director

d) is an activator and an ortho/para- director

e) is an activator and an ortho/meta/para director

f) none of the above

We have to remember that Cl is a halogen, therefore a lone pair of electrons will be involved in the resonance (see figure 1). So, we will have activation in the ortho and para positions.  We have to take into account that the halogens have a high electronegativity. This means that inductively they are electron-withdrawing. So, we will have opposite phenomenons that will result in very weak activation

A tissue sample at 275 K is submerged in 2 kg of liquid nitrogen at 70 K for cryopreservation. The final temperature of the nitrogen is 75 K. What is the heat capacity of the sample in J/K

Answers

Answer:

heat capacity of the sample = 37.8 J/K

Explanation:

Step 1: Data given

Temperature of the sample = 275 K

The mass of liquid nitrogen = 2kg

temperature of liquid nitrogen = 70 K

The final temperature of the nitrogen is 75 K

Step 2: Calculate heat

Q = m*c*ΔT

⇒with m = the mass of liquid nitrogen = 2 kg = 2000 grams

⇒with c= the specific heat of the liquid nitrogen = 1.04 J/g*K

⇒with ΔT = the change of temperature of liquid nitrogen = T2 - T1 = 75 - 70 = 5K

Q = 2000 grams * 1.04 J/g*K * 5K

Q = 10400 J

Step 3: Calculate the heat capacity of the sample

heat capacity of the sample = 10400 J / 275 K

heat capacity of the sample = 37.8 J/K

At a particular temperature, an equilibrium mixture the reaction below was found to contain 0.171 atm of I2, 0.166 atm of Cl2 and 9.81 atm of ICl. Calculate the value of the equilibrium constant, Kp at this temperature.I2(g) + Cl2(g) <=> 2 ICl(g)

Answers

Answer: 3390

Explanation:

Since this problem already gives is the equilibrium values, all we have to do is to plug them into the formula for [tex]K_{p}[/tex].

[tex]K_{p} =\frac{[ICl]^2}{[I_{2}][Cl_{2}] }[/tex]

[tex]K_{p} =\frac{(9.81)^2}{(0.171)(0.166)} =3390[/tex]

The enthalpy change for the complete burning of one mole of a substance
is the enthalpy of _______

thermochemical equation

combustion

released

vaporization

fusion

absorbed

heat

Answers

Answer:

combustion

Explanation:

The enthalpy change for the complete burning of one mole of a substance

is the enthalpy of __combustion_____ .

Consider the reaction of aqueous potassium sulfate with aqueous g silver nitrate based on the solubility rule predict the product likely to be precipitate write a balanced molecular equation describing the reaction.

Answers

Answer:

K₂SO₄(aq)  + 2AgNO₃ (aq) →  2KNO₃(aq) + Ag₂SO₄ (s) ↓

2Ag⁺ (aq) + SO₄⁻²(aq) ⇄ Ag₂SO₄ (s) ↓

Explanation:

Our reactants are: K₂SO₄ and AgNO₃

By the solubility rules, we know that sulfates are insoluble when they react to Ag⁺, Pb²⁺, Ca²⁺, Ba²⁺, Sr²⁺, Hg⁺

We also determine, that salts from nitrate are all soluble.

The reaction is:

K₂SO₄(aq)  + 2AgNO₃ (aq) →  2KNO₃(aq) + Ag₂SO₄ (s) ↓

2Ag⁺ (aq) + SO₄⁻²(aq) ⇄ Ag₂SO₄ (s) ↓

Methane (CH4) and the perchlorate ion (ClO4â) are both described as tetrahedral. Methane and the perchlorate ion are both described as tetrahedral. What does this indicate about their bond angles?

a. Bond angles in the methane molecule are greater than those in the perchlorate ion.
b. Bond angles in the perchlorate ion are greater than those in the methane molecule.
c. Bond angles in the methane molecule and in the perchlorate ion are approximately equal

Answers

Answer:

c

Explanation:

The correct answer would be that the bond angles in the methane molecule and in the perchlorate ion are approximately equal.

A tetrahedral shape or geometry means both molecules are sp3 hybridized with the hydrogen atoms and the oxygen ions surrounding the central carbon and chlorine atoms respectively being as far away from one another as possible. This results in an approximate angle of 109.5 degrees in both molecules.

The correct option is c.

Please what's the missing minor products? And kindly explain in your own words how they were formed.​ Thank you!

Answers

Answer:

it's a two step elimination reaction

Explanation:

it follows a carbocationic pathway. When carbocation is stable, the equation is favourable, that is, double bond is formed by expelling hydrogen atom.

