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3 years ago

Which unit can be used to express the rate of a reaction?

2 answers:

4 0

<span>The rate of reaction may be expressed as a unit of quantity divided by a unit of time. The only expression that has a quantity divided by time is the first one mL / s (i.e. milliliter per second), so the answer is the first option, mL/s.</span><span />

almond37 [142]3 years ago

4 0

Answer: mL / s

Explanation:

1) The rate of reaction is a speed. It is defined as the change in amount per time unit.

2) In form of equation reaction rate = Δ amount / time

3) Usually, it is more interesting to express the reaction rate in terms of the change of concentration, which is

reaction rate = Δ concentration / time

In those cases, the concentration is expressed in molar units per liter, so the rate is expressed as:

M / s = [mole/liter] / s

Which is what we usually find, but it is not mandatory.

4) So, other units can be used, like moles or volume (even pressure) to express the amount of a spices.

5) That is why rate of reation can be expressed as mL / s (mililiters per second).

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Which plants do siamangs rely on most as a source of their food

sp2606 [1]

They seem to eat a lot ranging from woody plants to vines, though in the rainforest most of their diet is made up of Malaysia and plant material, to be more specific

So I would go with leaves, 50%

Flowers, buds, and insects, 10%

Fruits, 40%

--They also really like figs

3 0

3 years ago

Be sure to answer all parts. Consider the reaction A + B → Products From the following data obtained at a certain temperature, d

worty [1.4K]

Answer : The order of reaction with respect to A is, first order reaction.

The order of reaction with respect to B is, zero order reaction.

The overall order of reaction is, first order reaction.

Explanation :

Rate law is defined as the expression which expresses the rate of the reaction in terms of molar concentration of the reactants with each term raised to the power their stoichiometric coefficient of that reactant in the balanced chemical equation.

For the given chemical equation:

$A+B\rightarrow Products$

Rate law expression for the reaction:

$\text{Rate}=k[A]^a[B]^b$

where,

a = order with respect to A

b = order with respect to B

Expression for rate law for first observation:

$3.20\times 10^{-1}=k(1.50)^a(1.50)^b$ ....(1)

Expression for rate law for second observation:

$3.20\times 10^{-1}=k(1.50)^a(2.50)^b$ ....(2)

Expression for rate law for third observation:

$6.40\times 10^{-1}=k(3.00)^a(1.50)^b$ ....(3)

Dividing 1 from 2, we get:

$\frac{3.20\times 10^{-1}}{3.20\times 10^{-1}}=\frac{k(1.50)^a(2.50)^b}{k(1.50)^a(1.50)^b}\\\\1=1.66^b\\b=0$

Dividing 1 from 3, we get:

$\frac{6.40\times 10^{-1}}{3.20\times 10^{-1}}=\frac{k(3.00)^a(1.50)^b}{k(1.50)^a(1.50)^b}\\\\2=2^a\\a=1$

Thus, the rate law becomes:

$\text{Rate}=k[A]^1[B]^0$

$\text{Rate}=k[A]$

Thus,

The order of reaction with respect to A is, first order reaction.

The order of reaction with respect to B is, zero order reaction.

The overall order of reaction is, first order reaction.

7 0

3 years ago

A 25.0-mL sample containing Cu2+ gave an instrument signal of 25.2 units (corrected for a blank). When exactly 0.500 mL of 0.027

irga5000 [103]

Answer:

The molar concentration of Cu²⁺ in the initial solution is 6.964x10⁻⁴ M.

Explanation:

The first step to solving this problem is calculating the number of moles of Cu(NO₃)₂ added to the solution:

$C = \frac{n}{V} \\0.0275 = \frac{n}{0.0005} \\$

n = 1.375x10⁻⁵ mol

The second step is relating the number of moles to the signal. We know the the n calculated before is equivalent to a signal increase of 19.9 units (45.1-25.2):

1.375x10⁻⁵ mol _________ 19.9 units

x _________ 25.2 units

x = 1.741x10⁻⁵mol

Finally, we can calculate the Cu²⁺ concentration :

C = 1.741x10⁻⁵mol / 0.025 L

C = 6.964x10⁻⁴ M

7 0

3 years ago

Sulfonation of benzene has the following mechanism: (1) 2 H2SO4 ⇌ H3O+ + HSO4− + SO3 [fast] (2) SO3 + C6H6 → H(C6H5+)SO3− [slow]

ziro4ka [17]

Question is incomplete, complete question is as follows :

Complete Question : .Sulfonation of benzene has the following mechanism:

(1) 2 H2SO4 ⇌ H3O+ + HSO4− + SO3

[fast]

(2) SO3 + C6H6 → H(C6H5+)SO3−

[slow]

(3) H(C6H5+)SO3− + HSO4− → C6H5SO3− + H2SO4

[fast]

(4) C6H5SO3− + H3O+ → C6H5SO3H + H2O

[fast]

write the overall rate law for the initial rate of the reaction as a fraction.

Rate=k(________/_________)

Answer:

The overall rate law for the initial reaction is = $k_{overall} [H_{2}SO_{4}]^{2} [C_{6}H_{6}]$

Explanation :

Frist of all, all the common terms are cancelled out and written the overall reaction.

As we know that the rate depednant step is the slowest step of the reaction, rate law is :

rate = $k_{2} [SO_{3}][C_{6}H_{6}]$

But the problem is that SO3 cannot be written in the overall rate law because it is an intermediate.

Rate law for synthesis of S03 is as follows :

rate = $k_{1}[H_{2}SO_{4}]^{2}$

Hence when we substitute equation 2 in equation one,

Rate comes out to be = $k_{overall} [H_{2}SO_{4}]^{2} [C_{6}H_{6}]$

6 0

3 years ago

1. Which statement describes the properties of both elements that have just one valence electron and elements that have seven va

FromTheMoon [43]

Q1. They are highly reactive. Q2. High reactivity, nonmetallic. Q3. Oxygen has an ion charge of -2. Q4. LiCl I believe. Q5. How electrons are shared. Q6 1. Q7. Share 2 valence electrons, I believe.

3 0

3 years ago

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