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2 years ago

A mixture of reactants and products for the reaction shown below is at

Chemistry

1 answer:

12345 [234]2 years ago

8 0

Answer:

D. The concentrations of N2, H2, and NH3 would remain the same.

Explanation:

In reversible reactions, equilibrium is a point in which the rate of forward reaction is equal to that of reverse reaction.
Given the equation;

N₂(g)+ 3H₂(g) → 2NH₃(g) is a equilibrium at a constant volume of 1.0 L

Addition of Argon gas which an inert gas dose not affect the equilibrium of the reaction.
This is because the addition of an inert gas at constant volume only causes an increase in the total pressure of the system while the concentrations of the reactants and products remain constant, therefore not affecting the equilibrium.

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Which is the best interpretation of the two flat portions of the graph

Fed [463]

Answer:

Explanation:

5 0

2 years ago

Read 2 more answers

The osmotic pressure of a solution containing 2.04 g of an unknown compound dissolved in 175.0 mLof solution at 25 ∘C is 2.13 at

kherson [118]

Answer: The molecular formula of the compound is $C_4H_{10}O_4$

Explanation:

To calculate the concentration of solute, we use the equation for osmotic pressure, which is:

$\pi=iMRT$

Or,

$\pi=i\times \frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}\times RT$

where,

$\pi$ = osmotic pressure of the solution = 2.13 atm

i = Van't hoff factor = 1 (for non-electrolytes)

Given mass of compound = 2.04 g

Volume of solution = 175.0 mL

R = Gas constant = $0.0821\text{ L atm }mol^{-1}K^{-1}$

T = temperature of the solution = $25^oC=[273+25]=298K$

Putting values in above equation, we get:

$2.13atm=1\times \frac{2.04\times 1000}{\text{Molar mass of compound}\times 175.0}\times 0.0821\text{ L.atm }mol^{-1}K^{-1}\times 298K\\\\\text{Molar mass of compound}=\frac{1\times 2.04\times 1000\times 0.0821\times 298}{2.13\times 175.0}=133.9g/mol$

Calculating the molecular formula:

The chemical equation for the combustion of compound having carbon, hydrogen and oxygen follows:

$C_xH_yO_z+O_2\rightarrow CO_2+H_2O$

where, 'x', 'y' and 'z' are the subscripts of carbon, hydrogen and oxygen respectively.

We are given:

Mass of $CO_2=36.26g$

Mass of $H_2O=14.85g$

We know that:

Molar mass of carbon dioxide = 44 g/mol

Molar mass of water = 18 g/mol

For calculating the mass of carbon:

In 44 g of carbon dioxide, 12 g of carbon is contained.

So, in 36.26 g of carbon dioxide, $\frac{12}{44}\times 36.26=9.89g$ of carbon will be contained.

For calculating the mass of hydrogen:

In 18 g of water, 2 g of hydrogen is contained.

So, in 14.85 g of water, $\frac{2}{18}\times 14.85=1.65g$ of hydrogen will be contained.

Mass of oxygen in the compound = (22.08) - (9.89 + 1.65) = 10.54 g

To formulate the empirical formula, we need to follow some steps:

Step 1: Converting the given masses into moles.

Moles of Carbon = $\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{9.89g}{12g/mole}=0.824moles$

Moles of Hydrogen = $\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{1.65g}{1g/mole}=1.65moles$

Moles of Oxygen = $\frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{10.54g}{16g/mole}=0.659moles$

Step 2: Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 0.659 moles.

For Carbon = $\frac{0.824}{0.659}=1.25\approx 1$

For Hydrogen = $\frac{1.65}{0.659}=2.5$

For Oxygen = $\frac{0.659}{0.659}=1$

Converting the mole fraction into whole number by multiplying the mole fraction by '2'

Mole fraction of carbon = (1 × 2) = 2

Mole fraction of oxygen = (2.5 × 2) = 5

Mole fraction of hydrogen = (1 × 2) = 2

Step 3: Taking the mole ratio as their subscripts.

The ratio of C : H : O = 2 : 5 : 2

The empirical formula for the given compound is $C_2H_5O_2$

For determining the molecular formula, we need to determine the valency which is multiplied by each element to get the molecular formula.

The equation used to calculate the valency is:

$n=\frac{\text{Molecular mass}}{\text{Empirical mass}}$

We are given:

Mass of molecular formula = 133.9 g/mol

Mass of empirical formula = 61 g/mol

Putting values in above equation, we get:

$n=\frac{133.9g/mol}{61g/mol}=2$

Multiplying this valency by the subscript of every element of empirical formula, we get:

$C_{(2\times 2)}H_{(5\times 2)}O_{(2\times 2)}=C_4H_{10}O_4$

Hence, the molecular formula of the compound is $C_4H_{10}O_4$

4 0

3 years ago

Which is the correct equilibrium constant expression for the following reaction? 2brcl 3 (g) br 2 (g) 3cl 2 (g)

marishachu [46]

Kc = concentrations of product / concentrations of reactant

Kc = [Br₂] [Cl₂]₃ / [BrCl₃]₂

What is the equilibrium constant?

The relationship between a reaction's products and reactants with regard to a certain unit is expressed by the equilibrium constant(K) This article introduces the mathematics needed to determine the partial pressure equilibrium constant as well as how to formulate expressions for equilibrium constants. By allowing a single reaction to reach equilibrium and then measuring the concentrations of each chemical participating in that reaction, one can determine the numerical value of an equilibrium constant. it is the ratio of product concentrations to reactant concentrations. The equilibrium constant for a given reaction is unaffected by the initial concentrations because the concentrations are measured at equilibrium.

To learn more about the equilibrium constant, visit:

brainly.com/question/19340344

#SPJ4

6 0

1 year ago

A naturally occurring element consists of three isotopes. The data for the isotopes are:

max2010maxim [7]

(46.972* 69.472% + 48.961*21.667% + 49.954*8.8610%)/100% =

= 46.972* 0.69472 + 48.961*0.21667 + 49.954*0.088610 =47.667 u

7 0

2 years ago

Assuming the relative rate of secondary hydrogen atom abstraction for the chlorination of propane is 3.9 times faster than the r

Ghella [55]

Answer:

% of n-propyl chloride = 43.48 %

Explanation:

There are 2 secondary hydrogens and 6 primary hydrogens

The rate of abstraction of seondary hydrogen = 3.9 X rate of abstraction of primary hydrogen

probability of formation of isopropyl chloride = 3.9 X 1 (relative rate X relative number of secondary hydrogens)

Probability of formation of n-propyl chloride = 1 X 3 (relative rate X relative number of primary hydrogens)

Total probability = 3.9

% of n-propyl chloride = 3 X 100 / 6.9 = 43.48 %

7 0

3 years ago