Treatment of 1 mole of dimethyl sulfate with 2 moles of sodium acetylide results in the formation of propyne as the major product.
A) Draw a reasonable mechanism accounting for the formation of the byproduct 2-butyne.
B) 2-Butyne is observed as a minor product of this reaction. Draw a mechanism accounting for the formation of this minor product and explain how your proposed mechanism is consistent with the observation that acetylene is present among the reaction products.
C) Predict the major and minor products that are expected if diethyl sulfate is used in place of dimethyl sulfate.

Answer:

THE MOLAR MASS OF THE GAS IS 147.78 G/MOLE

Explanation:

Using PV = nRT

n = Mass / molar mass

P = 732.6 mmHg = 1 atm = 760 mmHg

So therefore 732.6 mmHg will be equal to 732.6 / 760 = 0.964 atm

P = 0.964 atm

V = 275 mL = 275 *10 ^-3 L

R = 0.082 Latm/ mol K

T = -28 C = 273 - 28 K = 245 K

mass =  1.95 g

molar mass = unknown

Having known the other variables in the formula, the molar mass of the gas can be obtained.

PV = m R T/ molar mass

Molar mass = m RT / PV

Molar mass = 1.95 * 0.082 * 245 / 0.964 * 275 *10^-3

Molar mass = 39.1755 / 265.1 *10^-3

Molar mass = 39.1755 / 0.2651

Molar mass = 147.78 g/mol

The molar mass of the gas is 147.78 g/mol

Answer:

0.0913 M

Explanation:

We'll begin by writing the balanced equation for the reaction.

This is given below:

H2C3H2O4 + 2NaOH —> C3H2Na2O4 + 2H2O

From the balanced equation above, we obtained the following:

The mole ratio of the acid (nA) = 1

The mole ratio of the base (nB) = 2

Data obtained from the question include:

Molarity of base, NaOH (Mb) = 0.0990 M

Volume of base, NaOH (Vb) = 20.52 mL

Volume of acid, H2C3H2O4 (Va) = 11.13 mL

Molarity of acid, H2C3H2O4 (Ma) =..?

The molarity of the acid, H2C3H2O4 can be obtained as follow:

MaVa/MbVb = nA/nB

Ma x 11.13 / 0.0990 x 20.52 = 1/2

Cross multiply

Ma x 11.13 x 2 = 0.0990 x 20.52 x 1

Divide both side by 11.13 x 2

Ma = (0.0990 x 20.52)/ (11.13 x 2)

Ma = 0.0913 M

Therefore, the molarity of malonic acid, H2C3H2O4 solution is 0.0913 M

Answer:

The concentration of one or more of the products is small.

The reaction will not proceed very far to the right.

The reaction will generally form more reactants than products  

Explanation:

We often write

K =[Products]/[Reactants]

Thus, if K is small

A. is wrong. Usually, if K < 1, the concentration of reactants is greater than that of the products.

Explanation:

Hi there!

I attached a photo of a unit summary that states the difference between s-t and v-t graph.

Hope this helps ;) ❤❤❤

Answer:

C. 1.07 M.

Explanation:

Hello,

In this case, we can define the molarity of the bleach as shown below:

[tex]M=\frac{moles_{NaClO}}{V_{solution}}[/tex]

In such a way, given the mass of bleach in a 1-L solution, we can compute the density:

[tex]\rho = \frac{1100g}{1L}=1100g/L =1.1 kg/L=1.1g/mL[/tex]

In such a way, we can use the previously computed density to compute the volume of the solution, assuming a 100-g solution given the by-mass percent:

[tex]V_{solution}=100g*\frac{1mL}{1.1g} *\frac{1L}{1000mL} =0.091L[/tex]

Afterwards, using the by-mass percent of bleach we compute the mass:

[tex]m_{NaClO}=100g*0.0725=7.25g[/tex]

And the moles:

[tex]n_{NaClO}=7.25g*\frac{1mol}{74.44g} =0.097mol[/tex]

Therefore, the molarity turns out:

[tex]M=\frac{0.097mol}{0.091L}\\ \\M=1.07M[/tex]

Thus, answer is C. 1.07 M.

Regards.

Answer: The question is is not complete...here is the complete question.

Which of the following reagents should be used to convert 3-Hexyne to E-3-hexene

Option B.

Na, liquid NH3.

