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2 years ago

72. how is it possible to recognize the end point of a titration?

1 answer:

3 0

When the solution used begins to change color and stays the new color for at least 30 seconds we are able to tell that the solution has reached the end point of titration

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What type of scientific inquiry is friction ridge analysis

Ilia_Sergeevich [38]

Answer:

Towards this goal, this project aims to develop a statistical measure of the uncertainty of the decisions made on the friction ridge evidence (i.e., evidential value of fingerprint comparison), which ultimately can be referred to as a scientific basis of the identification decisions made in friction ridge analysis.

Explanation:

8 0

3 years ago

Read 2 more answers

Calcium has an atomic number of 20. How many electrons does calcium lose when forming an ionic bond with another ion?

oksano4ka [1.4K]

2 electrons. it has electron configuration of 2.8.8.2 hence the last 2 will be lost for it to gain stable.
if you need more explanation on periodic table you can reach me at [email protected]

5 0

3 years ago

Calculate the pH of the solutions: [H^+]= 1.6 x 10^-3 M

Yuliya22 [10]

Answer:

A) pH = 2.8

B) pH = 5.5

C) pH = 8.9

D) pH = 13.72

Explanation:

a) [H⁺] = 1.6 × 10⁻³ M

pH = -log [H⁺]

pH = -log [1.6 × 10⁻³ ]

pH = 2.8

b) [H⁺] = 3 × 10⁻⁶

pH = -log [H⁺]

pH = -log [3 × 10⁻⁶ ]

pH = 5.5

c) [OH⁻] = 8.2 × 10⁻⁶

pOH = -log[OH]

pOH = -log[8.2 × 10⁻⁶]

pOH = 5.1

pH + pOH = 14

pH = 14 - pOH

pH = 14 - 5.1

pH = 8.9

d) [OH⁻] = 0.53 M

pOH = -log[OH]

pOH = -log[0.53]

pOH = 0.28

pH + pOH = 14

pH = 14 - pOH

pH = 14 - 0.28

pH = 13.72

5 0

3 years ago

Substance ΔG°f(kJ/mol) M3O4(s) −8.80 M(s) 0 O2(g) 0 Consider the decomposition of a metal oxide to its elements, where M represe

baherus [9]

Answer:

The equilibrium constant is $K =0.02867$

The equilibrium pressure for oxygen gas is $P_{O_2} = 0.09367\ atm$

Explanation:

From the question we are told that

The equation of the chemical reaction is

$M_2 O_3 _{(s)} ----> 2M_{(s)} + \frac{3}{2} O_2_{(g)}$

The Gibbs free energy for $M_3 O_4_{(s)}$ is $\Delta G^o_{1} = -8.80 \ kJ/mol$

The Gibbs free energy for $M{(s)}$ is $\Delta G^o_{2} = 0 \ kJ/mol$

The Gibbs free energy for $O_2{(s)}$ is $\Delta G^o_{3} = 0 \ kJ/mol$

The Gibbs free energy of the reaction is mathematically represented as

$\Delta G^o_{re} = \sum \Delta G^o _p - \sum G^o _r$

$\Delta G^o_{re} = \sum \Delta G^o _1 - \sum( G^o _2 +G^o _3)$

Substituting values

From the balanced equation

$\Delta G^o_{re} =[ (2 * 0) + (\frac{3}{2} * 0 )] - [1 * - 8.80]$

$\Delta G^o_{re} = 8.80 kJ/mol =8800J/mol$

The Gibbs free energy of the reaction can also be represented mathematically as

$\Delta G^o_{re} = -RTln K$

Where R is the gas constant with a value of $R = 8.314 J/mol \cdot K$

T is the temperature with a given value of $T = 298 K$

K is the equilibrium constant

Now equilibrium constant for a reaction that contain gas is usually expressed in term of the partial pressure of the reactant and products that a gaseous in state

The equilibrium constant for this chemical reaction is mathematically represented as

$K_p =[ P_{O_2}]^{\frac{3}{2} }$

Where $[ P_{O_2}]$ is the equilibrium pressure of oxygen

The p subscript shows that we are obtaining the equilibrium constant using the partial pressure of gas in the reaction

Now equilibrium constant the subject on the second equation of the Gibbs free energy of the reaction

$K = e^{- \frac{\Delta G^o_{re}}{RT} }$

Substituting values

$K= e^{\frac{8800}{8.314 * 298} }$

$K =0.02867$

Now substituting this into the equation above to obtain the equilibrium of oxygen

$0.02867 = [P_{O_2}]^{[\frac{3}{2} ]}$

multiplying through by $1 ^{\frac{2}{3} }$

$P_{O_2} = [0.02867]^{\frac{2}{3} }$

$P_{O_2} = 0.09367\ atm$

3 0

3 years ago

What is the role of gravity in maintaining the atmospheric layer around Earth ? Please put a simple explanation Thank you !

Delvig [45]

Answer:

Explanation:

First of all, we need to understand what gravity is. Gravity in its simplest form can be best visualized as a force of attraction, that pulls all objects that have a mass towards itself.

It is usually exerted by an object with a larger mass and felt by objects with smaller masses.

Putting this into context, the earth exerts a gravitational pull on all substances with masses, which are around it. This includes the oceans, birds, airplanes, clouds, and even gasses in the atmosphere. The atmospheric gasses may be light enough to float, but they can't go very far, since gravity is constantly tugging at them. This keeps the gasses that make up our atmosphere, within a reasonable radius, which is very close to the earth. In a nutshell, gravity prevents them from floating away into outer space.

8 0

3 years ago