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3 years ago

why is it easier for water to flow through rock and sediment with larger pores than through rock and sediment with smaller pores

Chemistry

2 answers:

andrey2020 [161]3 years ago

6 0

Answer:

Because of its permeability.

Explanation:

It is on quizlet. I don't think it is right though, I am sorry.

joja [24]3 years ago

6 0

Answer:

Larger pores mean less friction between flowing water and the sides of the pores. Smaller pores mean more friction between flowing water and pore walls and more twists and turns for the water to navigate. A permeable material has a greater number of larger, well-connected pore spaces. An impermeable material has fewer, smaller pores that are poorly connected.

Explanation:

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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

Which form of matter has no definite shape and no definite volume

Sonbull [250]

Answer:

Gas has no definite shape and volume

4 0

3 years ago

How many moles of carbon are in 3.7 moles of c8h11no2

sleet_krkn [62]

Answer:- 29.6 moles of carbon.

Solution:- We have been given with 3.7 moles of $C_8H_11NO_2$ and asked to calculate the moles of C.

Looking at the formula of the compound, there are 8 carbons in it means 1 mol of he compound has 8 moles of C. So, if we multiply the given moles of the compound by 8 then we get the moles of C.

$3.7molC_8H_11NO_2(\frac{8molC}{1molC_8H_11NO_2})$