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3 years ago

How well did tossing the pennies simulate half lives?

1 answer:

5 0

Its a 50% chance that approx 1/2 of the pennies will land on tails. The next toss will result the same. and so on and so on. showing how a reaction would slowly eliminate 1/2 of the remaining lives per reaction, until nothing is left. I think it is a good stimulation.

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B. The equilibrium constant for the reaction 2H2(g) + S2(g) 2H2S(g) is Keq = . (3 points)

eimsori [14]

Answer:

i. Keq=4157.99.

ii. More hydrogen sulfide will be produced.

Explanation:

Hello,

i. In this case, for the concentrations at equilibrium on the given chemical reaction, the equilibrium constant results:

$Keq=\frac{[H_2S]^2}{[H_2]^2[S_2]} =\frac{(0.97M)^2}{(0.051M)^2(0.087)} =4157.99$

ii. Now, by means of the Le Chatelier's principle, the addition of a reactant shifts the reaction towards products, it means that more hydrogen sulfide will be produced in order to reach equilibrium.

Best regards.

4 0

3 years ago

The results of Rutherford's gold foil experiment gave him the evidence to arrive at two conclusions: (1) an atom was much more t

Vedmedyk [2.9K]

Answer: The results of Rutherford's gold foil experiment gave him the evidence to arrive at two conclusions: (1) an atom was much more than just empty space and scattered electrons and (2) an atom consists of a positive charge at the center where most of its mass is placed.

Explanation:

Ernest Rutherford performed an experiment in which he passed alpha particles through a thin gold foil sheet. Through this sheet some of the particles passed on to the other side but some of them were reflected back.

This experiment was done by Rutherford to prove than an atom contains a tiny and heavy nucleus. He concluded that an atom has more than empty space and electrons were present in scattered form.

Also, he concluded that an atom consists of a positive charge at the center where most of its mass is placed.

Thus, we can conclude that the results of Rutherford's gold foil experiment gave him the evidence to arrive at two conclusions: (1) an atom was much more than just empty space and scattered electrons and (2) an atom consists of a positive charge at the center where most of its mass is placed.

7 0

3 years ago

The pressure on a 200 milliliter sample of CO2 (g) at constant temperature is increased from

balu736 [363]

The answer for the following problem is mentioned below.

Therefore the final volume of the gas is 100 ml.

Explanation:

Given:

Initial pressure ( $P_{1}$ ) = 600 mm of Hg

Final pressure ( $P_{2}$ ) = 1200 mm of Hg

Initial volume ( $V_{1}$ ) = 200 ml

To find:

Final volume ( $V_{2}$ )

We know;

According to the ideal gas equation,

P × V = n × R × T

Where;

P represents the pressure of the gas

V represents the volume of the gas

n represents the no of moles of the gas

R represents the universal gas constant

T represents the temperature of the gas

So,

From the above mentioned equation,

P × V = constant

$\frac{P_{1} }{P_{2} }$ = $\frac{V_{1} }{V_{2} }$

Where,

( $P_{1}$ ) represents the initial pressure of the gas

( $P_{2}$ ) represents the final pressure of the gas

( $V_{1}$ ) represents the initial volume of the gas

( $V_{2}$ ) represents the final volume of the gas

So;

$\frac{600}{1200}$ = $\frac{V_{2} }{200}$

$V_{2}$ = 100 ml

Therefore the final volume of the gas is 100 ml.

5 0

3 years ago

Identify and label the Brønsted-Lowry acid, its conjugate base, the Brønsted-Lowry base, and its conjugate acid in each of the f

julia-pushkina [17]

Explanation:

As per Brønsted-Lowry concept of acids and bases, chemical species which donate proton are called Brønsted-Lowry acids.

The chemical species which accept proton are called Brønsted-Lowry base.

(a) $HNO_3 + H_2O \rightarrow H_3O^+ + NO_3^-$

$HNO_3$ is Bronsted lowry acid and $NO_3^-$ is its conjugate base.

$H_2O$ is Bronsted lowry base and $H_3O^+$ is its conjugate acid.

(b)

$CN^- + H_2O \rightarrow HCN + OH^-$

$CN^-$ is Bronsted lowry base and HCN is its conjugate acid.

$H_2O$ is Bronsted lowry acid and $OH^-$ is its conjugate base.

(c)

$H_2SO_4 + Cl^- \rightarrow HCl + HSO_4^-$

$H_2SO_4$ is Bronsted lowry acid and $HSO_4^-$ is its conjugate base.

Cl^- is Bronsted lowry base and HCl is its conjugate acid.

(d)

$HSO_4^-+OH^- \rightarrow SO_4^{2-}+H_2O$

$HSO_4^-$ is Bronsted lowry acid and $SO_4^{2-}$ is its conjugate base.

OH^- is Bronsted lowry base and $H_2O$ is its conjugate acid.

(e)

$O_{2-}+H_2O \rightarrow 2OH^-$

$O_{2-}$ is Bronsted lowry base and OH- is its conjugate acid.

$H_2O$ is Bronsted lowry acid and OH- is its conjugate base.

6 0

3 years ago

Who is all bor.ed bc im bor.ed who wan.na talk UwU

emmainna [20.7K]

Answer:

me ig

Explanation:

4 0

2 years ago

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