In addition to displacing halide ions, the acetylide ion also adds to carbonyl groups. 2-Methyl-3-butyn-2-ol (MBI) is an acetylenic alcohol used in the manufacture of products for the agrochemical and specialty chemical industry. It can be synthesized by the addition of acetylene to acetone to form the alkoxide ion and, as a second step, protonation of the alkoxide ion to produce the alcohol. Complete the mechanism for 2-methyl-3-butyn-2-ol production by drawing in the products of each step and the missing curved arrows. Sodium amide deprotonates the terminal alkyne to form sodium ethynide. Draw all missing reactants and/or products in the appropriate boxes by placing atoms on the canvas and connecting them with bonds. Add charges where needed. Electron flow arrows should start on an atom or a bond and should end on an atom, bond, or location where a new bond should be create

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Answer 1

Answer:

Answer:

Explanation:

check below for explanation.

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Calculate the electric double layer thickness of a alumina colloid in a dilute (0.1 mol/dm3) CsCI electrolyte solution at 30 °C. How are these numbers affected by the addition of 0.1 mol/dm3 of KCL? At what distance from the particle surface (r) has the potential decayed to 1% of its initial value?

Answers

Explanation:

The given data is as follows.

Concentration = 0.1 $mol/dm^(3)$

= 0.1 \frac{mol dm^{3}}{dm^{3}} \frac{10^{3}}{dm^{3}} \times \frac{6.022 \times 10^{23}}{1 mol} ions

= $6.022 * 10^(25) ions/m^(3)$

T = $30^(o)C$ = (30 + 273) K = 303 K

Formula for electric double layer thickness ( $\lambda_(D)$ ) is as follows.

$\lambda_(D)$ = $(1)/(k) = \sqrt (\varepsilon \varepsilon_(o) K_(g)T)/(2 n^(o) z^(2) \varepsilon^(2))$

where, $n^(o)$ = concentration = $6.022 * 10^(25) ions/m^(3)$

Hence, putting the given values into the above equation as follows.

$\lambda_(D)$ = $\sqrt (\varepsilon \varepsilon_(o) K_(g)T)/(2 n^(o) z^(2) \varepsilon^(2))$

= $\sqrt (78 * 8.854 * 10^(-12) c^(2)/Jm * 1.38 * 10^(-23)J/K * 303 K)/(2 * 6.022 * 10^(25) ions/m^(3) * (1)^(2) * (1.6 * 10^(-19)C)^(2))$

= $9.669 * 10^(-10)$ m

or, = $9.7 A^(o)$

= 1 nm (approx)

Also, it is known that $\lambda_(D)$ = $\sqrt (1)/(n^(o))$

Hence, we can conclude that addition of 0.1 $mol/dm^(3)$ of KCl in 0.1 $mol/dm^(3)$ of NaBr " $\lambda_(D)$ " will decrease but not significantly.

The rate constant for the second-order reaction 2NOBr(g) ¡ 2NO(g) 1 Br2(g) is 0.80/M ? s at 108C. (a) Starting with a concentration of 0.086 M, calculate the concentration of NOBr after 22 s. (b) Calculate the half-lives when [NOBr]0 5 0.072 M and [NOBr]0 5 0.054 M.

Answers

Explanation:

2NOBr(g) --> 2NO(g) 1 Br2(g)

Rate constant, k = 0.80

a) Initial concentration, Ao = 0.086 M

Final Concentration, A = ?

time = 22s

These parameters are connected with the equation given below;

1 / [A] = kt + 1 / [A]o

1 / [A] = 1 / 0.086 + (0.8 * 22)

1 / [A] = 11.628 + 17.6

1 / [A] = 29.228

[A] = 0.0342M

b) t1/2 = 1 / ([A]o * k)

when [NOBr]0 5 0.072 M

t1/2 = 1 / (0.072 * 0.80)

t1/2 = 1 / 0.0576 = 17.36 s

when [NOBr]0 5 0.054 M

t1/2 = 1 / (0.054 * 0.80)

t1/2 = 1 / 0.0432 = 23.15 s

Answer:

(a)

$0.0342M$

(b)

$t_(1/2)=17.36s\nt_(1/2)=23.15s$

Explanation:

Hello,

(a) In this case, as the reaction is second-ordered, one uses the following kinetic equation to compute the concentration of NOBr after 22 seconds:

$(1)/([NOBr])=kt +(1)/([NOBr]_0)\n(1)/([NOBr])=(0.8)/(M*s)*22s+(1)/(0.086M)=(29.3)/(M)\n$

$[NOBr]=(1)/(29.2/M)=0.0342M$

(b) Now, for a second-order reaction, the half-life is computed as shown below:

$t_(1/2)=(1)/(k[NOBr]_0)$

Therefore, for the given initial concentrations one obtains:

$t_(1/2)=(1)/((0.80)/(M*s)*0.072M)=17.36s\nt_(1/2)=(1)/((0.80)/(M*s)*0.054M)=23.15s$

Best regards.

