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2 years ago

Consider the following system at equilibrium:

Chemistry

1 answer:

alexgriva [62]2 years ago

6 0

Answer:

A - Increase (R), Decrease (P), Decrease(q), Triple both (Q) and (R)

B - Increase(P), Increase(q), Decrease (R)

C - Triple (P) and reduce (q) to one third

Explanation:

<em>According to Le Chatelier principle, when a system is in equilibrium and one of the constraints that affect the rate of reaction is applied, the equilibrium will shift so as to annul the effects of the constraint.</em>

P and Q are reactants, an increase in either or both without an equally measurable increase in R (a product) will shift the equilibrium to the right. Also, any decrease in R without a corresponding decrease in either or both of P and Q will shift the equilibrium to the right. Hence, Increase(P), Increase(q), and Decrease (R) will shift the equilibrium to the right.

In the same vein, any increase in R without a corresponding increase in P and Q will shift the equilibrium to the left. The same goes for any decrease in either or both of P and Q without a counter-decrease in R will shift the equilibrium to the left. Hence, Increase (R), Decrease (P), Decrease(q), and Triple both (Q) and (R) will shift the equilibrium to the left.

Any increase or decrease in P with a commensurable decrease or increase in Q (or vice versa) with R remaining constant will create no shift in the equilibrium. Hence, Triple (P) and reduce (q) to one third will create no shift in the equilibrium.

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1) How many molecules are there in 985 mL of nitrogen at 0.0° C and 1.00 x 10-6 mm Hg?

RSB [31]

Answer : The number of molecules present in nitrogen gas are, $3.48\times 10^{13}$

Explanation :

First we have to calculate the moles of nitrogen gas by using ideal gas equation.

$PV=nRT$

where,

P = Pressure of $N_2$ gas = $1.00\times 10^{-6}mmHg=1.32\times 10^{-9}atm$ (1 atm = 760 mmHg)

V = Volume of $N_2$ gas = 985 mL = 0.982 L (1 L = 1000 mL)

n = number of moles $N_2$ = ?

R = Gas constant = $0.0821L.atm/mol.K$

T = Temperature of $N_2$ gas = $0.0^oC=273+0.0=273K$

Now put all the given values in above equation, we get:

$(1.32\times 10^{-9}atm)\times 0.982L=n\times (0.0821L.atm/mol.K)\times 273K$

$n=5.78\times 10^{-11}mol$

Now we have to calculate the number of molecules present in nitrogen gas.

As we know that 1 mole of substance contains $6.022\times 10^{23}$ number of molecules.

As, 1 mole of $N_2$ gas contains $6.022\times 10^{23}$ number of molecules

So, $5.78\times 10^{-11}$ mole of $N_2$ gas contains $(5.78\times 10^{-11})\times (6.022\times 10^{23})=3.48\times 10^{13}$ number of molecules

Therefore, the number of molecules present in nitrogen gas are, $3.48\times 10^{13}$

8 0

3 years ago

Identify the balanced chemical equation that represents a single displacement reaction.

UNO [17]

3H₂SO₄ + 2Al₂(SO₄)₃ → Al₂(SO₄)₃ + 3H₂ is the balanced chemical equation that represents a single displacement reaction.

<h3>What is a balanced chemical equation?</h3>

The equation in which the number of atoms of all the molecules is equal on both sides of the equation is known as a balanced chemical equation.

3H₂SO₄ + 2Al₂(SO₄)₃ → Al₂(SO₄)₃ + 3H₂

In this type of reaction, one substance is replacing another:

A + BC → AC + B

In a single displacement reaction, atoms replace one another based on the activity series. Elements that are higher in the activity series. Also, if the element that is to replace the other in a compound is more reactive the reaction will occur. If it is less reactive, there will be no reation.

In the first equation, fluorine is more reactive than bromine. Therefore, bromine cannot replace bromine.

In the second equation, the displacement is between hydrogen and aluminium. Hydrogen is lower in the activity series, this implies that aluminium will replace it.

Hence, 3H₂SO₄ + 2Al₂(SO₄)₃ → Al₂(SO₄)₃ + 3H₂ is the balanced chemical equation that represents a single displacement reaction.

Learn more about the displacement reaction here:

brainly.com/question/15052184

#SPJ1

8 0

2 years ago

__________ is an important factor in the process of maturation because it provides a biological framework of how much one is cap

Anon25 [30]

It is definitely not A. B is an effect. I would say C because D is more of a conservative answer , C is more of a liberal answer, and we currently live in a liberally swayed world. They are probably looking for C. It is not in your nature to be bad.

7 0

2 years ago

Read 2 more answers

Temperate deciduous trees lose their leaves in Fall. Explain why trees in temperate rainforest and tropical rainforest don’t los

andrey2020 [161]

Answer:

So trees in temperate don't lose their leaves because the weather events aren't harsh enough.

Trees in tropical rainforest don't lose their leaves because they are a different type of tree known as evergreens that are green all year round.

Explanation:

Ok so first we'll define some things

Deciduous Trees= Trees that lose all of their leaves for part of the year.

Trees shed their leaves trees to try and survive harsh weather events.

Temperate deciduous trees lose their leaves in fall to better survive the winter conditions of extreme cold and reduced daylight.

Temperate rainforests = An area that doesn't experience extremely cold or extremely hot temperatures or what we would call harsh weather events.

Broad-leaved trees in tropical rainforests are known evergreen, they are known as this as they are green all year round.

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3 years ago

Can someone answer all please.

boyakko [2]

Answer:

poszukaj w necie , proszę a nie prosisz innych o gotowca

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3 years ago