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Zielflug [23.3K]

3 years ago

9

The reaction described by the equation O 3 ( g ) + NO ( g ) ⟶ O 2 ( g ) + NO 2 ( g ) has, at 310 K, the rate law rate of reactio

n = k [ O 3 ] [ NO ] k = 3.0 × 10 6 M − 1 ⋅ s − 1 Given that [ O 3 ] = 7.0 × 10 − 4 M and [ NO ] = 5.0 × 10 − 5 M at t = 0 , calculate the rate of the reaction at t = 0 .

1 answer:

UNO [17]3 years ago

6 0

Answer:

Explanation:

Equation of the reaction:

O3(g) + NO(g) ⟶ O2(g) + NO2(g)

From the Rate Law:

R = k[O3][NO]

The order of NO is 1 and that of O3 is also 1, therefore order of reaction = 1 + 1

= 2

Given :

Concentration of O3, [O3] = 7.0×10^-4M

Concentration of NO, [NO] = 5.0×10^-5M

Rate constant, k = 3.0 × 10^6M^-1s^-1

R = 3.0×10^6 × 7.0×10^-4 × 5.0×10^-5

= 0.105 M/s

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NikAS [45]

Answer:

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3 years ago

Using the Brønsted-Lowry concept of acids and bases, identify the Brønsted-Lowry acid and base in each of the following reaction

Luden [163]

Answer:

1. HSO³⁻(aq) + H₂O(l) → H₂SO₃(aq) + OH⁻(aq)

<u>The Brønsted-Lowry acid is H₂O and the Brønsted-Lowry base is HSO³⁻</u>

<u />

2. (CH₃)₃N(g) + BCl₃(g) → (CH₃)₃NBCl₃(s)

<u>There are no Brønsted-Lowry acids and bases in this reaction.</u>

Explanation:

According to the Brønsted-Lowry concept, when an acid (HA) and a base (B) undergoes a chemical reaction, the acid (HA) loses a proton and forms its conjugate base (A⁻), whereas the base gains (B) the proton to form its conjugate acid (HB⁺).

<em>The chemical equation for this reaction is:</em>

HA + B ⇌ A⁻ + HB⁺

Given reactions:

1. HSO³⁻(aq) + H₂O(l) → H₂SO₃(aq) + OH⁻(aq)

<u>The Brønsted-Lowry acid is H₂O and the Brønsted-Lowry base is HSO³⁻</u>

Reason: In this reaction, the acid H₂O loses a proton and forms its conjugate base, OH⁻. Whereas, the base HSO³⁻ gains a proton to form its conjugate acid, H₂SO₃.

2. (CH₃)₃N(g) + BCl₃(g) → (CH₃)₃NBCl₃(s)

<u>There are no Brønsted-Lowry acids and bases in this reaction.</u>

Reason: In this reaction, there is no exchange of proton between the acid and the base.

4 0

3 years ago

A mileage test is conducted for a new car model. Thirty randomly selected cars are driven for a month and the mileage is measure

Leni [432]

Answer: $(27.81,\ 29.39)$

Explanation:

Given : Sample size : n= 30 , it means it is a large sample (n≥ 30), so we use z-test .

Significance level : $\alpha: 1-0.95=0.05$

Critical value: $z_{\alpha/2}=1.96$

Sample mean : $\overline{x}=28.6$

Standard deviation : $\sigma=2.2$

The formula to find the confidence interval is given by :-

$\overline{x}\pm z_{\alpha/2}\dfrac{\sigma}{\sqrt{n}}$

i.e. $28.6\pm (1.96)\dfrac{2.2}{\sqrt{30}}$

i.e. $28.6\pm 0.787259889321$

$\approx28.6\pm 0.79=(28.6-0.79,28.6+0.79)=(27.81,\ 29.39)$

Hence, the 95% confidence interval for the mean mpg in the entire population of that car model = $(27.81,\ 29.39)$

4 0

3 years ago

In acidic aqueous solution, the purple complex ion Co(NH3)5Br2+ undergoes a slow reaction in which the bromide ion is replaced b

kirill115 [55]

Answer:

A) 0.065 M is its molarity after a reaction time of 19.0 hour.

B) In 52 hours $[Co(NH_3)5Br]^{2+}$ will react 69% of its initial concentration.

Explanation:

$Co(NH_3)_5(H_2O)_3+[Co(NH_3)5Br]^{2+}(Purple)(aq)+H_2O(l)\rightarrow [Co(NH_3)_5(H_2O)]^{3+}(Pinkish-orange)(aq)+Br^-(aq)$

The reaction is first order in $[Co(NH_3)5Br]^{2+}$ :

Initial concentration of $[Co(NH_3)5Br]^{2+}$ = $[A_o]=0.100 M$

a) Final concentration of $[Co(NH_3)5Br]^{2+}$ after 19.0 hours= $[A]$

t = 19.0 hour = 19.0 × 3600 seconds ( 1 hour = 3600 seconds)

Rate constant of the reaction = k = $6.3\times 10^{-6} s^{-1}$

The integrated law of first order kinetic is given as:

$[A]=[A_o]\times e^{-kt}$

$[A]=0.100 M\times e^{-6.3\times 10^{-6} s^{-1}\times 19.0\times 3600 s}$

$[A]=0.065 M$

0.065 M is its molarity after a reaction time of 19.0 h.

b)

Initial concentration of $[Co(NH_3)5Br]^{2+}$ = $[A_o]=x$

Final concentration of $[Co(NH_3)5Br]^{2+}$ after t = $[A]=(100\%-69\%) x=31\%x=0.31x$

Rate constant of the reaction = k = $6.3\times 10^{-6} s^{-1}$

The integrated law of first order kinetic is given as:

$[A]=[A_o]\times e^{-kt}$

$0.31x=x\times e^{-6.3\times 10^{-6} s^{-1}\t}$

t = 185,902.06 s = $\frac{185,902.06 }{3600} hour = 51.64 hours$ ≈ 52 hours

In 52 hours $[Co(NH_3)5Br]^{2+}$ will react 69% of its initial concentration.

6 0

4 years ago

Are Zombies Living or Non-living Reasoning and Evidence?

Thepotemich [5.8K]

Answer:

I believe that zombies are not living

Explanation:

because zombies don't grow and develop because they are practically mindless. Also, zombies definitely respond to the environment (like when they see a human) but they don't adapt to it

7 0

3 years ago