What is the reaction of Dimethylamamine + acetic acid

Answers

Answer 1
DMA is prepared commercially by the reaction of dimethylamine with acetic anhydride or acetic acid. Dehydration of the salt of dimethylamine and acetic acid also furnishes this compound: CH3CO2H·HN(CH3)2 → H2O + CH3CON(CH3) Dimethylacetamide can also be produced by the reaction of dimethylamine with methyl acetate.

Related Questions

Please help:
An object rolled 15m/s for 3 seconds. How far did it roll?

Answers

45m!

explanation:
15x3=45 :)
9 m/s is the answer,.,,,.,,

if a lake has excess levels of phosphates and nirates in its water what will most likely result ?

Answers

Answer:

Eutrophication

Explanation:

Eutrophication results when there is excess levels of phosphates and nitrates in its water of lake.

When there is high level of nutrition level in the water body of lake, then the productivity of water plants increases thereby leading to algal boom and hence Eutrophication.

A gas in a sealed container has a pressure of 125 atm at a temperature of 303 K. If the pressure in the container is increased to 200 atm, what is the new temperature if the volume remains constant?
(Show work pls :)!)

Answers

Answer:

[tex]T_2=484.8K[/tex]

Explanation:

Hello there!

In this case, according to the the variable temperature and pressure and constant volume, it turns out possible for us to calculate the new temperature via the Gay-Lussac's law as shown below:

[tex]\frac{T_2}{P_2} =\frac{T_1}{P_1}[/tex]

Thus, by solving for the final temperature, T2, we obtain:

[tex]T_2 =\frac{T_1P_2}{P_1}[/tex]

So we plug in the given data to obtain:

[tex]T_2 =\frac{303K*200atm}{125atm}\\\\T_2=484.8K[/tex]

Best regards!

Imine formation from an aldehyde and an amine proceeds reversibly under slightly acidic conditions. The reaction is reversible due to acid-catalyzed hydrolysis of the imine. Imine formation generates an equivalent of water. Given these facts, explain why the imine can be isolated from the reaction mixture.

Answers

Answer:

Imine can be isolated from the reaction mixture as water is continuously removed from the reaction chamber

Explanation:

In this reaction, a non -aqueous solvent  is not used (not mentioned in the question). Thus, we can say that there is continuous removal water under suitable reacting conditions and hence the imine formed is left behind.

An isolated atom of a certain element emits light of wavelength 655 nm when the atom falls from its fifth excited state into its second excited state. The atom emits a photon of wavelength 435 nm when it drops from its sixth excited state into its second excited state. Find the wavelength of the light radiated when the atom makes a transition from its sixth to its fifth excited state.

Answers

Answer:

[tex]\frac{1}{\lambda_{6-5}} \approx 752nm[/tex]

Explanation:

From the question we are told that

Light wavelength  [tex]\lambda_l=655nm[/tex]

Light wavelength atom fall [tex]x_L=5th-2nd[/tex]

Photon wavelength  [tex]\lambda_p=435nm[/tex]

Photon wavelength atom fall [tex]x_P=^th-2nd[/tex]

Generally the equation for the reciprocal of wavelength of emitted photon is is mathematically given by

[tex]\frac{1}{\lambda}=R(\frac{1}{ \lambda _f^2}-\frac{1}{\lambda _i^2} )[/tex]

Therefore for initial drop of 5th to 2nd

[tex]\frac{1}{\lambda_{5-2}}=R(\frac{1}{2^2}-\frac{1}{5^2} )[/tex]

Therefore for initial drop of 6th to 2nd

[tex]\frac{1}{\lambda_{6-2}}=R(\frac{1}{2^2}-\frac{1}{6^2} )[/tex]

Generally we subtract (5th to 2nd) from (6th to 2nd)

[tex]\frac{1}{\lambda_{5-2}}- \frac{1}{\lambda_{6-2}}=R(\frac{1}{2^2}-\frac{1}{5^2} )-\frac{1}R(\frac{1}{2^2}-\frac{1}{6^2} )[/tex]

[tex]\frac{1}{\lambda_{5-2}}- \frac{1}{\lambda_{6-2}}=R(\frac{1}{5^2}-\frac{1}{6^2} )[/tex]

[tex]\frac{1}{\lambda_{5-2}}- \frac{1}{\lambda_{6-2}}=\frac{1}{\lambda_{5-6}}[/tex]

[tex]\frac{1}{\lambda_{5-6}}=\frac{1}{4350nm}-\frac{1}{655nm}[/tex]

[tex]\frac{1}{\lambda_{5-6}}=1.33*10^{-3}[/tex]

Therefore for 6th to 5th stage is mathematically given by

[tex]\frac{1}{\lambda_{6-5}}=(1.33*10^{-3})^{-1}[/tex]

[tex]\frac{1}{\lambda_{6-5}}=751.879nm[/tex]

[tex]\frac{1}{\lambda_{6-5}} \approx 752nm[/tex]

When 148. mg of a certain molecular compound X are dissolved in 75.g of cyclohexane (CH), the freezing point of the solution is measured to be 59 'C. Calculate the molar mass of X If you need any additional information on cyclohexane, use only what you find in the ALEKS Data resource. Also, be sure your answer has a unit symbol, and is rounded to l significant digit. XS ?

