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Rina8888 [55]
2 years ago
11

Given that this reaction is exothermic, what direction will the equilibrium shift when the temperature of the reaction is decrea

sed? 3H₂(g) + N₂(g) ⇔ 2NH₃(g)
A. Left (towards the reactants)
B. Right (towards the products)
C. No Shift
Chemistry
1 answer:
Svetradugi [14.3K]2 years ago
7 0

Equilibrium will shift towards the products when temperature is decreased in an exothermic reaction of the  formation of ammonia.

<h3>What is an exothermic reaction?</h3>

An exothermic reaction is a reaction in which heat content of the reactants is greater than the heat content of product.

In an exothermic reaction, heat is given off.

For an exothermic reaction in equilibrium, increasing temperature shifts equilibrium to the towards the left, towards the reactants.

On the other, equilibrium will shift towards the products when temperature is decreased.

Therefore, equilibrium will shift towards the products when temperature is decreased in the reaction of the  formation of ammonia.

Learn more about exothermic reactions at: brainly.com/question/13892884

#SPJ1

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The answer is 9.5 one as it is more basic so it contains more OH ion
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3 years ago
What volume of stock solution and water must you add to prepare 36.25ml of a 1.25M solution
Serjik [45]

Given :

Number of moles , n = 36.25 mol .

Molarity , M = 1.25 M .

To Find :

The volume of water required .

Solution :

Moarity is given by :

M=\dfrac{n}{V}

So , V=\dfrac{n}{M}

Here , n is number of moles and M is molarity .

Putting all values in above equation , we get :

V=\dfrac{36.25}{1.25}\\\\V=29\ L

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5 0
3 years ago
A 3.82L balloon filled with gas is warmed from 204.9K to 304.8 K.
likoan [24]

Given :

A 3.82L balloon filled with gas is warmed from 204.9K to 304.8 K.

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The volume of the gas after it is heated.

Solution :

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Now, we know by ideal gas equation at constant pressure :

\dfrac{V_1}{V_2} = \dfrac{T_1}{T_2}\\\\\dfrac{3.82}{V_2}= \dfrac{204.9}{304.8}\\\\V_2 = \dfrac{304.8}{204.9} \times 3.82\\\\V_2 = 5.68 \ L

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3 0
3 years ago
After a 4.626g sample of silver oxide is heated, 4.306g of silver metal remains. What is the empirical formula of the compound?
saw5 [17]

Answer:

The empirical formula is Ag2O.

The empirical formula is Ag2O.Explanation:

The empirical formula is Ag2O.Explanation:The empirical formula is the simplest whole-number ratio of atoms in a compound.

The empirical formula is Ag2O.Explanation:The empirical formula is the simplest whole-number ratio of atoms in a compound.The ratio of atoms is the same as the ratio of moles. So our job is to calculate the molar ratio of Ag to 2O.

do the steps ...

To get this into an integer ratio, we divide both numbers by the smaller value.

From this point on, I like to summarize the calculations in a table.

ElementAgMass/gXMolesXllRatiomllIntegers

—————————————————−———mAgXXXm7.96Xm0.07377Xll2.00mmm2

mlOXXXXl0.59mm0.0369Xml1mmmml1

There are 2 mol of Ag for 1 mol of O.

3 0
3 years ago
Read 2 more answers
A mouse is placed in a sealed chamber with air at 769.0 torr. This chamber is equipped with enough solid KOH to absorb any CO2 a
svlad2 [7]

<u>Answer:</u> The amount of oxygen gas consumed by mouse is 0.202 grams.

<u>Explanation:</u>

We are given:

Initial pressure of air = 769.0 torr

Final pressure of air = 717.1 torr

Pressure of oxygen = Pressure decreased = Initial pressure - Final pressure = (769.0 - 717.1) torr = 51.9 torr

To calculate the amount of oxygen gas consumed, we use the equation given by ideal gas which follows:

PV=nRT

where,

P = pressure of the gas = 51.9 torr

V = Volume of the gas = 2.20 L

T = Temperature of the gas = 292 K

R = Gas constant = 62.364\text{ L. Torr }mol^{-1}K^{-1}

n = number of moles of oxygen gas = ?

Putting values in above equation, we get:

51.9torr\times 2.20L=n\times 62.364\text{ L. Torr }mol^{-1}K^{-1}\times 292K\\\\n=\frac{51.9\times 2.20}{62.364\times 292}=0.0063mol

To calculate the mass from given number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}

Moles of oxygen gas = 0.0063 moles

Molar mass of oxygen gas = 32 g/mol

Putting values in above equation, we get:

0.0063mol=\frac{\text{Mass of oxygen gas}}{32g/mol}\\\\\text{Mass of oxygen gas}=(0.0063mol\times 32g/mol)=0.202g

Hence, the amount of oxygen gas consumed by mouse is 0.202 grams.

5 0
3 years ago
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