Explanation:

3-Hexyne to E-3-hexene can be converted with by using the reagent of Na, NH3 (birch reduction) and this can be done by hdrogenation of H2.

The reagents NaNH3 convert 3-Hexyne to E-3-hexene because it is a reducing agents that convert or has the ability to reduce alkynes to trans alkenes.

3 Hexyne is an alkynes and it is converted to E- 3 hexene Na and NH3.

Explanation:

Answer:

1) before the addition of barium hydroxide

pH = -log[H⁺] = -log (0.276) = 0.559≈0.56

2)after the addition of barium hydroxide

pH = -log [H⁺] = -log(0.0857) = 1.067

3)at equivalent point, the solution will be neutral

pH = 7.0

4) after adding 40.7mL barium hydroxide

Explanation:

equation of reaction

2HCl(aq) + Ba(OH)₂(aq) ------->BaCl₂(aq) + 2H₂O(l)

1) Before the addition of barium hydroxide

concentration of HBr = 0.276M

[H⁺] = 0.276M

pH = -log[H⁺] = -log (0.276) = 0.559≈0.56

2) After adding 16.7mL barium hydroxide

moles of [OH⁻] = 16.7mL × 0.105 × 2

=3.507m mol = 3.507 × 10³mol

moles  of [H⁺] = 25.5mL × 0.276M

=7.038m mol = 7.038 × 10³mol

moles of  [H⁺] remaining = (7.038 - 3.421)m mol

= 3.617m mol = 3.617 × 10³mol

[H⁺]= [tex]\frac{3.617}{25.5 + 16.7}[/tex] = 0.0857

pH = -log [H⁺] = -log(0.0857) = 1.067

3) At equivalent point, the solution will be neutral

pH = 7.0

4) After adding 40.7mL barium hydroxide

moles of [OH⁻] = 40.7mL × 0.105M × 2

=8.547

moles of [OH⁻] remaining = 8.547 - 7.038

= 1.509m mol = 1.509 × 10³mol

pOH= -log[OH⁻]= 2.82

pH = 14 - 2.82 = 11.18

Answer:

16 J/K.mol

Explanation:

From the question,

ΔS = ΔH/T............... Equation 1

Where ΔH = Heat change, T = Temperature

But,

ΔH = n(Hv).................. Equation 2

Where n = number of  mole, Hv = heat of vaporization.

Given: Hv = 28.0 kJ/mol, n = 473/85 = 5.59 mole.

Substitute these values into equation 2

ΔH = 28/5.59

ΔH = 5.01 kJ.

Also: T = 273+39.8 = 312.8 J

Substitute into equation 1

ΔS = 5.01/312.8

ΔS = 0.016 kJ/K

ΔS = 16 J/K.mol

Answer:

Molar mass of the gas is 0.0961 g/mol

Explanation:

The effusion rate of an unknown gas = 11.1 min

rate of [tex]H_{2}[/tex] effusion = 2.42 min

molar mass of hydrogen = 1 x 2 = 2 g/m

molar mas of unknown gas = ?

From Graham's law of diffusion and effusion, the rate of effusion and diffusion is inversely proportional to the square root of its molar mass.

from

[tex]\frac{R_{g} }{R_{h} }[/tex] = [tex]\sqrt{\frac{M_{h} }{M_{g} } }[/tex]

where

[tex]R_{h}[/tex] = rate of effusion of hydrogen gas

[tex]R_{g}[/tex] = rate of effusion of unknown gas

[tex]M_{h}[/tex] = molar mass of H2 gas

[tex]M_{g}[/tex] = molar mass of unknown gas

substituting values, we have

[tex]\frac{11.1 }{2.42 }[/tex] = [tex]\sqrt{\frac{2 }{M_{g} } }[/tex]

4.587 = [tex]\sqrt{\frac{2 }{M_{g} } }[/tex]

[tex]\sqrt{M_{g} }[/tex] = [tex]\sqrt{2}[/tex]/4.587

[tex]\sqrt{M_{g} }[/tex] = 0.31

[tex]M_{g}[/tex] = [tex]0.31^{2}[/tex] = 0.0961 g/mol

The molar mass of the unknown gas will be "0.0961 g/mol".

