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Anton [14]
3 years ago
7

Ethyl trichloroacetate is significantly more reactive toward hydrolysis than ethyl acetate. Explain this observation. The three

chlorine atoms withdraw electron density via induction. This effect renders the carbonyl group less electrophilic. The three chlorine atoms add electron density via induction. This effect renders the carbonyl group less electrophilic. The three chlorine atoms withdraw electron density via induction. This effect renders the carbonyl group more electrophilic. The three chlorine atoms add electron density via induction. This effect renders the carbonyl group more electrophilic.
Chemistry
1 answer:
AleksAgata [21]3 years ago
8 0

Answer:

The three chlorine atoms withdraw electron density via induction. This effect renders the carbonyl group more electrophilic.

Explanation:

Ethyl trichloroacetate(C4H5Cl3O2) is significantly more reactive towards hydrolysis than ethyl acetate. This is because the three chlorine atoms withdraw electron density via induction and renders the carbonyl group C=O more electrophlic(tendency to attract more electrons).

Hydrolysis is a chemical reaction which involves water molecules being added to a substance.

The increased electrophilic feature enhances its reaction with water.

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If the average atomic mass of hydrogen in nature is 1.0079, what does that tell you about the percent composition of H-1 and H-2
vovangra [49]

Answer:

That the isotope H-1 is the most abundant in nature.

Explanation:

Hello!

In this case, since the average atomic mass of an element is computed considering the mass of each isotope and the percent abundance each, for hydrogen we would set up something like this:

m_H=m_{H_1}*\%abund_{H_1}+m_{H_2}*\%abund_{H_2}

Moreover, since the isotope notation H-1 and H-2 means that the atomic mass of H-1 is 1 amu, that of H-2 is 2 amu and the average one is 1.0079 amu, we can infer that the most of the hydrogen in nature is H-1 as the most of it composes the average hydrogen atom.

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4 0
2 years ago
In a coffee-cup calorimeter, 50.0ml of 0.100M AgNO3 and50.0ml of 0.100M HCl are mixed to yield the following reaction:
Marina CMI [18]

Answer:

-66.88KJ/mol

Explanation:

It is possible to obtain the heat involved in a reaction using a calorimeter. Formula is:

q = -C×m×ΔT

<em>Where q is heat of reaction, C is specific heat capacity (4.18J/°Cg), m is mass of solution (100.0g) and ΔT is temperature change (23.40°C-22.60°C = 0.80°C)</em>

Replacing:

q = -4.18J/°Cg×100.0g×0.80°C

q = -334.4J

Now, in the reaction:

Ag⁺ + Cl⁻→ AgCl

<em>AgNO₃ as source of Ag⁺ and HCl as source of Cl⁻</em>

Moles that react are:

0.050L× (0.100mol /L) = 0.0050moles

If 0.0050 moles produce -334.4J. Heat of reaction is:

-334.4J / 0.0050moles = -66880J/mol = <em>-66.88KJ/mol</em>

5 0
3 years ago
A 250-ml sample of na2so4 is reacted with an excess of bacl2. if 5.28 g baso4 is precipitated, what is the molarity of the na2so
Oksanka [162]
To get the molarity you need to follow this equation
                       moles of solute
Molarity (M = -----------------------
                        Liters of solution

But before you apply that equation you need to find the moles of solute and the liters of solution. Follow this equation

Na2SO4 + BaCl2 = BaSO4 + 2 NaCl

Solution

Moles of BaSO4 = 5.28 g 
                               ---------------
                                233.43 g / mol
                             =  0.0226  moles
Moles of NaSO4 = 0.0226 
                  0.0226 mole
Molarity = -----------------
                  0.250 L
              =  0.0905 mol / L

So the answer is 0.0905 mol / L
3 0
3 years ago
D‑Glucose and L‑glucose are
tensa zangetsu [6.8K]

Answer:

<h3>A. Epimers </h3>

Explanation:

<h2>Hope it help ❤️❤️❤️❤️❤️❤️❤️❤️❤️❤️❤️❤️❤️❤️❤️</h2>
3 0
3 years ago
Read 2 more answers
A student sets up the following equation to convert a measurement.
dybincka [34]

Answer:

\frac{1 m}{100 cm}

Explanation:

The final answer has a different set of units. In particular, meters (m) changes to centimeters (cm). To make this change, you need to multiply the first value by proportions.

When writing these proportions, it is important that they are arranged in a way that allows for the cancellation of units. For instance, since m is located in the denominator, it must be located in the numerator of the conversion.

<u>Proportion:</u>

1 m = 100 cm

The full expression:

<h3>-1.7*10^5\frac{V}{m}  ·  \frac{1 m}{100 cm}  =  ? \frac{V}{cm}</h3><h2>                 ^</h2>

As you can see, the old unit (m) cancels out and you are left with cm in the denominator.

7 0
2 years ago
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