Answers

Answer: The molar mass of X is 61.3g/mol

Explanation:

Depression in freezing point is given by:

[tex]\Delta T_f=K_f\times m[/tex]

[tex]\Delta T_f=T_f^0-T_f=(6.55^0C-5.9^0C)=0.65^0C[/tex] = Depression in freezing point

[tex]K_f[/tex] = freezing point constant = [tex]20.2^0C/mol[/tex]

m= molality  = [tex]\frac{\text{Mass of solute}}{\text{Molar mass of solute}\times \text{Mass of solvent in Kg}}[/tex]

where,  

Now put all the given values in this formula, we get

[tex]0.65^0C=20.2°C/m\times frac{0.148g}}{M\times 0.075kg}[/tex]

[tex]M=61.3g/mol[/tex]

Thus molar mass of X is 61.3g/mol

What is the greatest concentration of pollution

Answers

Answer:

B

Explanation:The poisonous substances, present in the environment can easily get into the trophic level as living organism depends on each other and environment for food and nutrition. These poisonous substances may not be broken down in the body or excreted easily, efficiently and quickly. Instead, they accumulate in the tissues, and as the living organism eats more, the concentration of these substances increases and they pass from one trophic level to the next. The tertiary consumer being at the top of trophic levels receives the maximum pollutant. This phenomenon is known as biological magnification.

So, the correct answer is option B.

Gelatinization happens with what type of Carbohydrate?

Answers

it happens with starch granules!

What is the main function of burette?

Answers

Answer:

I  think it A

Explanation:

Answer:

its d i think

Explanation:

lave
Use the drop-down menus to identify the parts of a
wave.
A
B:
C
D

Answers

this doesn’t make sense

Answer:

Use the drop-down menus to make your selections.

What is the difference between wave A and wave B?

✔ Wave A has higher amplitude than wave B.

What is the difference between wave C and wave D?

✔ Wave C has a lower frequency than wave D.

Explanation:

Select the statement that is FLASE about this experiment. Group of answer choices Changing temperature will affect the redox potential. Changing concentration will affect the redox potential. Changing metal solution will affect the redox potential. To ensure stable reading, insert the conductivity meter in the solution for at least 60 seconds. To ensure consistent data, collect the redox potential values at least three times.

Answers

Answer:

To ensure stable reading, insert the conductivity meter in the solution for at least 60 seconds.

Explanation:

The False statement is ; To ensure stable reading, insert the conductivity meter in the solution for at least 60 seconds.

All other statement will affect the experiment's redox potential but  inserting the conductivity meter in the solution for at least 60 seconds will not affect the Redox potential  .

Redox potential is used to describe/measure a system's oxidizing capacity

How many grams of hydroxide pellets, NaOh are required to prepare 50.0ml of a 0.150 M solution ?

A.0.300g
B.3.00g
C.2.00g
D.200.g
E. 0.025 M

Answers

the correct answer is E10.
(O.0500L) x (0.150mo/L)x(39.997 15 g NaOh/mol)=0.300 g NaOh



ANSWER A

A 31.25 mL aliquot of weak base that has a concentration of 0.683 M will be titrated with 0.434 M HCl. Calculate the pH of of the solution upon neutralization of half of the weak base. The Kb of the base is 1.5×10-6.

Answers

Answer:

pH = 8.18

Explanation:

The weak base, X, reacts with HCl as follows:

X + HCl → HX⁺ + Cl⁻

Where 1 mole of X with 1 mole of HCl produce 1 mole of HX⁺ (The conjugate acid of the weak base).

