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2 years ago

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"The combustion of ethylene proceeds by the reaction C2H4(g) + 3O2(g) → 2CO2(g) + 2H2O(g) When the rate of disappearance of O2 i

s 0.23 M s-1, the rate of disappearance of C2H4 is ________ M s-1."

1 answer:

Julli [10]2 years ago

7 0

Answer:

The rate of disappearance of $C_2H_4$ is 0.0766 M/s.

Explanation:

$C_2H_4(g)+3O_2(g)\rightarrow 2CO_2(g)+2H_2O(g)$

Rate of the reaction = R

$R=-\frac{1}{1}\frac{d[C_2H_4]}{dt}=-\frac{1}{3}\frac{d[O_2]}{dt}$

[te]R=\frac{1}{2}\frac{d[CO_2]}{dt}=\frac{1}{2}\frac{d[H_2O]}{dt}[/tex]

Rate of disappearance of $O_2$ =0.23 M/s

Rate of disappearance of $O_2=3\times R$

$0.23 M/s=3\times R$

R =0.07666 M/s

Rate of disappearance of $C_2H_4=1\times R$

$\frac{d[C_2H_4]}{dt}=1\times 0.23 m/s=0.0766 M/s$

The rate of disappearance of $C_2H_4$ is 0.0766 M/s.

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The MSDS for chloroform indicates that it is a clear liquid that has a pleasant smell and substantial vapor pressure. People sho

aleksandr82 [10.1K]

Answer:C.He should locate the chloroform stored in a dark container in chemical storage and should take it to the fume hood to pour.

Explanation:

The statement of the question clearly states that chloroform is sensitive to light and it's vapour is toxic.

If a substance is sensitive to light, then it must be stored in a dark bottle. This is because. If a substance that is sensitive to light is stored in a transparent container, it may be decomposed by light.

Being a substance whose fumes are toxic, Malik should pour the liquid in a fumes hood so that he does not inhale the fumes.

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2 years ago

An atom of lead has a radius of and the average orbital speed of the electrons in it is about . Calculate the least possible unc

victus00 [196]

The question is incomplete. Here is the complete question.

An atom of lead has a radius of 154 pm and the average orbitalspeed of the electron in it is about 1.8x $10^{8}$ m/s. Calculate the least possible uncertainty in a measurement of the speed of an electron in an atom of lead. Write your answer as a percentage of the average speed, and round it to significant 2 digits.

Answer: v% = 0.21 m/s

Explanation: To calculate the uncertainty, use <u>Heisenberg's Uncertainty Principle</u>, which states that: ΔpΔx≥ $\frac{h}{4\pi }$

where h is <u>Planck's constant</u> and it is equal to 6.626. $10^{-34}$ m²kg/s.

Since p (momentum) is p = m.v:

mΔv.Δx ≥ $\frac{h}{4\pi }$

Δv = $\frac{h}{4\pi.x.m }$

Given that: r = x = 1.54. $10^{-10}$ m and mass of an electron is m=9.1. $10^{-31}$ kg

Δv = $\frac{6.626.10^{-34} }{4.3.14.1.54.10^{-10}.9.1.10^{-31}}$

Δv = 0.0376. $10^{7}$

As percentage of average speed:

Δv. $\frac{1}{v}$ .100% = $\frac{0.0376.10^{7} }{1.8.10^{8} }$ .10² = 0.021.10 = 0.21%

The least possible uncertainty in a speed of an electron is 0.21%.

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2 years ago

In each row, check the box under the compound that can reasonably be expected to be more acidic in aqueous solution, e.g. have t

vredina [299]

Answer:

$HCH_{3}SO_{2}$

$H_{3}PO_{3}$

$HClO_{2}$

Explanation:

Every acid (HA) tends to disolve into proton ( $H^{+}$ ) and anion ( $A^{-}$ ) in aqueous solution. Acid strength can be determined by measuring this tendency to separate into proton an anion. Strength of an acid can be quantified by its acid dissociation value - Ka. A strong acid will have a tendency to easily release proton and will have larger Ka value and smaller logarithmic value (pKa = - logKa) similar to calculating pH of the solution. So the easiest way to resolve this issue is by looking for Ka or pKa value of the acid (This table may be useful in more complex tasks and is attached below). However, stronger acid can be determined elsehow.

a) Carbon is element 14 with 4 valent electrons and sulfur is element 16 with 6 valence electrons. Thus, sulfur has stronger electronegativity (tendency to attract bonded electrons towards itself). This means that sulfur will hold oxygen tighter to itself so the hydrogen bond to it can be more easily separated from it. $HCH_{3}SO_{2}$ is more acidic in aqueous solution.

b) In $H_{3}PO_{4}$ , phosphorus holds one double bond with oxygen and three OH group equally. To show an acidic tendency, phosphorus would need to let go one hydrogen out of one of OH groups. In $H_{3}PO_{3}$ , phosporus holds two double bong with oxygen, one OH and one hydrogen, all single and lonely, ready to leave phosphorus and show acidic characteristics in aqueous solution. Thus, $H_{3}PO_{3}$ is more acidic compound.

C) In all Cl acids, the electron density is placed around Cl so the more oxygen around Cl, the more acidic will be the chemical. This is comparable to an oxidation state - the bigger oxidation state, the stronger acid will be:

$HClO_{4} ^{+7} >HClO_{3}^{+5} >HClO_{2}^{+3} >HClO_{}^{+1}$

$HClO_{2}$ can reasonably be expected to be more acidic in aqueous solution.

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2 years ago

Determine the mass in grams of 3.00 × 10²¹ atoms of arsenic. (The mass of one mole of arsenic is 74.92 g.)

Tpy6a [65]

Answer:

The answer to your question is: 0.373 g

Explanation:

Data

mass = ? g

atoms = 3 x 10 ²¹

AM = 74.92 g

Process

1 mol of As ------------------ 6.023 x 10²³ atoms

x ------------------ 3 x 10 ²¹ atoms

x = 4.98 x 10⁻³ moles

1 mol ------------------------ 74.92 g

4.98 x 10⁻³ moles------- x

x = (4.98 x 10⁻³ x 74.92)/1

x = 0.373 g of As

3 0

2 years ago

Uranium–232 has a half–life of 68.9 years. A sample from 206.7 years ago contains 1.40 g of uranium–232. How much uranium was or

givi [52]

<u>Answer:</u> The initial amount of Uranium-232 present is 11.3 grams.

<u>Explanation:</u>

All the radioactive reactions follows first order kinetics.

The equation used to calculate half life for first order kinetics:

$t_{1/2}=\frac{0.693}{k}$

We are given:

$t_{1/2}=68.9yrs$

Putting values in above equation, we get:

$k=\frac{0.693}{68.9}=0.0101yr^{-1}$

Rate law expression for first order kinetics is given by the equation:

$k=\frac{2.303}{t}\log\frac{[A_o]}{[A]}$

where,

k = rate constant = $0.0101yr^{-1}$

t = time taken for decay process = 206.7 yrs

$[A_o]$ = initial amount of the reactant = ?

[A] = amount left after decay process = 1.40 g

Putting values in above equation, we get:

$0.0101yr^{-1}=\frac{2.303}{206.7yrs}\log\frac{[A_o]}{1.40}$

$[A_o]=11.3g$

Hence, the initial amount of Uranium-232 present is 11.3 grams.

4 0

1 year ago

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