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3 years ago

What distinguishes effusion from diffision? How are these processes similar?

Chemistry

1 answer:

zvonat [6]3 years ago

7 0

Answer:

In effusion, a substance escapes through a tiny pinhole, or other hole whereas diffusion is the spreading out of a substance within a dispersing medium.

Effusion and diffusion are similar in terms of rates and speeds.

Explanation:

According to effusion, if there is a small hole in a container, a gas will leak from it. So, in this process a substance escapes through a tiny pinhole, or other hole.

According to diffusion, the two solutions will reach the lowest possible concentration of both, if they are combined and not mixed. So, diffusion is the spreading out of a substance within a dispersing medium.

Effusion and diffusion are similar in terms of rates and speeds.

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A generator does work on an electric heater by forcing an electric current through it. Suppose 1 kJ of work is done on the heate

azamat

Answer:

The change in internal energy of the heater is 0 kJ

Explanation:

∆U = Q - W

Q is quantity of heat transferred = 1 kJ

W is work done on the heater = 1 kJ

Change in internal energy (∆U) = 1 - 1 = 0 kJ

3 0

3 years ago

Read 2 more answers

A student sets up an experiment according to the table below. What question was the student most likely investigating?

user100 [1]

Answer:

Explanation:

7 0

3 years ago

The equation for the reaction between copper and nitric acid is vCu + wHNO3--->xCu(NO3)2 + yNO + zH2O Balance the equation an

lana [24]

Answer:

The coefficients of the balanced equation are:

v= 1
w= 4
x= 1
y= 2
z=2

Explanation:

The Law of Conservation of Matter, also called the Law of Conservation of Mass or the Lomonósov-Lavoisier Law, postulates that "mass is neither created nor destroyed, it is only transformed". This means that the reagents interact with each other and form new products with different physical and chemical properties than the reagents, but the amount of matter or mass before and after a transformation (chemical reaction) is always the same. In other words, then the mass before the chemical reaction is equal to the mass after the reaction.

Then, you must balance the following chemical equation:

v Cu + w HNO₃ → x Cu(NO₃)₂ + y NO + z H₂O

For that, you must first look at the subscripts next to each atom to find the number of atoms in the equation. If the same atom appears in more than one molecule, you must add its amounts

Left side: 1 copper, 1 hydrogen, 1 nitrogen and 3 oxygen.
Right side: 1 copper, 2 hydrogen, 3 nitrogen and 8 oxygen.

The coefficients located in front of each molecule indicate the amount of each molecule for the reaction. This coefficient can be modified to balance the equation, just as you should never alter the subscripts.

By multiplying the coefficient mentioned by the subscript, you get the amount of each element present in the reaction.

Generally, hydrogen and oxygen balance in the end. So you balance nitrogen first, because copper is already balanced (there is the same amount on both sides of the reaction). If w = 2, then:

Left side: 1 copper, 2 hydrogen, 2 nitrogen and 6 oxygen.
Right side: 1 copper, 2 hydrogen, 3 nitrogen and 8 oxygen.

But you see then that the oxygen is unbalanced and you have less quantity in the reagents. Then w must be greater. Being w = 4:

Left side: 1 copper, 4 hydrogen, 4 nitrogen and 10 oxygen.
Right side: 1 copper, 2 hydrogen, 3 nitrogen and 8 oxygen.

Going back to the idea of balancing nitrogen, being y = 2:

Left side: 1 copper, 4 hydrogen, 4 nitrogen and 10 oxygen.
Right side: 1 copper, 2 hydrogen, 4 nitrogen and 9 oxygen.

Balancing the hydrogen, being z = 2:

Left side: 1 copper, 4 hydrogen, 4 nitrogen and 10 oxygen.
Right side: 1 copper, 4 hydrogen, 4 nitrogen and 10 oxygen.

Since you have the same amount of each element on each side of the reaction, the reaction is balanced. Then, the balanced equation is:

Cu + 4 HNO₃ → Cu(NO₃)₂ + 2 NO + 2 H₂O

Finally:

v= 1
w= 4
x= 1
y= 2
z=2

6 0

3 years ago

Given the unbalanced equation below, answer the following: Calculate the number of liters of 3.00 M lead (II) iodide solution pr

mr_godi [17]

The number of liters of 3.00 M lead (II) iodide : 0.277 L

<h3>Further explanation</h3>

Reaction(balanced)

Pb(NO₃)₂(aq) + 2KI(aq) → 2KNO₃(aq) + PbI₂(s)

moles of KI = 1.66

From the equation, mol ratio of KI : PbI₂ = 2 : 1, so mol PbI₂ :

$\tt \dfrac{1}{2}\times 1.66=0.83$

Molarity shows the number of moles of solute in every 1 liter of solute or mmol in each ml of solution

$\large \boxed {\bold {M ~ = ~ \dfrac {n} {V}}}$

Where

M = Molarity

n = Number of moles of solute

V = Volume of solution

So the number of liters(V) of 3.00 M lead (II) iodide-PbI₂ (n=0.83, M=3):

$\tt V=\dfrac{n}{M}\\\\V=\dfrac{0.83}{3}\\\\V=0.277~L$

5 0

3 years ago

From the list of the instructions and tips below, select the statement that would NOT lead to a successful recrystallization O D

Pachacha [2.7K]

Answer:

C. Precipitate the crystals as fast as possible

Explanation:

C. Precipitate the crystals as fast as possible

Precipitation if done fast , can lead to the formation of impure crystals , as impurities also get stuck inside the crystals. So it must be done slowly to obtain the pure crystals . Slower the crystals form , purer they are .

All the other options can lead to a successful recrystallization.

5 0

3 years ago