How will you solve this?

Alexxandr [17]

Physical property of gas: There are a lot, here are some examples: => Air => Oxygen => Carbon dioxide => Flourine => Phosphine => Argon => Carbon monoxide This examples are all gases and each of these gases has its own physical property. The only thing that they makes them the same are that, they are gas with lesser molecules. Usually gas density is lesser compare to the solid or liquid and most gases has undefined melting point.

7 0

3 years ago

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The equilibrium constant for the reaction

FinnZ [79.3K]

Answer: The concentrations of $Cl_2$ at equilibrium is 0.023 M

Explanation:

Moles of $Cl_2$ = $\frac{\text {given mass}}{\text {Molar mass}}=\frac{10g}{71g/mol}=0.14mol$

Volume of solution = 1 L

Initial concentration of $Cl_2$ = $\frac{0.14mol}{1L}=0.14M$

The given balanced equilibrium reaction is,

$COCl_2(g)\rightleftharpoons CO(g)+Cl_2(g)$

Initial conc. 0.14 M 0 M 0M

At eqm. conc. (0.14-x) M (x) M (x) M

The expression for equilibrium constant for this reaction will be,

$K_c=\frac{[CO]\times [Cl_2]}{[COCl_2]}$

Now put all the given values in this expression, we get :

$4.63\times 10^{-3}=\frac{x)^2}{(0.14-x)}$

By solving the term 'x', we get :

x = 0.023 M

Thus, the concentrations of $Cl_2$ at equilibrium is 0.023 M

7 0

3 years ago

Imagine the reaction A + B ⇌ C + D is at equilibrium, where the forward and reverse reactions are at equal rates. What would hap

kondaur [170]

Answer:

Rate of forward reaction will increase.

Explanation:

Effect of change in reaction condition on equilibrium is explained by Le Chatelier's principle. According to this principle,

If an equilibrium condition of a dynamic reversible reaction is disturbed by changing concentration, temperature, pressure, volume, etc, then reaction will move will in a direction which counteract the change.

In the given reaction,

A + B ⇌ C + D

If concentration of A is increase, then reaction will move in a direction which decreases the concentration of A to reestablish the equilibrium.

As concentration A decreases in forward direction, therefore, rate of forward reaction will increase.

3 0

3 years ago

Answer these please ASAP need help no idea how to do these

STALIN [3.7K]

Answer:

Explanation:

Cu:

Number of moles = Mass / molar masa

2 mol = mass / 64 g/mol

Mass = 128 g

Mg:

Number of moles = Mass / molar masa

0.5 mol = mass / 24 g/mol

Mass = g

Cl₂:

Number of moles = Mass / molar masa

Number of moles = 35.5 g / 24 g/mol

Number of moles = 852 mol

H₂:

Number of moles = Mass / molar mass

8 mol = Mass / 2 g/mol

Mass = 16 g

P₄:

Number of moles = Mass / molar masa

2 mol = mass / 124 g/mol

Mass = 248 g

O₃:

Number of moles = Mass / molar masa

Number of moles = 1.6 g /48 g/mol

Number of moles = 0.033 mol

H₂O

Number of moles = Mass / molar masa

Number of moles = 54 g / 18 g/mol

Number of moles = 3 mol

CO₂

Number of moles = Mass / molar masa

2 mol = mass / 124 g/mol

Mass = 248 g

NH₃

Number of moles = Mass / molar masa

Number of moles = 8.5 g / 17 g/mol

Number of moles = 0.5 mol

CaCO₃

Number of moles = Mass / molar masa

Number of moles = 100 g / 100 g/mol

Number of moles = 1 mol

a)

Given data:

Mass of iron(III)oxide needed = ?

Mass of iron produced = 100 g

Solution:

Chemical equation:

F₂O₃ + 3CO → 2Fe + 3CO₂

Number of moles of iron:

Number of moles = mass/ molar mass

Number of moles = 100 g/ 56 g/mol

Number of moles = 1.78 mol

Now we compare the moles of iron with iron oxide.

Fe : F₂O₃

2 : 1

1.78 : 1/2×1.78 = 0.89 mol

Mass of F₂O₃:

Mass = number of moles × molar mass

Mass = 0.89 mol × 159.69 g/mol

Mass = 142.124 g

100 g of iron is 1.78 moles of Fe, so 0.89 moles of F₂O₃ are needed, or 142.124 g of iron(III) oxide.

b)

Given data:

Number of moles of Al = 0.05 mol

Mass of iodine = 26 g

Limiting reactant = ?

Solution:

Chemical equation:

2Al + 3I₂ → 2AlI₃

Number of moles of iodine = 26 g/ 254 g/mol

Number of moles of iodine = 0.1 mol

Now we will compare the moles of Al and I₂ with AlI₃.

Al : AlI₃

2 : 2

0.05 : 0.05

I₂ : AlI₃

3 : 2

0.1 : 2/3×0.1 = 0.067

Number of moles of AlI₃ produced by Al are less so it will limiting reactant.

Mass of AlI₃:

Mass = number of moles × molar mass

Mass = 0.05 mol × 408 g/mol

Mass = 20.4 g

26 g of iodine is 0.1 moles. From the equation, this will react with 2 moles of Al. So the limiting reactant is Al.

c)

Given data:

Mass of lead = 6.21 g

Mass of lead oxide = 6.85 g

Equation of reaction = ?

Solution:

Chemical equation:

2Pb + O₂ → 2PbO

Number of moles of lead = mass / molar mass

Number of moles = 6.21 g/ 207 g/mol

Number of moles = 0.03 mol

Number of moles of lead oxide = mass / molar mass

Number of moles = 6.85 g/ 223 g/mol

Number of moles = 0.031 mol

Now we will compare the moles of oxygen with lead and lead oxide.

Pb : O₂

2 : 1

0.03 : 1/2×0.03 = 0.015 mol

Mass of oxygen:

Mass = number of moles × molar mass

Mass = 0.015 mol × 32 g/mol

Mass = 0.48 g

The mass of oxygen that took part in equation was 0.48 g. which is 0.015 moles of oxygen. The number of moles of Pb in 6.21 g of lead is 0.03 moles. So the balance equation is

2Pb + O₂ → 2PbO

6 0

3 years ago

Which statement describes a metallic bond? A. Electrons are mobile within a metal. B. Electrons transition from an inner shell t

k0ka [10]

A- electrons good luck

8 0

3 years ago