Calculate the heat change in kilojoules for condensation of 195 g of steam at 100 ° C

Answers

Answer:

Q = 81.59kJ

Explanation:

Hello,

The heat of condensation is the energy required to to convert the steam into water.

Mass = 195g

Specific heat capacity of water = 4.184J/g°C

Initial temperature(T1) = 100°C

Final temperature(T2) = 0°C

Heat energy (Q) = ?

Heat energy (Q) = mc∇T

M = mass of the substance

C = specific heat capacity of the substance

∇T = T2 - T1 = change in temperature of the substance

Q = 195 × 4.184 × (0 - 100)

Q = -81588J

Q = -81.588kJ

The heat required for the condensation of 195g of steam is 81.59kJ

The following reactions all have K < 1. 1) HCOO- (aq) + C6H5COOH (aq) HCOOH (aq) + C6H5COO- (aq) 2) C9H7O4- (aq) + C6H5COOH (aq) C6H5COO- (aq) + HC9H7O4 (aq) 3) HCOOH (aq) + C9H7O4- (aq) HC9H7O4 (aq) + HCOO- (aq) Arrange the substances based on their relative acid strength.

Answers

Answer:

Explanation:

C₉H₇O₄⁻ = weakest base

C₆H₅COO⁻ = strongest base

HCOO⁻ = intermediate base

HCOOH = not a Bronsted-Lowry base

HC₉H₇O₄ = not a Bronsted-Lowry base

C₆H₅COOH = not a Bronsted-Lowry base

Tubes through which water flows as it is brought from 0.8 MPa, 150C to 240C at essentially constant pressure in the boiler of a power plant. The total mass flow rate of the water is 100 kg/s. Combustion gases passing over the tubes cool from 1067 to 547C at essentially constant pressure. The combustion gases can be modeled as air as an ideal gas. There is no significant heat transfer from the boiler to its surroundings. Surrounding (dead state) temperature and pressure are given as 25C and 1 atm, respectively. Determine i) the exergetic efficiency of the boiler ii) rate of exergy destruction as kW iii) mass flow rate of the combustion gases as kg/s

Answers

Answer:

The correct answer is i) 50.2 % ii) 13440.906 kW and iii) 71.986 kg/s.

Explanation:

In order to find the mass flow rate of the combustion of gases, there is a need to use the energy balance equation:  

Mass of water × specific heat of water (T2 -T1)w = mass of gas × specific heat of gas (T2-T1)g

100 × 4.18 × [(240 + 273) - (150 + 273)] = mass of gas × 1.005 × [(1067+273) - (547+273)]

Mass of gas = 71.986 kg/s

The entropy generation of water can be determined by using the formula,  

(ΔS)w = mass of water × specific heat of water ln(T2/T1)w

= 100 × 4.18 ln(513/423)

= 80.6337 kW/K

Similarly the entropy generation of water will be,

(ΔS)g = mass of gas × specific heat of gas ln(T2/T1)g

= 71.986 × 1.005 ln (820/1340)

= -35.53 kW/K

The rate of energy destruction will be,  

Rate of energy destruction = To (ΔS)gen

= T₀ [(ΔS)w + (ΔS)g]

= (25+273) [80.6337-53.53)

Rate of energy destruction = 13440.906 kW

The availability of water will be calculated as,  

= mass of water (specific heat of water) [(T₁-T₂) -T₀ ln T₁/T₂]

= 100 × 4.8 [(513-423) - 298 ln 513/423]

= 13591.1477 kW

The availability of gas will be calculated as,  

= mass of gas (specific heat of gas) [(T₁-T₂) - T₀ ln T₁/T₂]

= 71.986 × 1.005 × [(1340-820) - 298 ln 1340/820]

= 27031.7728 kW

The exergetic efficiency can be calculated as,  

= Gain of availability / loss of availability  

= 13591.1477/27031.7728

= 0.502

The exergetic efficiency is 50.2%.  

What is Key for the reaction 2503(9) = 2802(9) + O2(g)?

Answers

Answer:

Option C. Keq = [SO2]² [O2] /[SO3]²

Explanation:

The equilibrium constant keq for a reaction is simply the ratio of the concentration of the products raised to their coefficient to the concentration of the reactants raised to their coefficient.

Now, let us determine the equilibrium constant for the reaction given in the question.