Given:

Effusion rate of unknown gas,

Effusion rate of [tex]H_2[/tex],

Molar mass of hydrogen,

              [tex]= 2 \ g/m[/tex]

According to the Graham's law, we get

→    [tex]\frac{R_g}{R_h} = \sqrt{\frac{M_h}{M_g} }[/tex]

By substituting the values, we get

→   [tex]\frac{11.1}{2.42} = \sqrt{\frac{2}{M_g} }[/tex]

→ [tex]4.587=\sqrt{\frac{2}{M_g} }[/tex]

→ [tex]\sqrt{M_g} = \sqrt{\frac{2}{4.587} }[/tex]

   [tex]\sqrt{M_g} = 0.31[/tex]

       [tex]M_g = 0.0961 \ g/mol[/tex]

Thus the above solution is right.          

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Answer: The percentage composition of nitrogen , hydrogen and oxygen is 22.2 % , 1.59 % and 76.2% respectively.

Explanation:

Percentage composition is defined as the ratio of mass of substance to the total mass in terms of percentage.

Percentage composition=[tex]\frac{\text {mass of the element}}{\text {Total mass of the substance}}\times 100\%[/tex]

a) [tex]{\text {percentage composition of nitrogen}}=\frac{\text {mass of nitrogen}}{\text {Total mass}}\times 100\%[/tex]

[tex]{\text {percentage composition of nitrogen}}=\frac{9.8g}{9.8+0.7+33.6}\times 100\%=22.2\%[/tex]

b) [tex]{\text {percentage composition of hydrogen}}=\frac{\text {mass of hydrogen}}{\text {Total mass}}\times 100\%[/tex]

[tex]{\text {percentage composition of hydrogen}}=\frac{0.7}{9.8+0.7+33.6}\times 100\%=1.59\%[/tex]

c) [tex]{\text {percentage composition of oxygen}}=\frac{\text {mass of oxygen}}{\text {Total mass}}\times 100\%[/tex]

[tex]{\text {percentage composition of oxygen}}=\frac{33.6}{9.8+0.7+33.6}\times 100\%=76.2\%[/tex]

The percentage composition of nitrogen , hydrogen and oxygen is 22.2 % , 1.59 % and 76.2% respectively.

Answer:

Hey there!

That would be the alkaline earth metals.

Hope this helps :)

Answer: alkaline earth metals

Explanation:

Answer:

Answers are in the explanation.

Explanation:

Given concentrations are:

In the reaction:

2SO₂(g) + O₂(g) ⇄ 2SO₃(g)

Kc is defined as:

Kc = 15 = [SO₃]² / [O₂] [SO₂]²

Where concentrations of each species are equilbrium concentrations.

Also, you can define Q (Reaction quotient) as:

Q = [SO₃]² / [O₂] [SO₂]²

Where concentrations of each species are ACTUAL concentrations.

If Q > Kc, the reaction will shift to the left until Q = Kc;

If Q < Kc, the reaction will shift to the right until Q = Kc

If Q = Kc, there is no net reaction because reaction would be en equilibrium.

Replacing with given concentrations:

Answer:

D

Explanation:

Answer:

The halogens are the ortho and para directing groups. Whenever they react with other benzene compounds they will attach to the ortho or para positions of the benzene ring.

Major products which are obtained by reacting these given compounds are given in attached pictures with complete reactions.      

HBr will always be the side product of the bromine reactions along with the major compound.                                      

Explanation:

Answer:

They are called nodes.

Explanation:

Answer:

i guess this is the ans nodes

hope this helps

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Answer:

Lewis acid

Explanation:

In chemistry, a Lewis acid is any chemical specie that accepts a lone pair of electrons while a Lewis base is any chemical specie that donates a lone pair of electrons.

If we look at the formation of PF6^-, the process is as follows;

PF5 + F^- -----> PF6^-

We can see that PF5 accepted a lone pair of electrons from F^- making PF5 a lewis acid according to our definition above.

Hence in the formation of PF6^-, PF5 acts a Lewis acid.

Answer:

f is =2,2-dimethyl butane

g is = 2,2-dimethyl propane

h is = 3,3-diethyl pentane

Explanation:

please give me brainliest

Answer:

3.00 L

Explanation:

PV = nRT

(82 atm × 101325 Pa/atm) V = (10 mol) (8.314 J/mol/K) (27 + 273) K

V = 0.00300 m³

V = 3.00 L

Answer:

It usually consists of six steps: question, observation or investigation, hypothesis, experiment, analysis of data (reviewing what happened during the experiment), and conclusion. Scientific inquiry, on the other hand, is non-linear, which means it does not follow a consistent step-by-step process.

