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4 years ago

Write a net ionic equation to show that triethanolamine, C6H15O3N, behaves as a Bronsted-Lowry base in water.

Chemistry

1 answer:

Romashka [77]4 years ago

7 0

Answer:

C₆H₁₅O₃N + H₂O ⇄ C₆H₁₅O₃NH⁺ + OH⁻

Explanation:

A Bronsted-Lowry base is defined as a substance than can accept a proton, H⁺.

When triethanolamine, C₆H₁₅O₃N, reacts with water, H₂O, as follows:

As you can see, in this reaction, triethanolamine is able to accept a proton that is given for the water.

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For the product of the reaction below, which proton is removed irreversibly by NaNH2 base, thus preventing any isomerization of

Serggg [28]

Answer:

The Highly acidic proton joined to one of the carbon in the ALKYNE bond.

(Kindly Check the attachment for the drawing because the solution will need us to draw).

Explanation:

So, let us start by defining some major key terms in this particular Question given above;

(1). ISOMERIZATION: isomerization can simply be defined as the kind is of chemical rearrangement whichay lead to the breaking and the formation of new bonds.

(2). NaNH2 BASE: Sodium amide is a Chemical compound which has a Molar mass of 39.01 g/mol and Heat capacity (C) of 66.15 J/mol K. It is also known as sodamide. It is a good nucleophile.

(3). ALKYNE BOND: it is a C-C joined together by three bonds.

The chemical reaction given in the Question is given in the attachment too.

Therefore, The Highly acidic proton joined to one of the carbon in the ALKYNE bond is removed irreversibly by NaNH2 base.

7 0

3 years ago

The decomposition of \rm XY is second order in \rm XY and has a rate constant of6.96Ã10â3M^{-1} \cdot s^{-1} at a certain temper

LUCKY_DIMON [66]

Explanation:

Half life is the amount of time taken by a radioactive material to decay to half of its original value.

Half life for second order kinetics is given by:

$t_{1/2}=\frac{1}{k\times a_0}$

k = rate constant =?

$a_0$ = initial concentration

a = concentration left after time t

Integrated rate law for second order kinetics is given by:

$\frac{1}{a}=kt+\frac{1}{a_0}$

a) Initial concentration of XY = $a_o=0.100 M$

Rate constant of the reaction = k = $6.96\times 10^{-3} M^{-1} s^{-1}$

Half life of the reaction is:

$t_{1/2}=\frac{1}{6.96\times 10^{-3} M^{-1} s^{-1}\times 0.100 M}$

$=1,436.78 s$

1,436.78 seconds is the half-life for this reaction.

b) Initial concentration of XY = 0.100 M

Final concentration after time t = 12.5% of 0.100 M = 0.0125 M

$\frac{1}{0.0125 M}=6.96\times 10^{-3} M^{-1} s^{-1}\times t+\frac{1}{0.100M}$

Solving for t;

t = 10,057.47 seconds

In 10,057.47 seconds the concentration of XY will become 12.5% of its initial concentration.

c) Initial concentration of XY = 0.200 M

Final concentration after time t = 12.5% of 0.200 M = 0.025 M

$\frac{1}{0.025 M}=6.96\times 10^{-3} M^{-1} s^{-1}\times t+\frac{1}{0.200M}$

Solving for t;

t = 5,028.73 seconds

In 5,028.73 seconds the concentration of XY will become 12.5% of its initial concentration.

d) Initial concentration of XY = 0.160 M

Final concentration after time t = $6.20\times 10^{-2} M$

$\frac{1}{6.20\times 10^{-2} M}=6.96\times 10^{-3} M^{-1} s^{-1}\times t+\frac{1}{0.200M}$

Solving for t;

t = 1,419.40 seconds

In 1,419.40 seconds the concentration of XY will become $6.20\times 10^{-2} M$ .

e) Initial concentration of $SO_2Cl_2= 0.050 M$

Final concentration after time t = x

t = 55.0 s

$\frac{1}{x}=6.96\times 10^{-3} M^{-1} s^{-1}\times 55.0 s+\frac{1}{0.050 M}$

Solving for x;

x = 0.04906 M

The concentration after 55.0 seconds is 0.04906 M.

f) Initial concentration of XY= 0.050 M

Final concentration after time t = x

t = 500 s

$\frac{1}{x}=6.96\times 10^{-3} M^{-1} s^{-1}\times 500 s+\frac{1}{0.050 M}$

Solving for x;

x = 0.04259 M

The concentration after 500 seconds is 0.0.04259 M.

7 0

3 years ago

You removed a Coke can from the refrigerator. When you come back a few minutes later, you realize the can has begun to "sweat".

Vinvika [58]

Answer:

Explanation:

8 0

3 years ago

What is the percent composition of sulfur in H2SO4?

Triss [41]

Answer:

D (32.7%)

Explanation:

4 0

3 years ago

What happens to iron oxide during decomposition

elena55 [62]

Answer:

it gets reduced from a +3 oxidation to a 0.

Explanation:

the decomposition of iron oxide to elemenTal can be represented by the following equation: iron oxide (Fe203), the oxidation state of iron is +3 while that of oxygen is -2. therefore, the above reaction is a redox (reduction oxidation reaction)

4 0

3 years ago