If Nurse Antonio adds that 7 grams of NaCI to water to make 1 liter of solution, what is the molar concentration of the solution? Use dimensional analysis and show how you completed the unit conversion.

Answers

Given :

Nurse Antonio adds that 7 grams of NaCI to water to make 1 liter of solution.

To Find :

The molar concentration of the solution .

Solution :

Molecular mass of NaCl , $m=58.44\ g/mol$ .

Now , number of moles is given by :

$n=\frac{\text{Given weight}}{\text{Molecular Mass}}\n\nn=(7)/(58.44)\ mole\n\nn=0.12\ moles$

Molarity is given by :

$M=\frac{\text{Number of moles}}{\text{Volume ( in Liters)}}\n\nM=(0.12)/(1)\ M\n\nM=0.12\ M$

Hence , this is the required solution .

Solve the following problem:

Answers

Answer:

Option 3.

Explanation:

Isomerism is a phenomenon where by two or more compounds have the same molecular formula but different structural patterns.

Geometric Isomerism is a type of Isomerism that occurs within a double bond i.e Geometric isomers have different arrangement within the double bond.

Considering the options given above,

The 1st option is exactly the same as the compound only, it is inverted.

The 2nd option is still the same as the compound, only it is laterally inverted.

The 3rd option satisfy geometric Isomerism as the arrangement differ from the compound in the double bond.

The 4th option is entirely a saturated compound in which geometric Isomerism is not possible.

Complete combustion of 4.20 g of a hydrocarbon produced 12.9 g of CO2 and 6.15 g of H2O. What is the empirical formula for the hydrocarbon?

Answers

We calculate first the number of moles of CO2 and H2O produced by dividing the given masses by the molar masses of CO2 and H2O.
moles CO2 = (12.9 g CO2) x (1 mole CO2 / 12 g CO2) = 1.075 moles.
moles H2O = (6.15 g H2O) x (1 mole H2O / 18 g H2O) = 0.36 moles
Then, we count the number of C, H, and O moles. This gives us 1.075 moles C, 2.5 moles O and 0.72 moles H. The empirical formula is,
C1.075H0.72O2.5
Simplifying,
C4H3O10

How many iron atoms are in 0.32 mol of Fe2031? 3.9x 1023 jron atoms O 3.9 iron atoms O 6.02 x 1023 iron atoms 1.9 x 1023 iron atoms O 11x 10-24 iron atoms

Answers

Answer: $3.9* 10^(23)$ iron atoms

Explanation:

According to avogadro's law, 1 mole of every substance weighs equal to the molecular mass and contains avogadro's number $6.023* 10^(23)$ of particles.

$1 molecule of [tex]Fe_2O_3$ contains= 2 atoms of iron

$1 mole of [tex]Fe_2O_3$ contains= $2* 6.023* 10^(23)=12.05* 10^(23)$ atoms of iron

thus 0.32 moles of $Fe_2O_3$ contains= $(12.05* 10^(23))/(1)* 0.32=3.9* 10^(23)$ atoms of iron

Thus the sample would have $3.9* 10^(23)$ iron atoms.

Other Questions

A 99.8 mL sample of a solution that is 12.0% KI by mass (d: 1.093 g/mL) is added to 96.7 mL of another solution that is 14.0% Pb(NO3)2 by mass (d: 1.134 g/mL). How many grams of PbI2 should form?Pb(NO3)2(aq) + 2 KI(aq) PbI2(s) + 2 KNO3(aq)

In the laboratory a "coffee cup" calorimeter, or constant pressure calorimeter, is frequently used to determine the specific heat of a solid, or to measure the energy of a solution phase reaction. A student heats 62.08 grams of magnesium to 97.96 °C and then drops it into a cup containing 77.81 grams of water at 23.19 °C. She measures the final temperature to be 35.60 °C. The heat capacity of the calorimeter (sometimes referred to as the calorimeter constant) was determined in a separate experiment to be 1.79 J/°C. Assuming that no heat is lost to the surroundings calculate the specific heat of magnesium.

A solid that forms from a solution during a chemical reaction is called a(an)

Find the initial velocity for an enzymatic reaction when Vmax = 6.5 x 10–5 mol•sec–1 , [S] = 3.0 x 10–3 M, and KM = 4.5 x 10–3 M. A) not enough information is given to make this calculation B) 2.6 x 10–5 mol•sec–1 C) 1.4 x 10–2 mol•sec–1 D) 8.7 x 10–3 mol•sec–1 E) 3.9 x 10–5 mol•sec–1