Now, using H-H equation for bases:

pOH = pKb + log [XH⁺] / [X]

Where pOH is the pOH of the buffer (pH = 14 -pOH)

pKb is -log Kb = 5.824

And [X] [HX⁺] are the molar concentrations of each specie

Now, at the neutralization of the half of HX⁺, the other half is as X, that means:

[X] = [HX⁺]

And:

pOH = pKb + log [HX⁺] / [X]

pOH = 5.824 + log 1

pOH = 5.824

pH = 14-pOH

pH = 8.18

the number of particles in a mole if substance

A.avografo number

b.molar volume

c.molar mass

d.molar ratio

Answers

its avagadro number
represented by Na
6.02*10^23

Which of the following behaviors might indicate a patient is drug seeking?
A. A patient wants to avoid a specific medication because of
potential side effects.
B. A patient explains that she is from out of town and needs a
specific medicine because she left her prescription at home.
C. A patient fears her new prescription will conflict with another
medication she's currently taking.
D. A patient thinks she needs a smaller dose of her prescription
because it gives her headaches.
SUBMIT

Answers

The answer would be D sorry if wrong

Answer:

B is correct :)

Explanation:

Trust me I just took the test

Identify the term that matches each electrochemistry definition.

a. The electrode where reduction occurs ___________
b. An electrochemical cell powered by a spontaneous redox reaction ___________
c. The electrode where oxidation occurs__________
d. An electrochemical cell that takes in energy to carry out a nonspontaneous redox reaction ____________
e. A chemical equation showing either oxidation or reduction ___________

Answers

Answer: a. Cathode

b. Galvanic cell

c. Anode

d. Electrolytic cell

e. half reaction

Explanation:

Galvanic cell or Electrochemical cell is defined as a device which is used for the conversion of the chemical energy produced in a spontaneous redox reaction into the electrical energy.

Electrolytic cell is a device where electrical energy is used to drive a non spontaneous chemical reaction.

In the electrochemical cell, the oxidation occurs at an anode which is a negative electrode and the reduction occurs at the cathode which is a positive electrode.  Thus the electrons are produced at anode and travel towards cathode.

The balanced two-half reactions will be:

Oxidation half reaction : [tex]M\rightarrow M^{n+}+ne^-[/tex]

Reduction half reaction : [tex]N^{n+}+ne^-\rightarrow N[/tex]

Thus the overall reaction will be: [tex]M+N^{n+}\rghtarrow M^{n+}+N[/tex]

The half-life of argon-39 is 269 years. It decays into krypton-39. After 1,076 years, what fraction of the original amount of argon-39 in a sample will still be argon

1/16
1/4
1/2
1/8

Answers

Answer:

1/16

Explanation:

From the question given above, the following data were obtained:

Half-life (t½) = 269 years

Time (t) = 1076 years

Fraction remaining =?

Next, we shall determine the number of half-lives that has elapsed. This can be obtained as follow:

Half-life (t½) = 269 years

Time (t) = 1076 years

Number of half-lives (n) =?

n = t / t½

n = 1076 / 269

n = 4

Thus, 4 half-lives has elapsed.

Finally, we shall determine the fraction of the original amount remaining. This can be obtained as follow:

Let N₀ be the original amount.

Let N be the amount remaining.

Number of half-lives (n) = 4

Fraction remaining (N/N₀ ) =?

N = 1/2ⁿ × N₀

N = 1/2⁴ × N₀

N = 1/16 × N₀

Divide both side by N₀

N/N₀ = 1/16

Thus, the fraction of the original amount remaining is 1/16

what is the difference between an acid and a base?

Answers

Answer:

A substance, mostly liquid that donates a proton or accepts an electron pair in reactions. An acid increases the concentration of H+ ions. A base is a substance that releases hydroxide (OH-) ions in aqueous solution, donates electrons and accepts protons.

[tex]{\underline{\underline{\bf{\red{\:\:Acid\:\:}}}}}[/tex]

Proton donor and electron pair acceptorIt can increase the concentration of H+ ion in aq. solution.pH value is <7Turn blue litmus paper into redAcid reacts with bases to form saltsAcid taste sour

[tex]{\underline{\underline{\bf{\orange{\:\:Base\:\:}}}}}[/tex]

Proton acceptor and electron pair donorIt can increase the concentration of OH- ion in aq. solution.pH value is >7Turn red litmus paper into blueBases reacts with acids to form saltsBases taste soapy

You and your science group have been handed five mineral samples to identity. First, you on the minerals by color None of the
minerals are white. You and your group know that you probably do not have

A) Calcite

B) gypsum

C) Calcite and Talc

D) feldspar and calcite

Answers

Answer: A

Explanation:

calcite for sure, but feldspar is also white so it could be d.

