What is the density of 96 ml of a liquid that has a mass of 90.5 g?

COCl2(g) decomposes according to the equation above. When pure COCl2(g) is injected into a rigid, previously evacuated flask at

mestny [16]

Answer: The value of $K_p$ for the reaction at 690 K is 0.05

Explanation:

We are given:

Initial pressure of $COCl_2$ = 1.0 atm

Total pressure at equilibrium = 1.2 atm

The chemical equation for the decomposition of phosgene follows:

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

Initial: 1 - -

At eqllm: 1-x x x

We are given:

Total pressure at equilibrium = [(1 - x) + x+ x]

So, the equation becomes:

$[(1 - x) + x+ x]=1.2\\\\x=0.2atm$

The expression for $K_p$ for above equation follows:

$K_p=\frac{p_{CO}\times p_{Cl_2}}{p_{COCl_2}}$

$p_{CO}=0.2atm\\p_{Cl_2}=0.2atm\\p_{COCl_2}=(1-0.2)=0.8atm$

Putting values in above equation, we get:

$K_p=\frac{0.2\times 0.2}{0.8}\\\\K_p=0.05$

Hence, the value of $K_p$ for the reaction at 690 K is 0.05

3 0

1 year ago

Explain how a solution can be both dilute and saturated.

svlad2 [7]

Dilution is when you decrease the concentration of a solution by adding a solvent. As a result, if you want to dilute salt water, just add water. ... Add more solute until it quits dissolving. That point at which a solute quits dissolving is the point at which it's saturated.

4 0

2 years ago

Read 2 more answers

g You observed the formation of several precipitates in the Reactions in Solution lab exercise. Identify the precipitate in each

RUDIKE [14]

Answer:

For a: Lead iodide is a yellow precipitate.

For b: Barium sulfate is a white precipitate.

For c: Ferric hydroxide is a brown precipitate.

For d: Copper (II) hydroxide is a blue precipitate.

Explanation:

Precipitation reaction is defined as the reaction where a solid precipitate (solid substance) is formed at the end of the reaction. It is insoluble in water.

For the given options:

For (a):

The chemical reaction between KI and lead (II) nitrate follows:

$2KI(aq)+Pb(NO_3)_2(aq)\rightarrow PbI_2(s)+2KNO_3(aq)$

The iodide of lead is generally insoluble in water. Thus, lead iodide is a yellow precipitate.

For b:

The chemical reaction between barium chloride and sulfuric acid follows:

$BaCl_2(aq)+H_2SO_4(aq)\rightarrow BaSO_4(s)+2HCl(aq)$

The sulfate of barium is insoluble in water. Thus, barium sulfate is a white precipitate.

For c:

The chemical reaction between NaOH and ferric chloride follows:

$3NaOH(aq)+FeCl_3(aq)\rightarrow Fe(OH)_3(s)+3NaCl(aq)$

The hydroxide of iron is insoluble in water. Thus, ferric hydroxide is a brown precipitate.

For d:

The chemical reaction between NaOH and copper sulfate follows:

$CuSO_4+2NaOH\rightarrow Cu(OH)_2+Na_2SO_4$

The hydroxide of copper is insoluble in water. Thus, copper (II) hydroxide is a blue precipitate.

6 0

1 year ago

It took many of the metals several minutes to begin to react. Once the reaction was initiated, several of these metals continued

notsponge [240]

Answer:

Passivation of Oxide layers of the metals.

Explanation:

Passivation is a non-electrolytic finishing process that makes most metals rust-resistant. The prosses removes free iron from the surface by using either nitric or citric acid. When this happens, it results to an inert, protective oxide layer that is very slow or less likely to chemically react with air and cause corrosion.

Passivity caused many of the metals several minutes to begin to react. Once the finishing process that makes metals less likely to react was eroded, reaction was initiated vigorously.

8 0

2 years ago

You have two containers of equal volume. one is full of helium gas. the other holds an equal mass of nitrogen gas both gases hav

alexira [117]

Let's assume that both He and N₂ have ideal gas behavior.

Then we can use ideal gas law,
PV = nRT
Where, P is the pressure of gas, V is the volume, n is moles of gas, R is universal gas constant and T is the temperature in Kelvin.

The P and V are same for the both gases.
R is a constant.

The only variables are n and T.

Let's say temperature of He is T₁ and temperature of N₂ is T₂.

n = m/M where n is moles, m is mass and M is molar mass.

Molar mass of He is 4 g/mol and molar mass of N₂ is 28 g/mol

Since mass (m) of both gases are same,
moles of He = m/4
moles of N₂ = m/28

Let's apply the ideal gas equation for both gases.
For He gas,
PV = (m/4)RT₁ (1)

For N₂ gas,
PV = (m/28)RT₂ (2)

(1) = (2)
(m/4)RT₁ = (m/28)RT₂ 
 T₁/4 = T₂/28
T₁ = T₂/7
 7T₁ = T₂

Hence, the temperature of N₂ gas is higher by 7 times than the temperature of He gas.

8 0

2 years ago