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Svetllana [295]
3 years ago
7

The natural abundance of 2h is 0.015% and that of 18o is 0.20%. How many 2h218o molecules are there in 1.0 moles of water?

Chemistry
1 answer:
Ulleksa [173]3 years ago
6 0

Answer:

\boxed{2.7 \times 10^{13}}

Explanation:

Probability of one ²H is 0.00015 = 1.5× 10⁻⁴

Probability of two ²H is 1.5× 10⁻⁴ × 1.5 × 10⁻⁴ = 2.25 × 10⁻⁸

Probability of two ²H plus one ¹⁸O is 2.25 × 10⁻⁸ × 0.0020

= 4.5× 10⁻¹¹

Molecules in 1.0 mol water = 1.0 × 6.022 × 10²³ × 4.5 × 10¹¹

= 2.7 × 10¹³

There are 2.7 × 10¹³ molecules of  ²H₂¹⁸O in 1.0 mol of water.

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How many neutrons does an element have if its atomic number is 41 and its mass number is 170?
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atomic number = protons.

protons+neutrons=atomic mass.

170 (mass) - 41 (proton/number) = 129

that should be ur answer. I hope this helps!

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How many valence electrons does each of the following atoms have?
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Which term refers to the process by which ions that have entered solution are kept in solution
saw5 [17]

Answer:

dissociation. *electrolyte. hydration. ionization.

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What is the first thing you must do to solve a stoichiometry problem
belka [17]

Answer:

the first step in any stoichiometric problem is to always ensure that the chemical reaction you are dealing with is balanced,

Explanation:

3 0
2 years ago
An equilibrium mixture of PCl5(g), PCl3(g), and Cl2(g) has partial pressures of 217.0 Torr, 13.2 Torr, and 13.2 Torr, respective
djverab [1.8K]

Answer:

The new partial pressures after equilibrium is reestablished for PCl_3:

P_1'=6.798 Torr

The new partial pressures after equilibrium is reestablished Cl_2:

P_2'=26.398 Torr

The new partial pressures after equilibrium is reestablished for PCl_5:

P_3'=223.402 Torr

Explanation:

PCl_3(g) + Cl_2(g)\rightleftharpoons PCl_5(g)

At equilibrium before adding chlorine gas:

Partial pressure of the PCl_3=P_1=13.2 Torr

Partial pressure of the Cl_2=P_2=13.2 Torr

Partial pressure of the PCl_5=P_3=217.0 Torr

The expression of an equilibrium constant is given by :

K_p=\frac{P_1}{P_1\times P_2}

=\frac{217.0 Torr}{13.2 Torr\times 13.2 Torr}=1.245

At equilibrium after adding chlorine gas:

Partial pressure of the PCl_3=P_1'=13.2 Torr

Partial pressure of the Cl_2=P_2'=?

Partial pressure of the PCl_5=P_3'=217.0 Torr

Total pressure of the system = P = 263.0 Torr

P=P_1'+P_2'+P_3'

263.0Torr=13.2 Torr+P_2'+217.0 Torr

P_2'=32.8 Torr

PCl_3(g) + Cl_2(g)\rightleftharpoons PCl_5(g)

At initail

(13.2) Torr     (32.8) Torr                        (13.2) Torr

At equilbriumm

(13.2-x) Torr     (32.8-x) Torr                        (217.0+x) Torr

K_p=\frac{P_3'}{P_1'\times P_2'}

1.245=\frac{(217.0+x)}{(13.2-x)(32.8-x)}

Solving for x;

x = 6.402 Torr

The new partial pressures after equilibrium is reestablished for PCl_3:

P_1'=(13.2-x) Torr=(13.2-6.402) Torr=6.798 Torr

The new partial pressures after equilibrium is reestablished Cl_2:

P_2'=(32.8-x) Torr=(32.8-6.402) Torr=26.398 Torr

The new partial pressures after equilibrium is reestablished for PCl_5:

P_3'=(217.0+x) Torr=(217+6.402) Torr=223.402 Torr

5 0
3 years ago
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