3. Convert the word equations below to symbolic:
Sodium + chlorine sodium chloride
• Calcium + bromine ---- calcium bromide
• Potassium + water → potassium hydroxide + hydrogen
ot

Answers

Answer:

1)2Na + Cl2 ----> 2NaCl

2)Ca + Br2 ---->CaBr2

3)K + H2O -----> KOH + H2

- Calculate the Standard Enthalpy of the reaction below:
NH3(g) + HCI (g) → NH4Cl(s)
Using the following Enthalpy of Reactions:
2HCI(g) → H2(g) + Cl2(g)
AH = +184.6 KJ
2H2(g) + 1/2 N2(g) + 1/2 Cl2(g) → NH4Cl(s) deltaH = -314.4 kJ
N2(g) + 3 H2(g) → 2 NH3(g)
deltaH = +184.6 kJ​

Answers

Answer:

Explanation:

We have the three equations:

[tex]NH_{3(g)} + HCl_{(g)} => NH_4Cl_{(s)} ..... \Delta H = ? (1)\\2HCl_{(g)} => H_{2(g)} + Cl_{2(g)} .... \Delta H = +184.6 kJ (2)\\2H_{2(g)} + 1/2N_{2(g)} + 1/2Cl_{2(g)} => NH_4Cl_{(s)} ..... \Delta H = -314.2 kJ (3)\\ N_{2(g)} + 3H_{2(g)} => 2NH_{3(g)} .... \Delta H = +184.6kJ (4)[/tex]

(can you double check that it is 184.6kJ for both equations 2 and 4 because it seems unlikely). We need to solve for equation 1 by addition and changing equations 2, 3 and 4. After possibly some trial and error, we can find that if we flip equations 4, multiply equation 3 by 2, add the equations together, and then finally divide by 2, we can get equation 1. We will get the answer of -314.2 kJ. However, I am again skeptical about the delta H values for equation 2 and 4 so double check that. This method might be super confusing and it is really hard to explain. So what I would suggest you to watch videos on Hess' law.

How many grams of H2SO4 will react 650.0 grams of NaOH?

Answers

Answer:

H2SO4 + 2NAOH➡NA2SO4+2H2O

H2SO4 = 1*2 +32+16*3= 98g

NAOH = (11+16+1)=28

Explanation:

To calculate the number of Mole = Mole *Molar Mass

To calculate for gram = Mole/ Molar Mass

20.5g NaoH/1 * 1 mole NaOH/28g NaoH = 0.73M NaoH

0.73M NaoH /1 * 1 M H2SO4 /2M NaoH = 0.37M H2SO4

0.37 M H2SO4 /1 * 98g H2SO4 /1M H2SO4 = 0.00367g H2SO4

:- 0.00367g of Sulphuric Acid will be needed to complete the reaction.

What is the balanced net ionic reaction for the following -
CaCl2 + Na2S --> NaCl + CaS?​

Answers

Answer:

CaCl2 + Na2S --> 2NaCl + CaS?​

Explanation:

CaCl2 + Na2S --> 2NaCl + CaS?​

does anyone know thisss???

Answers

B............................

pH de un jugo de naranja con [H+]=5.6×10-4M.​

Answers

Explanation:

[tex]pH = - log[H {}^{ + } ] \\ = - log(5.6 \times {10}^{ - 4} ) \\ = 3.25[/tex]

water and Air are needed for iron rust answer true or false​

Answers

Answer:True

Explanation:

PLZ GIVE ME BRAINLIEST, I NEED IT

Answer: True


Explanation: also true

Which phase change is exothermic?

Answers

Answer:

B. Solidification

Hope this helped! :)

c. Melting ! pls lmk if that was correct !:)

How many moles of BaS would be used to make 1200 mL of a 10.OM solution?

Answers

200 bas would be used to make 1200 ml of soulution

The reduction of iron(III) oxide () to pure iron during the first step of steelmaking, ()()() is driven by the high-temperature combustion of coke, a purified form of coal: ()()() Suppose at the temperature of a blast furnace the Gibbs free energies of formation of and are and , respectively. Calculate the maximum mass of pure iron that can be produced by the combustion of of coke. Round your answer to significant digits.

Answers

Answer:

hello your question is incomplete attached below is the complete question

answer : To 2 significant digits = 5500 kg

Explanation:

Given

C(s) + O2 (g)  ----------> CO2  ( g )

mass of CO2 = 8.9 * 10^6 g

number of moles =  8.9 * 10^6 / 12  = 7.416 * 10 ^5  mol

same amount of moles is needed by O2 hence

attached below is the detailed  solution

Calculate the Standard Enthalpy of the reaction below:
NH3(g) + HCl (g) → NH4Cl(s)
Using the following Enthalpy of Reactions:
2HCl(g) → H2(g) + Cl2(g)
AH = +184.6 kJ
2H2(g) + 12 N2(g) + 2 Cl2(g) NH4Cl(s) AH = -314.4 kJ
N2(g) + 3 H2(g) → 2 NH3(g)
AH = +184.6 kJ​

Answers

Answer:

(we use hess's law) it is so simple but the second reaction is not correct please right it

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