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3 years ago

List of things that you’ll use in the scientific method

Chemistry

1 answer:

MrRissso [65]3 years ago

8 0

Explanation:

Make an observation

ask a question

form a hyposisese

make a prediction

test the prediction

use the results to fix the prediction and hypotheses

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Some SbCl5 is allowed to dissociate into SbCl3 and Cl2 at 521 K. At equilibrium, [SbCl5] = 0.195 M, and [SbCl3] = [Cl2] = 6.98×1

Brilliant_brown [7]

Answer:

a) The equilibrium will shift in the right direction.

b) The new equilibrium concentrations after reestablishment of the equilibrium :

$[SbCl_5]=(0.370-x) M=(0.370-0.0233) M=0.3467 M$

$[SbCl_3]=(6.98\times 10^{-2}+x) M=(6.98\times 10^{-2}+0.0233) M=0.0931 M$

$[Cl_2]=(6.98\times 10^{-2}+x) M=(6.98\times 10^{-2}+0.0233) M=0.0931 M$

Explanation:

$SbCl_5(g)\rightleftharpoons SbCl_3(g) + Cl_2(g)$

a) Any change in the equilibrium is studied on the basis of Le-Chatelier's principle.

This principle states that if there is any change in the variables of the reaction, the equilibrium will shift in the direction to minimize the effect.

On increase in amount of reactant

$SbCl_5(g)\rightleftharpoons SbCl_3(g) + Cl_2(g)$

If the reactant is increased, according to the Le-Chatlier's principle, the equilibrium will shift in the direction where more product formation is taking place. As the number of moles of $SbCl_5$ is increasing .So, the equilibrium will shift in the right direction.

$SbCl_5(g)\rightleftharpoons SbCl_3(g) + Cl_2(g)$

Concentration of $SbCl_5$ = 0.195 M

Concentration of $SbCl_3$ = $6.98\times 10^{-2} M$

Concentration of $Cl_2$ = $6.98\times 10^{-2} M$

On adding more $[SbCl_5$ to 0.370 M at equilibrium :

$SbCl_5(g)\rightleftharpoons SbCl_3(g) + Cl_2(g)$

Initially

0.370 M $6.98\times 10^{-2}M$

At equilibrium:

(0.370-x)M $(6.98\times 10^{-2}+x)M$

The equilibrium constant of the reaction = $K_c$

$K_c=2.50\times 10^{-2}$

The equilibrium expression is given as:

$K_c=\frac{[SbCl_3][Cl_2]}{[SbCl_5]}$

$2.50\times 10^{-2}=\frac{(6.98\times 10^{-2}+x)M\times (6.98\times 10^{-2}+x)M}{(0.370-x) M}$

On solving for x:

x = 0.0233 M

The new equilibrium concentrations after reestablishment of the equilibrium :

$[SbCl_5]=(0.370-x) M=(0.370-0.0233) M=0.3467 M$

$[SbCl_3]=(6.98\times 10^{-2}+x) M=(6.98\times 10^{-2}+0.0233) M=0.0931 M$

$[Cl_2]=(6.98\times 10^{-2}+x) M=(6.98\times 10^{-2}+0.0233) M=0.0931 M$

3 0

3 years ago

Which of these answers best describe a fossil ?

Mekhanik [1.2K]

You would have to show me the answers

3 0

3 years ago

For a half-reaction in an acidic solution, which substances should you add to balance the equation?

solmaris [256]

Hello!

I believe the correct answer to this question is H+ and H2O.

I hope you found this helpful! :)

8 0

3 years ago

mixture or pure substance: 1.blood 2.dyes 3.self-raising flour 4.muesli 5.copper wire 6.distilled water 7.table salt 8.milk 9.br

jenyasd209 [6]

Answer: Mixture: Blood , Self raising flour,muesli ,dyes, milk, tea, air, bronze

Pure substance: Copper wire, distilled water, table salt, oxygen.

Explanation:

Mixture is a substance which is made up two or more number of compounds which chemically inactive and retain their distinct chemical properties.

Blood , Self raising flour,muesli ,dyes, milk, tea, air, bronze

Pure substance is defined as anything with uniform and unchanging composition is known s pure substance.

Copper wire, distilled water, table salt, oxygen.

5 0

3 years ago

If the formula mass of one molecule is X amu, the molar mass is X g/mol. Group of answer choices True False

Irina-Kira [14]

yes

Explanation:

the molar mass of a compound is g/mol

8 0

2 years ago