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3 years ago

What is the name of this organic compound?

Chemistry

1 answer:

Stella [2.4K]3 years ago

7 0

ch3 is:

A methyl group is an alkyl derived from methane, containing one carbon atom bonded to three hydrogen atoms — CH3. In formulas, the group is often abbreviated Me. Such hydrocarbon groups occur in many organic compounds

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Which ecosystem is the least stable?

Softa [21]

Answer:

tundra bc it has barley any food source and has vary little inhabitants

Explanation:

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3 years ago

Neutral grain spirit is essentially the same as

Dennis_Churaev [7]

The neutral grain spirit is essentially known as rectified spirits or ethyl alcohol.

<h3>Meaning of neutral grain spirit</h3>

neutral grain spirit can be defined as raw material that is very concentrated with alcohol used to produce strong alcoholic products.

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2 years ago

3. The following data of decomposition reaction of thionyl chloride (SO2Cl2) were collected at a certain temperature and the con

KonstantinChe [14]

Answer:

a) First-order.

b) 0.013 min⁻¹

c) 53.3 min.

d) 0.0142M

Explanation:

Hello,

In this case, on the attached document, we can notice the corresponding plot for each possible order of reaction. Thus, we should remember that in zeroth-order we plot the concentration of the reactant (SO2Cl2 ) versus the time, in first-order the natural logarithm of the concentration of the reactant (SO2Cl2 ) versus the time and in second-order reactions the inverse of the concentration of the reactant (SO2Cl2 ) versus the time.

a) In such a way, we realize the best fit is exhibited by the first-order model which shows a straight line (R=1) which has a slope of -0.0013 and an intercept of -2.3025 (natural logarithm of 0.1 which corresponds to the initial concentration). Therefore, the reaction has a first-order kinetics.

b) Since the slope is -0.0013 (take two random values), the rate constant is 0.013 min⁻¹:

$m=\frac{ln(0.0768)-ln(0.0876)}{200min-100min} =-0.0013min^{-1}$

c) Half life for first-order kinetics is computed by:

$t_{1/2}=\frac{ln(2)}{k}=\frac{ln(2)}{0.013min^{-1}} =53.3min$

d) Here, we compute the concentration via the integrated rate law once 1500 minutes have passed:

$C=C_0exp(-kt)=0.1Mexp(-0.013min^{-1}*1500min)\\\\C=0.0142M$

Best regards.

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3 years ago