The abiotic components of an

what is the ratio of the rate of effusion of helium (atomic mass 4.00 amu) to that of oxygen gas (molecular mass 32.0 amu)?

nignag [31]

Answer:

3 : 1

Explanation:

Let the rate of He be R1

Molar Mass of He (M1) = 4g/mol

Let the rate of O2 be R2

Molar Mass of O2 (M2) = 32g/mol

Recall:

R1/R2 = √(M2/M1)

R1/R2 = √(32/4)

R1/R2 = √8

R1/R2 = 3

The ratio of rate of effusion of Helium to oxygen is 3 : 1

8 0

3 years ago

.500 mol of br2 and .500 mol of cl2 are placed in a .500L flask and allowed to reach equilibrium. at equilibrium the flask was f

Tasya [4]

Answer : The value of $K_c$ for the given reaction is, 0.36

Explanation :

Equilibrium constant : It is defined as the equilibrium constant. It is defined as the ratio of concentration of products to the concentration of reactants.

The equilibrium expression for the reaction is determined by multiplying the concentrations of products and divided by the concentrations of the reactants and each concentration is raised to the power that is equal to the coefficient in the balanced reaction.

As we know that the concentrations of pure solids and liquids are constant that is they do not change. Thus, they are not included in the equilibrium expression.

The given equilibrium reaction is,

$Br_2(aq)+Cl_2(aq)\rightleftharpoons 2BrCl(aq)$

The expression of $K_c$ will be,

$K_c=\frac{[BrCl]^2}{[Br_2][Cl_2]}$

First we have to calculate the concentration of $Br_2,Cl_2\text{ and }BrCl$ .

$\text{Concentration of }Br_2=\frac{Moles}{Volume}=\frac{0.500mol}{0.500L}=1M$

$\text{Concentration of }Cl_2=\frac{Moles}{Volume}=\frac{0.500mol}{0.500L}=1M$

$\text{Concentration of }BrCl=\frac{Moles}{Volume}=\frac{0.300mol}{0.500L}=0.6M$

Now we have to calculate the value of $K_c$ for the given reaction.

$K_c=\frac{[BrCl]^2}{[Br_2][Cl_2]}$

$K_c=\frac{(0.6)^2}{(1)\times (1)}$

$K_c=0.36$

Therefore, the value of $K_c$ for the given reaction is, 0.36

6 0

3 years ago

Read 2 more answers

The pH of a 0.65M solution of hydrofluoric acid HF is measured to be 1.68. Calculate the acid dissociation constant Ka of hydrof

sergey [27]

Answer:

Kₐ = 6.7 x 10⁻⁴

Explanation:

First lets write the equilibrium expression, Ka , for the dissociation of hydrofluoric acid:

HF + H₂O ⇄ H₃O⁺ + F⁻

Kₐ = [ H₃O⁺ ] [ F⁻ ] /[ [ HF ]

Since we are given the pH we can calculate the [ H₃O⁺ ] ( pH = - log [ H₃O⁺ ] , and because the acid dissociates into a 1: 1 relation , we will also have [F⁻ ]. The [ HF ] is given in the question so we have all the information that is needed to compute Kₐ.

pH = -log [ H₃O⁺ ]

1.68 = - log [ H₃O⁺ ]

Taking antilog to both sides of this equation:

10^-1.68 = [ H₃O⁺ ] ⇒ 2.1 X 10⁻² M= [ H₃O⁺ ]

[ F⁻ ] = 2.1 X 10⁻² M

Solving for Kₐ :

Kₐ = ( 2.1 X 10⁻² ) x ( 2.1 X 10⁻² ) / 0.65 = 6.7 x 10⁻⁴

(Rounded to two significant figures, the powers of 10 have infinite precision )

4 0

3 years ago

What is the difference between a crystalline material and an amorphous material? Give an example of each type.

Igoryamba

Answer :

As we know that there are two types of solids.

(1) Amorphous Solids

(2) Crystalline Solids

Amorphous Solids : It is a type of solids in which the constituent particles of the matter are arranged in the random manner.

That means there is no proper arrangement of atoms in solid lattice but the atoms or molecules are closely spaced that means they can move freely from one place to another.

The examples of amorphous solid are, plastics, glass, rubber, metallic glass, polymers, gel, fused silica, pitch tar, thin film lubricants, wax.

Crystalline Solids : It is a type of solids where the constituent particles of the matter are arranged in the specific manner.

That means there is a proper arrangement of atoms in solid lattice. They do not have space between the molecules or atoms and they can not move freely from one place to another.

The examples of crystalline solids are, quartz, calcite, sugar, mica, diamonds, snowflakes, rock, calcium fluoride, silicon dioxide, alum.

7 0

3 years ago

How do you find the molecular formula from the empirical formula?

miskamm [114]

To find them you would have numbers of the elements in percentage or grams then you divide them by their molar mass to get their moles. From there you divide by the smallest number. Round it to two or one sig fig. If you have a number that is for ex. 2.5 you multiply it by 2 to make it whole as well the other whole numbers. Then to find the molecular formula the problem must give you another molar mass and using your empirical formula convert it to its molar mass then you divide them, larger number over smaller number. You should get a number round it to 1 sig fig. Now you use that number and multiply the subscripts on the empirical formula to get the molecular formula.

7 0

3 years ago