Explanation:

Hope it helps

Answer:

Use 62% - the equation is for the amount present at a given time. 0.62 = (1) e-kt -> ln(0.62)=-kt -> k = -ln(0.62)/t. I get k = .00117 hr-1 t(half) = 0.693/k = 590 hr.

HOPE THIS HELPS AND PLSSS MARK AS BRAINLIEST AND THNXX :)

The half-life is the time at which the substance's concentration is reduced by half of its initial amount. The half-life of a nuclide that lost its 38.0% mass is 590 hr. Thus, option C is correct.

Half-life is the time required by a substance to get reduced to half of its initial concentration. The half-life of the substance can be determined by the rate constant.

Given,

The initial quantity of substance (A₀) = 100

Remaining quantity (At) = 10 - 38 = 62

Time elapse (t) = 407 hours

The rate constant (k) is calculated as:

ln (At ÷ A₀) = - kt

ln (62 ÷ 100) ÷ 407 hour = - k

-0.47803580094 ÷ 407 = - k

k = 0.00117453513

Now, half-life from rate constant (k) is calculated as:

[tex]\rm t ^{\frac{1}{2}}[/tex] = 0.693 ÷ k

[tex]\rm t ^{\frac{1}{2}}[/tex] = 0.693 ÷ 0.00117453513

[tex]\rm t ^{\frac{1}{2}}[/tex] = 590 hours

Therefore, option C. 590 hours is the half-life of the substance.

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Answer

Hello

I think the reaction is like this FeO+OFe²O³

And the balance reaction is 2Fe+OFe²O³

Explanation:

At first we should find sth that has more atoms than the other then for example we realized that we have two atoms of Fe in Fe²O³ then put 2 before FeO and now we have 2 atoms of Fe in right side and 2 atoms of Fe in left then Oxygen in FeO change to 2 atoms of Oxygen and we have an other one in right side that they become 3 atoms of Oxygen and now we have 3 atoms of Oxygen in both right and left side.

Finally our reaction balanced.

Good luck

The Average atomic mass of phosphorus is 29.9.

The average atomic mass (sometimes called atomic weight) of an element is the weighted average mass of the atoms in a naturally occurring sample of the element.

Average masses are generally expressed in unified atomic mass units (u), where 1 u is equal to exactly one-twelfth the mass of a neutral atom of carbon-12.

An element can have differing numbers of neutrons in its nucleus, but it always has the same number of protons.

The versions of an element with different neutrons have different masses and are called isotopes.

The average atomic mass for an element is calculated by summing the masses of the element’s isotopes, each multiplied by its natural abundance on Earth i.e,

Average atomic mass of P = ∑(Isotope mass) (its abundance)

∴ Average atomic mass of P = (P-29 mass) (its abundance) + (P-30 mass)(its abundance) + (P-31 mass) (its abundance) + (P-33 mass) (its abundance)

Abundance of isotope = % of the isotope / 100.

∴ Average atomic mass of P = (29)(0.327) + (30)(0.4803) + (31)(0.184) + (33)(0.0087) = 29.88 a.m.u ≅ 29.9 a.m.u.

Hence , The Average atomic mass of phosphorus is 29.9.

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Answer:

a. increase

Explanation:

Catalysis is the process of increasing the rate of a chemical reaction by adding a substance known as a catalyst, which is not consumed in the catalyzed reaction.

By default, catalysts exists to speed up the rate of reactions. Increasing the amount of catalysts means that there would be an increase in the rate of reaction. The correct option is A.

✔ C5H4 has a molecular molar mass of :

M(C5H4) = 5 x M(C) + 4 x M(H)

✔ The molecular mass of C5H4 is therefore 64 g/mol.

Answer:

C10H8

Explanation:

I clicked on that answer, and it is correct.

