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2 years ago

9

Which of the graphs below shows a person walking at a constant rate with no change in speed or velocity?

2 answers:

charle [14.2K]2 years ago

8 0

$▪▪▪▪▪▪▪▪▪▪▪▪▪ {\huge\mathfrak{Answer}}▪▪▪▪▪▪▪▪▪▪▪▪▪▪$

The graph showing a person who is walking at the same rate with no change in speed or velocity is :

Person B

maria [59]2 years ago

7 0

Answer:

all of these maybe I think

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A student ran the following reaction in the laboratory at 311 K:CH4(g) + CCl4(g) 2CH2Cl2(g)When she introduced 4.10×10-2 moles o

statuscvo [17]

<u>Answer:</u> The value of $K_{eq}$ is 0.044

<u>Explanation:</u>

We are given:

Initial moles of methane = $4.10\times 10^{-2}mol=0.0410moles$

Initial moles of carbon tetrachloride = $6.51\times 10^{-2}mol=0.0651moles$

Volume of the container = 1.00 L

Concentration of a substance is calculated by:

$\text{Concentration}=\frac{\text{Number of moles}}{\text{Volume}}$

So, concentration of methane = $\frac{0.0410}{1.00}=0.0410M$

Concentration of carbon tetrachloride = $\frac{0.0651}{1.00}=0.0651M$

The given chemical equation follows:

$CH_4(g)+CCl_4(g)\rightleftharpoons 2CH_2Cl_2(g)$

<u>Initial:</u> 0.0410 0.0651

<u>At eqllm:</u> 0.0410-x 0.0651-x 2x

We are given:

Equilibrium concentration of carbon tetrachloride = $6.02\times 10^{-2}M=0.0602M$

Evaluating the value of 'x', we get:

$\Rightarrow (0.0651-x)=0.0602\\\\\Rightarrow x=0.0049M$

Now, equilibrium concentration of methane = $0.0410-x=[0.0410-0.0049]=0.0361M$

Equilibrium concentration of $CH_2Cl_2=2x=[2\times 0.0049]=0.0098M$

The expression of $K_{eq}$ for the above reaction follows:

$K_{eq}=\frac{[CH_2Cl_2]^2}{[CH_4]\times [CCl_4]}$

Putting values in above expression, we get:

$K_{eq}=\frac{(0.0098)^2}{0.0361\times 0.0603}\\\\K_{eq}=0.044$

Hence, the value of $K_{eq}$ is 0.044

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3 years ago

Which organism is able to cause an infection

spin [16.1K]

Answer: The answer is bacteria. The correct answer is The letter B or the seconed one!

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3 years ago

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Which statement describes one feature of Rutherford model of the atom

serious [3.7K]

$\huge{ \color{magenta}{ \fcolorbox{magenta}{black}{ \huge{ \color{white}{ \fcolorbox{aqua}{black}{♡Answer♡ }}}}}}$

<em><u>The Rutherford model shows that an atom is mostly empty space, with electrons orbiting a fixed, positively charged nucleus in set, predictable paths.</u></em>

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2 years ago

Consider the following reaction where K. = 154 at 298 K: 2NO(g) + Brz(9) 2NOBr(g) A reaction mixture was found to contain 2.69x1

bekas [8.4K]

Explanation:

$2NO(g) + Br_2(g)\rightleftharpoons 2NOBr(g)$

Equilibrium constant of reaction = $K=154$

Concentration of NO = $[NO]=\frac{2.69\times 10^{-2} mol}{1 L}=2.69\times 10^{-2} M$

Concentration of bromine gas = $[Br_2]=\frac{3.85\times 10^{-2} mol}{1 L}=3.85\times 10^{-2} M$

Concentration of NOBr gas = $[Br_2]=\frac{9.56\times 10^{-2} mol}{1 L}=9.56\times 10^{-2} M$

The reaction quotient is given as:

$Q=\frac{[NOBr]^2}{[NO]^2[Br_2]}=\frac{(9.56\times 10^{-2} M)^2}{(2.69\times 10^{-2} M)^2\times 3.85\times 10^{-2} M}$

$Q=328.06$

$Q>K$

The reaction will go in backward direction in order to achieve an equilibrium state.

1. In order to reach equilibrium NOBr (g) must be produced. False

2. In order to reach equilibrium K must decrease. False

3. In order to reach equilibrium NO must be produced. True

4. Q. is less than K . False

5. The reaction is at equilibrium. No further reaction will occur. False

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3 years ago

On a periodic table, element names are represented by:

alina1380 [7]

Answer: Symbols

Explanation:

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3 years ago