This is illustrated below:

2SO3(g) <==> 2SO2(g) + O2(g)

Reactant => SO3

Product => SO2, O2

Keq = concentration of products /concentration of reactants

Keq = [SO2]² [O2] /[SO3]²

Which of the following would describe a spontaneous process? (Select all that apply.) ΔGreaction < 0 ΔSuniverse < 0 ΔSuniverse > 0 ΔHreaction > 0 ΔHreaction < 0 ΔGreaction > 0

Answers

Answer:

ΔGreaction < 0

ΔSuniverse > 0

ΔHreaction < 0

Explanation:

A spontaneous process is one which can proceed without additional input of energy releasing free energy in the process and then moves to a lower more stable thermodynamical state.

For an isolated system, a spontaneous process proceeds with an increase in entropy.

The conditions for a spontaneous process at constant temperature and pressure, can be determined using the change in Gibbs free energy, which is given by: ∆G = ∆H - T∆S

Where ∆G is change in free energy; ∆H is change in enthalpy or Heat content; ∆S is change in entropy, T is temperature.

For a process to be spontaneous, the following conditions are necessary:

1. ∆G < 0; must be negative

2. ∆S > 0; there must be an increase in entropy

3. ∆H < 0; enthalpy change must be negative such that heat is lost to the surroundings

The above conditions ensures that ∆G is negative and the process is spontaneous.

Which of the following is evidence of a physical change?

A) burning
B) melting
C) decomposing
D) rusting

Answers

The answer is >>>>
C) decomposing
A RESPOSTA CORRETA É A c

When 1-iodo-1-methylcyclohexane is treated with NaOCH2CH3 as the base, the more highly substituted alkene product predominates. When KOC(CH3)3 is used as the base, the less highly substituted alkene predominates. Give the structures of the two products and offer an explanation.

Answers

Answer:

See explanation

Explanation:

In this case, we have 2 types of reactions. [tex]CH_3CH_2ONa[/tex] is a strong base but only has 2 carbons therefore we will have less steric hindrance in this base. So,  the base can remove hydrogens that are bonded on carbons 1 or 6, therefore, we will have a more substituted alkene (1-methylcyclohex-1-ene).

For the  [tex]KOC(CH_3)_3[/tex] we have more steric hindrance. So, we can remove only the hydrogens from carbon 7 and we will produce a less substituted alkene (methylenecyclohexane).

See figure 1

I hope it helps!

A solution of benzene in methanol has a transmittance of 93.0 % in a 1.00 cm cell at a wavelength of 254 nm. Only the benzene absorbs light at this wavelength, not the methanol. What will the solution's transmittance be if it is placed in a 10.00 cm long pathlength cell

Answers

Answer:

T = 48.39%

Explanation:

In this case we need to apply the Beer law which is the following:

A = CεL  (1)

Where:

A: Absorbance of solution

C: Concentration of solution

ε: Molar Absortivity (Constant)

L: Length of the cell

Now according to the given data, we have transmittance of 93% or 0.93. We can calculate absorbance using the following expression:

A = -logT (2)

Applying this expression, let's calculate the Absorbance:

A = -log(0.93)

A = 0.03152

Now that we have the absorbance, let's calculate the concentration of the solution, using expression (1).

A = CεL

C = A / εL

Replacing:

C = 0.03152 / 1 *ε   (3)

Now, we want to know the transmittance of the solution with a length of 10 cm. so:

A = CεL

Concentration and ε are constant, so:

A = (0.03152 / ε) * ε * 10

A = 0.3152

Now that we have the new absorbance, we can calculate the new transmittace:

T = 10^(-A)

T = 0.4839 ----> 48.39%

Telluric acid (H2TeH4O6) is a diprotic acid with Ka1 = 2.0x10-8 and Ka2 = 1.0x10-11. A 0.25 M H2TeH4O6 contains enough HCl so that the pH is 3.00. What is the concentration of HTeH4O6

Answers

Answer:

5x10⁻⁶ = [HTeH₄O₆⁺]

Explanation:

The first dissociation equilibrium of the telluric acid in water is:

H₂TeH₄O₆ + H₂O ⇄ HTeH₄O₆⁺ + H₃O⁺

Using H-H equation for telluric acid:

pH = pKa + log₁₀ [HTeH₄O₆⁺] / [H₂TeH₄O₆]

pKa of telluric acid is -logKa1

pKa = -log 2.0x10⁻⁸

pKa = 7.699

As concentration of [H₂TeH₄O₆] is 0.25M, replacing in H-H equation:

