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2 years ago

Explain how the lines seen in the spectroscope relate to the position of electrons in the metal atom?

Chemistry

1 answer:

mojhsa [17]2 years ago

7 0

Answer:

The lines fall somewhere in one or more of the colors in the spectroscope. When electrons in an atom are excited by shooting electricity through them for instance, some electrons are forced to temporatily jump to a higher energy level. As it is only temporary, the electron soon falls back down to its own level. However as it loses potential energy it gives off a brief burst of light energy which we can see. If the light is red then the electron didn't jump very high, and therefore lost only a little eneergy falling back down. If the light is violet or purple it fell from a high orbital and released a lot of energy falling back down. Everything else falls in between.

Explanation:

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For the reaction PCl5(g) <--> PCl3(g) Cl2(g) at equilibrium, which statement correctly describes the effects of increasing

xenn [34]

The given question is incomplete. The complete question is :

For the reaction $PCl_5(g)\rightleftharpoons PCl_3(g)+Cl_2(g)$ at equilibrium, which statement correctly describes the effects of increasing pressure and adding $PCl_5$ , respectively

a) Increasing pressure causes shift to reactants, adding $PCl_5$ causes shift to products.

b) Increasing pressure causes shift to products ,adding $PCl_5$ causes shift to reactants.

c) Increasing pressure causes shift to products, adding $PCl_5$ causes shift to products.

d) Increasing pressure causes shift to reactants,adding $PCl_5$ causes shift to reactants

Answer: Increasing pressure causes shift to reactants, adding $PCl_5$ causes shift to products.

Explanation:

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.

For the given equation:

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

a) If the pressure is increased, the volume will decrease according to Boyle's Law. Now, according to the Le-Chatlier's principle, the equilibrium will shift in the direction where decrease in pressure is taking place. As the number of moles of gas molecules is lesser at the reactant side. So, the equilibrium will shift in the left direction. i.e. towards reactants.

b) If $PCl_5$ is added, the equilibrium will shift in the direction where $PCl_5$ is decreasing. So, the equilibrium will shift in the right direction. i.e. towards products.

8 0

2 years ago

There are two types of nucleic acids, dna and rna. nearly all organisms use dna, not rna, as the central repository for genetic

Leya [2.2K]

Found the choices. Pls see attachment.

The statements that explains this phenomenon are:
1) DNA contains adenine as one of its nitrogenous bases.
2) DNA has a double-stranded structure that ensures an accurate mechanism of duplication.

6 0

2 years ago

Two weak acids, A and B, have pKa values of 4 and 6, respectively. Which statement is true?A) Acid A dissociates to a greater ex

zalisa [80]

Answer:

A) Acid A dissociates to a greater extent in water than acid B

Explanation:

A) Acid A dissociates to a greater extent in water than acid B

We are given the pKa for both acids, and we know

pKa = - log Ka

Taking antilog to both sides of the equation we can solve for Ka

⇒ -pKa = log Ka

-antilog pKa = Ka

10 ^-pka = Ka

So ka for acid A = 10⁻⁴

and

ka for acid B = 10⁻⁶

True the equilibrium constant for acid A is greater, so it dissociates more.

B) For solutions of equal concentration, acid B will have a lower pH.

We know the stronger acid is A, and it dissociates more. Since pH is the negative log of H₃O⁺ concentration, it follows that at equal concentrations the acid A will have at equilibrium a greater [H₃O⁺] and hence a lower pH

C) B is the conjugate base of A

False:

If B were the conjugate base of A, its Kb would have been given by:

Ka x Kb = Kw

Kb = 10⁻¹⁴/10⁻⁶ = 10⁻⁸ for the conjugate base of acid B

Kb = 10⁻¹⁴/10⁻⁴ = 10⁻¹⁰ for the conjugate base of acid A

which are not equal.

D) Acid A is more likely to be a polyprotic acid than acid B.

False

Just having the pkas for both acids one cannot know if any of the acids is polyprotic. We will need the formula for the acids.

E) The equivalence point of acid A is higher than that of acid B

False

The equivalence point depends on the the concentration of the acids and their volumes.

The equivalence point is reached in the titration when the number of equivalents of base equals the number of equivalents of acid:

# equivalents acid = # equivalents of base @ end point

and

# equivalents acid = Molarity of acid x Volume of acid

4 0

2 years ago

What mass of solid NaOH (97.0 % by mass) is required to prepare 1.00 L of a 10.0% solution of NaOH by mass? The density of the 1

Lelechka [254]

Answer: The mass of 97 % of NaOH solution required is 114.33 g

Explanation:

To calculate mass of a substance, we use the equation:

$\text{Density of substance}=\frac{\text{Mass of substance}}{\text{Volume of substance}}$

We are given:

Density of 10 % solution = 1.109 g/mL

Volume of 10% solution = 1 L = 1000 mL (Conversion factor: 1 L = 1000 mL)

Putting values in above equation, we get:

$1.109g/mL=\frac{\text{Mass of }10\%\text{ solution}}{1000mL}\\\\\text{Mass of }10\%\text{ solution}=1109g$

The mass of 10 % solution is 1109 g.

To calculate the mass of concentrated solution, we use the equation:

$c_1m_1=c_2m_2$

where,

$c_1\text{ and }m_1$ are the concentration and mass of concentrated solution.

$c_2\text{ and }m_2$ are the concentration and mass of diluted solution.

We are given:

$c_1=97\%\\m_1=?g\\c_2=10\%\\m_2=1109g$

Putting values in above equation, we get:

$97\times m_1=10\times 1109\\\\m_1=114.33g$

Hence, the mass of 97 % of NaOH solution required is 114.33 g

3 0

2 years ago

The process of joining many small molecules into larger molecules is called(1) neutralization (3) saponification(2) polymerizati

yulyashka [42]

(2) polymerization. polymerization is a process of reacting monomer molecules together in a chemical reaction to form polymer chains or three-dimensional networks.

8 0

2 years ago

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