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a_sh-v [17]
3 years ago
11

Define the term mole in terms of Avogardo's constant

Chemistry
1 answer:
lilavasa [31]3 years ago
3 0

Explanation: One mole of a substance is equal to 6.022 × 10²³ units of that substance (such as atoms, molecules, or ions). The number 6.022 × 10²³ is known as Avogadro's number or Avogadro's constant. The concept of the mole can be used to convert between mass and number of particles.. Created by Sal Khan.

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The common Eastern mole, a mammal, typically has a mass of 83 g, which corresponds to about 8.3 moles of atoms. (A mole of atoms
Alchen [17]

Answer:

1.6605 × 10⁻²³ g / atom , 9.99 amu

Explanation:

Atoms present in 1 mole = 6.022 × 10²³ atoms

Given that

Moles of common Eastern mole = 8.3 moles

So,

Atoms in 8.3 moles are:

8.3 moles = 6.022 × 10²³ × 8.3 atoms

Also,

Average mass = Mass of the sample / Number of atoms

Mass of sample = 83 g

Average mass = 83 g / 6.022 × 10²³ × 8.3 atoms = 1.6605 × 10⁻²³ g / atom

In amu :

1 amu = 1.66 × 10⁻²⁴ g / atom

So,

Average mass = 1.6605 × 10⁻²³ / 1.66 × 10⁻²⁴ amu = 9.99 amu

6 0
3 years ago
You can see, smell, and taste microorganisms.*<br> True<br> False
ludmilkaskok [199]
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8 0
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USPshnik [31]
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Which is an example of chemical change?
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4 0
3 years ago
Consider the following reaction at a high temperature. Br2(g) ⇆ 2Br(g) When 1.35 moles of Br2 are put in a 0.780−L flask, 3.60 p
UNO [17]

Answer : The equilibrium constant K_c for the reaction is, 0.1133

Explanation :

First we have to calculate the concentration of Br_2.

\text{Concentration of }Br_2=\frac{\text{Moles of }Br_2}{\text{Volume of solution}}

\text{Concentration of }Br_2=\frac{1.35moles}{0.780L}=1.731M

Now we have to calculate the dissociated concentration of Br_2.

The balanced equilibrium reaction is,

                              Br_2(g)\rightleftharpoons 2Br(aq)

Initial conc.         1.731 M      0

At eqm. conc.      (1.731-x)    (2x) M

As we are given,

The percent of dissociation of Br_2 = \alpha = 1.2 %

So, the dissociate concentration of Br_2 = C\alpha=1.731M\times \frac{1.2}{100}=0.2077M

The value of x = 0.2077 M

Now we have to calculate the concentration of Br_2\text{ and }Br at equilibrium.

Concentration of Br_2 = 1.731 - x  = 1.731 - 0.2077 = 1.5233 M

Concentration of Br = 2x = 2 × 0.2077 = 0.4154 M

Now we have to calculate the equilibrium constant for the reaction.

The expression of equilibrium constant for the reaction will be :

K_c=\frac{[Br]^2}{[Br_2]}

Now put all the values in this expression, we get :

K_c=\frac{(0.4154)^2}{1.5233}=0.1133

Therefore, the equilibrium constant K_c for the reaction is, 0.1133

7 0
3 years ago
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