3.00 = 7.699+ log₁₀ [HTeH₄O₆⁺] / [0.25M]

-4.699 = log₁₀ [HTeH₄O₆⁺] / [0.25M]

2x10⁻⁵ = [HTeH₄O₆⁺] / [0.25M]

5x10⁻⁶ = [HTeH₄O₆⁺]

17. Write the molecular balanced equation for the recovering of copper metal. 18. Write the complete ionic balanced equation for the recovering of copper metal. 19. Write the net ionic balanced equation for the recovering of copper metal. 20. What type of reaction is this

Answers

Answer:

Explanation:

17. it goes from solid copper to aqueous copper:

Cu(s) --> Cu₂(aq) + 2e⁻

18. complete ionic:

Cu(s) --> Cu₂(aq) + 2e⁻

19. net ionic, must include only reacting species, so

Cu(s) --> Cu₂(aq) + 2e⁻

20. this type of reaction is dissolution reaction(redox reaction)

copper reduced from Cu²⁺ to Cu.

A glass flask has a volume of 500 mL at a temperature of 20° C. The flask contains 492 mL of mercury at an equilibrium temperature of 20°C. The temperature is raised until the mercury reaches the 500 mL reference mark. At what temperature does this occur? The coefficients of volume expansion of mercury and glass are 18 ×10-5 K-1 (mercury) and 2.0 ×10-5 K-1 (glass).

Answers

Answer:

101.63° C

Explanation:

Volume expansivity γa = γr -  γ g = 18 × 10⁻⁵ - 2.0 × 10⁻⁵ = 16 × 10⁻⁵ /K

v₂ - v₁ / v₁θ = 16 × 10⁻⁵ /K

(500 - 492 ) mL / (492 × 16 × 10⁻⁵) = θ

θ = 101.63° C

With methyl, ethyl, or cyclopentyl halides as your organic starting materials and using any needed solvents or inorganic reagents, outline syntheses of each of the following. More than one step may be necessary and you need not repeat steps carried out in earlier parts of this problem. (a) CH3I (b) I (c) CH3OH (d) OH (e) CH3SH (f) SH (g) CH3CN (h) CN (i) CH3OCH3 (j) OMe

Answers

Answer:

In the attachment you can find all the possible chemical reactions.

Some reaction can not be obtained by using alkyl halides because halides are weak leaving group which can leave compound during reaction easily but hydroxyl groups is a strong nucleophile which can not leave compound easily. So we can obtain alcohol from ethyl bromide, but we can not obtain hydroxyl ion from ethyl bromide.  

Explanation:

The methyl of ethyl halides as the organic starting materials are using the needed solvents or the inorganic reagents. These can be not repeated in steps that arrive out in earlier parts.

The reaction can not be taken by the use of alkyl halides as the halides are the weakest leaving group which leave the compound during reaction easily.the hydroxyl group is the strong nucleophile that cannot leave the compound easily. Thus we can get alcohol from the ethyl bromide, but we can not obtain the hydroxyl ion from the ethyl bromide.  

Learn more about the methyl or the cyclopentyl.

brainly.com/question/12621202

Suppose 1.87g of nickel(II) bromide is dissolved in 200.mL of a 52.0mM aqueous solution of potassium carbonate. Calculate the final molarity of nickel(II) cation in the solution. You can assume the volume of the solution doesn't change when the nickel(II) bromide is dissolved in it.

Answers

Answer:

Molarity = 0.0428 M = 42.8 mM

Explanation:

Step 1: Data given

Mass of nickel(II) bromide = 1.87 grams

Molar mass of nickel(II) bromide = 218.53 g/mol

Volume = 200 mL = 0.200 L

Step 2: Calculate moles of nickel(II) bromide

Moles nickel (II) bromide = mass / molar mass

Moles nickel (II) bromide = 1.87 grams / 218.53 g/mol

Moles nickel (II) bromide = 0.00856 moles

Step 3: Calculate moles nickel (II) cation

For 1 mol NiBr2 we have 1 mol Ni^2+

For 0.00856 moles NiBr2 we have 0.00856 moles Ni^2+

Step 4: Calculate final molarity of Ni^2+

Molarity = moles / volume

Molarity = 0.00856 moles / 0.200 L

Molarity = 0.0428 M = 42.8 mM

Why polythene bags create big problem in garbage disposal ?

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Answer:

Polythene bags are so lightweight and aerodynamic, they are easily picked up and carried by the wind. They can escape from trash bins, recycle bins, garbage trucks, and landfills, and end up littering the landscape.

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