Answer:

Explanation:

CH₃CHOHCOOH    ⇄    CH₃CHOHCOO⁻   +    H⁺

ionisation constant = 1.36 x 10⁻⁴ .

molecular weight of lactic acid = 90 g

moles of acid used = 20 / 90

= .2222

it is dissolved in one litre so molar concentration of lactic acid formed

C = .2222M

Let n be the fraction of moles ionised  

CH₃CHOHCOOH    ⇄    CH₃CHOHCOO⁻   +    H⁺

C  - nC                                          nC                  nC

By definition of ionisation constant Ka

Ka = nC x nC / C - nC

= n²C ( neglecting n in the denominator )

n² x .2222 = 1.36 x 10⁻⁴

n = 2.47  x 10⁻²

nC = 2.47  x 10⁻² x .2222

= 5.5 x 10⁻³

So concentration of hydrogen or hydronium ion = 5.5  x 10⁻³ g ion per litre .

The concentration of hydrogen or hydronium ion = 5.5  x 10⁻³ g ion per liter .

Ionization of lactic acid can be represented as:

CH₃CHOHCOOH⇄ CH₃CHOHCOO⁻  + H⁺

Given:

ionization constant = 1.36 x 10⁻⁴

mass= 20.0 g

Now, Molecular weight of lactic acid = 90 g

[tex]\text{Number of moles}=\frac{20}{90} =0.22mol[/tex]

It is dissolved in 1.00L so molar concentration of lactic acid formed will be

C = 0.22M

Consider "n" to be the fraction of moles ionized  

CH₃CHOHCOOH    ⇄    CH₃CHOHCOO⁻   +    H⁺

C  - nC                                          nC                  nC

By definition of ionization constant Ka

[tex]K_a =\frac{nC*nC}{C-nC}[/tex]

[tex]K_a= n^2C[/tex] ( neglecting n in the denominator )

On substituting the values we will get:

[tex]n^2 *0.22 = 1.36 *10^{-4}\\\\n = 2.47 * 10^{-2}[/tex]

To find the concentration of hydronium ion in the solution,

[tex]nC = 2.47 *10^{-2} *0.22\\\\nC= 5.5 * 10^{-3}[/tex]

So, concentration of hydrogen or hydronium ion = 5.5  x 10⁻³ g ion per liter.  

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Answer:

8.96g \ cm3

Explanation:

(89.6\ 10) (g\ cm3) = 8.96g\cm3

Answer:

The water would act as a base and would produce an undesired product of ethanol (CH3OH) through a dissociation reaction. If doing a titration reaction, it will likely yield inaccurate results.

Answer:

V2 = 17371.43ml

Explanation:

We use Boyles laws

since temperature is constant

P1V1=P2V2

760 x 400 = 17.5 x V2

304000 = 17.5 x V2

V2 = 304000/17.5

V2 = 17371.43ml

The volume will the dry hydrogen gas occupy at the temperature of 20°C and vapor pressure at  760 torrs will be 18 ml.

What is vapor pressure?

The vapor pressure of a liquid is independent of the volume of liquid in the container, whether one liter or thirty liters; both samples will have the same vapor pressure at the same temperature.

The temperature has an exponential connection with vapor pressure, which means that as the temperature rises, the vapor pressure rises as well the equation is -

P1 V1 / T1 = P2 V2 / T1

here, P = pressure

       T = temperature

        V = volume

substituting the value in the equation,

400 ×760 / 20 = 17.5× V / 20

V = 400× 760 / 20 × 17.5 / 20

V = 18 ml

Therefore the volume of the hydrogen gas remaining at this temperature will be 18 ml.

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Answer:

3.97 L

Explanation:

Data obtained from the question include the following:

Initial volume (V1) = 3.5 L

Initial temperature (T1) = 19 °C

Final temperature (T2) = 58 °C

Final volume (V2) =..?

Next, we shall convert celsius temperature to Kelvin temperature. This can be done as shown below:

Temperature (K) = temperature (°C) + 273

T (K) = T (°C) + 273

Initial temperature (T1) = 19 °C

Initial temperature (T1) = 19 °C + 273 = 292 K

Final temperature (T2) = 58 °C

Final temperature (T2) = 58 °C + 273 = 331 K

Finally, we shall determine the new volume of the gas by using Charles' law equation as shown below:

Initial volume (V1) = 3.5 L

Initial temperature (T1) = 292 K

Final temperature (T2) = 331 K

Final volume (V2) =..?

V1 /T1 = V2 /T2

3.5 /292 = V2 /331

Cross multiply

292 x V2 = 3.5 x 331

Divide both side by 292

V2 = (3.5 x 331) / 292

V2 = 3.97 L

Therefore, the new volume of the gas is 3.97 L.