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2 years ago

How many carbon atoms are represented by the model below? A. 12 B. 5 C. 4 D. 6

Chemistry

2 answers:

Papessa [141]2 years ago

5 0

I think it’s d sorry if I didn’t help

solong [7]2 years ago

4 0

Answer:

D. 6

Explanation:

Each pointy end represents 1 carbon. So in total we have six carbons.

The name of this organic compound is hexane.

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This graph represents a population of molecules in a gas versus the distribution of the average velocity(speed) of its molecules

pshichka [43]

Answer:

Part A

Given that the graph is symmetrical and bell shaped, the average kinetic energy is given by the midline of graph, which corresponds to the common speed of the highest number of the population

Part B

The formula for the average kinetic energy, K.E. = (3/2)·(R/NA)·T

Therefore, the part of the graph that indicates the temperature of the sample is the average kinetic energy. K.E.

Part C

At a lower temperature, the heat is less evenly distributed and we have the distribution T2 higher than T1

Please see the attached graph created with MS Visio

Explanation:

3 0

2 years ago

Consider the reaction of A(g) + B(g) + C(g) => D(g) for which the following data were obtained:

Drupady [299]

Answer:

$r = k [A]^{2}[B]^{2}$

Explanation:

A + B + C ⟶ D

$\text{The rate law is } r = k [A]^{m}[B]^{n}[C]^{o}$

Our problem is to determine the values of m, n, and o.

We use the method of initial rates to determine the order of reaction with respect to a component.

(a) Order with respect to A

We must find a pair of experiments in which [A] changes, but [B] and C do not.

They would be Experiments 1 and 2.

[B] and [C] are constant, so only [A] is changing.

$\begin{array}{rcl}\dfrac{r_{2}}{r_{1}} & = & \dfrac{ k[A]_{2}^{m}}{ k[A]_{1}^{m}}\\\\\dfrac{2.50\times 10^{-2}}{6.25\times 10^{-3}} & = & \dfrac{0.100^{m}}{0.0500^{m}}\\\\4.00 & = & 2.00^{m}\\m & = & \mathbf{2}\\\end{array}\\\text{The reaction is 2nd order with respect to A}$

(b) Order with respect to B

We must find a pair of experiments in which [B] changes, but [A] and [C] do not. There are none.

They would be Experiments 2 and 3.

[A] and [C] are constant, so only [B] is changing.

$\begin{array}{rcl}\dfrac{r_{3}}{r_{2}} & = & \dfrac{ k[B]_{3}^{n}}{ k[B]_{2}^{n}}\\\\\dfrac{1.00\times 10^{-1}}{2.50\times 10^{-2}} & = & \dfrac{0.100^{n}}{0.0500^{n}}\\\\4.00 & = & 2.00^{n}\\n & = & \mathbf{2}\\\end{array}\\\text{The reaction is 2nd order with respect to B}$

(c) Order with respect to C

We must find a pair of experiments in which [C] changes, but [A] and [B] do not.

They would be Experiments 1 and 4.

[A] and [B] are constant, so only [C] is changing.

$\begin{array}{rcl}\dfrac{r_{4}}{r_{1}} & = & \dfrac{ k[C]_{4}^{o}}{ k[C]_{1}^{o}}\\\\\dfrac{6.25\times 10^{-3}}{6.25\times 10^{-3}} & = & \dfrac{0.0200^{o}}{0.0100^{o}}\\\\1.00 & = & 2.00^{o}\\o & = & \mathbf{0}\\\end{array}\\\text{The reaction is zero order with respect to C.}\\\text{The rate law is } r = k [A]^{2}[B]^{2}$