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Vedmedyk [2.9K]

2 years ago

8

In a labratory activity the density of a sample of vanadium is determined to be 6.9g/cm^3 at room temperature. what is the perce

nt error for the determined value?
answers:
a : 0.15%
b : 0.87%
c : 13%
d : 15%

PLEASE EXPLAIN! please

1 answer:

Snowcat [4.5K]2 years ago

7 0

Answer:

c : 13%

Explanation:

Data Give:

Experimental density of vanadium = 6.9 g/cm³

percent error = ?

Solution:

Formula used to calculate % error

% error = [experimental value -accepted value/accepted value] x 100

The reported accepted density value for vanadium = 6.11 g/cm³

Put value in the above equation

% error = [ 6.9 - 6.11 / 6.11 ] x 100

% error = [ 0.79 / 6.11 ] x 100

% error = [ 0.129] x 100

% error = 12.9

Round to the 2 significant figure

% error = 13 %

So, option c is correct

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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

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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 $Cl_2$ :

$P_2'=26.398 Torr$

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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 $PCl_5=P_3'=217.0 Torr$

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

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 $PCl_5$ :

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

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Answer:

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Step 1: Data given

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Mass of benzene = 40.00 mL * 0.87865 g/mL = 35.146 g

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Number of moles of CCl4 = 95.64 grams / 154g/mol = 0.62104 mol

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Total vapor pressure = 31.37 + 52.92 = 84.29 mmHg

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