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3 years ago

RATE LAW QUESTION !

Chemistry

1 answer:

vivado [14]3 years ago

6 0

In general, we have this rate law express.:

$\mathrm{Rate} = k \cdot [A]^x [B]^y$
we need to find x and y

ignore the given overall chemical reaction equation as we only preduct rate law from mechanism (not given to us).

then we go to compare two experiments in which only one concentration is changed

compare experiments 1 and 4 to find the effect of changing [B]
divide the larger [B] (experiment 4) by the smaller [B] (experiment 1) and call it Δ[B]

Δ[B]= 0.3 / 0.1 = 3

now divide experiment 4 by experient 1 for the given reaction rates, calling it ΔRate:

ΔRate = 1.7 × 10⁻⁵ / 5.5 × 10⁻⁶ = 34/11 = 3.090909...

solve for y in the equation $\Delta \mathrm{Rate} = \Delta [B]^y$

$3.09 = (3)^y \implies y \approx 1$

To this point, $\mathrm{Rate} = k \cdot [A]^x [B]^1$

do the same to find x.
choose two experiments in which only the concentration of B is unchanged:

Dividing experiment 3 by experiment 2:
Δ[A] = 0.4 / 0.2 = 2
ΔRate = 8.8 × 10⁻⁵ / 2.2 × 10⁻⁵ = 4

solve for x for $\Delta \mathrm{Rate} = \Delta [A]^x$

$4= (2)^x \implies x = 2$

the rate law is

Rate = k·[A]²[B]

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What’s the relationship between atomic radius and ease of oxidation on the activity series.

user100 [1]

The bigger the atomic radius the easier it is to oxidise the atom. Remember that an atom is oxidized by the loss of an electron.

Explanation:

The bigger the atomic radius the further away the valence electron are from the attractive force of the atomic nucleus. This means that the energy required to remove an electron from the valence shell is easier compared to an atom with a smaller atomic radius. This is because you need to overcome the attractive force of the nucleus on the electron for you to oxidize the atom.

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3 years ago

The half life of 226/88 Ra is 1620 years. How much of a 12 g sample of 226/88 Ra will be left after 8 half lives?

Aloiza [94]

Answer:

0.0468 g.

Explanation:

The decay of radioactive elements obeys first-order kinetics.

For a first-order reaction: k = ln2/(t1/2) = 0.693/(t1/2).

Where, k is the rate constant of the reaction.

t1/2 is the half-life time of the reaction (t1/2 = 1620 years).

∴ k = ln2/(t1/2) = 0.693/(1620 years) = 4.28 x 10⁻⁴ year⁻¹.

For first-order reaction: <em>kt = lna/(a-x).</em>

where, k is the rate constant of the reaction (k = 4.28 x 10⁻⁴ year⁻¹).

t is the time of the reaction (t = t1/2 x 8 = 1620 years x 8 = 12960 year).

a is the initial concentration (a = 12.0 g).

(a-x) is the remaining concentration.

∴ kt = lna/(a-x)

(4.28 x 10⁻⁴ year⁻¹)(12960 year) = ln(12)/(a-x).

5.54688 = ln(12)/(a-x).

Taking e for the both sides:

256.34 = (12)/(a-x).

8 0

3 years ago

belka [17]

Answer:

3. Inverse 1. Direct

Explanation:

P- pressure

V - volume

T - temperature

P1*V1 / T1 = P2*V2 / T2 ...... (1)

That's the general gas law with the combined ideas of charles, boyle & lussac.

Whenever you are restricted as "constant" temperature, volume, or pressure...cancel them off of your equation.

in this case 3. is indirectly telling us to cancel the temperature (T).

so we'll be left w P1*V1 = P2*V2

now notice that any relation ship that is multiplied like the one above consists of inversely related quantities. & so we conclude that-

P & V are inversely proportional or have an inverse relationship.

similarly in 1. we'll cancel p off of the general formula (1)

to be left with V1/T1 = V2/T2

also note that quantities involved in division are directly related to each other & hence the answer.

5 0

2 years ago

A sample of 3.62 moles of diphosphorous trioxide is

Sindrei [870]

$H_3PO_3$ $2H_3PO_3$ $2H_3PO_3$ $P_2O_3$ Answer:

Explanation:

This question is about stoichiometry. From the balanced equation $P_2O_3 + 3H_2O$ ⇒ $2H_3PO_3$ , we see that 3 moles of water is needed to react with 1 mole of $P_2O_3$ .

This means that, to fully react 3.62 moles of $P_2O_3$ , we would need 3*3.62 or 10.86 moles of water. However, we only have 6.31 moles, so water is the limiting reactant.

Since 3 moles of water react with 1 mole of $P_2O_3$ , 6.31 moles of water can fully react with 6.31÷3 or 2.1033 moles of $P_2O_3$ .

From the balanced equation, we see that every mole of $P_2O_3$ reacted gets you 2 moles of $2H_3PO_3$ . Therefore, 2.1033 moles of $P_2O_3$ would give you approximately 4.21 moles of $H_3PO_3$ .

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2 years ago

A 46.9 gram sample of a substance has a volume of about 3.5 centimeters3. It is solid at a room temperature of 23ºC. Out of the

Masteriza [31]

Answer:- C. Hafnium.

Solution:- Mass of the sample is 46.0 g and it's volume is $3.5cm^3$ .

From mass and volume, we can calculate it's density using the formula:

$density=\frac{mass}{volume}$

$density=\frac{46.9g}{3.5cm^3}$

$density=\frac{13.4g}{cm^3}$

On the basis of the density, this substance could either be mercury or hafnium. Since the substance is a solid at room temperature where as mercury is liquid. So, it can't be mercury.

The right choice is C) Hafnium.

3 0

3 years ago