Atmospheric pressure is the sum of the pressure created by all gases in the

The following equilibrium is formed when copper and bromide ions are placed in a solution:

JulsSmile [24]

Answer:

A)

1. Reaction will shift rightwards towards the products.

2. It will turn green.

3. The solution will be cooler..

B) It will turn green.

Explanation:

Hello,

In this case, for the stated equilibrium:

$heat + Cu(H_2O)_6 ^{+2} (blue) + 4Br^- \rightleftharpoons 6H_2O + CuBr_4^{-2} (green)$

In such a way, by thinking out the Le Chatelier's principle, we can answer to each question:

A)

1. If potassium bromide, which adds bromide ions, is added more reactant is being added to the solution, therefore, the reaction will shift rightwards towards the products.

2. The formation of the green complex is favored, therefore, it will turn green.

3. The solution will be cooler as heat is converted into "cold" in order to reestablish equilibrium.

B) In this case, as the heat is a reactant, if more heat is added, more products will be formed, which implies that it will turn green.

Regards.

3 0

2 years ago

∆G° for the process benzene (l) benzene (g) is 3.7 Kj/mol at 60 °C , calculate the vapor pressure of benzene at 60 °C [R=0.0821

Alex787 [66]

Answer:

4) 0.26 atm

Explanation:

In the process:

Benzene(l) → Benzene(g)

ΔG° for this process is:

ΔG° = -RT ln Q

<em>Where Q = P(Benzene(g)) / P°benzene(l) P° = 1atm</em>

ΔG° = 3700J/mol = -8.314J/molK * (60°C + 273.15) ln P(benzene) / 1atm

1.336 = ln P(benzene) / 1atm

0.26atm = P(benzene)

Right answer is:

6 0

2 years ago

a closed flask of air (0.250 L) contains 5.00 "puffs" of particles. The pressure probe on the flask reads 93 kPa. A student uses

Sergio039 [100]

Answer: New pressure inside the flask would be 148.8 kPa.

Explanation: The combined gas law equation is given by:

$PV=nRT$

As the flask is a closed flask, so the volume remains constant. Temperature is constant also.

So, the relation between pressure and number of moles becomes

$P=n\\or\\\frac{P}{n}=constant$

$\frac{P_1}{n_1}=\frac{P_2}{n_2}$

Initial conditions:

$P_1=93kPa\\n_1=5\text{ puffs}$

Final conditions: When additional 3 puffs of air is added

$P_2=?kPa\\n_2=8\text{ puffs}$

Putting the values, in above equation, we get

$\frac{93}{5}=\frac{P_2}{8}\\P_2=148.8kPa$

3 0

2 years ago

ASAPPP EASY!!!!!!!!!!

zhenek [66]

Did you know conventional argult culture has increased greenhouse gas emissions, soil erosion, water pollution, and threatened humans health. Let’s stop this from harming our environment and take action about this today. Organic farming has a smaller carbon footprint, conserves and builds soil, replenishes natural ecosystems for cleaner water and air, all with a toxic pesticide residues.

5 0

2 years ago

please help me with Chem I ONLY HAVE 5 MINUTES if methane gas (CH4) flows at a rate of 0.25L/s, how many grams of methane gas wi

Nookie1986 [14]

Answer:

643g of methane will there be in the room

Explanation:

To solve this question we must, as first, find the volume of methane after 1h = 3600s. With the volume we can find the moles of methane using PV = nRT -<em>Assuming STP-</em>. With the moles and the molar mass of methane (16g/mol) we can find the mass of methane gas after 1 hour as follows:

<em>Volume Methane:</em>

3600s * (0.25L / s) = 900L Methane

<em>Moles methane:</em>

PV = nRT; PV / RT = n

<em>Where P = 1atm at STP, V is volume = 900L; R is gas constant = 0.082atmL/molK; T is absolute temperature = 273.15K at sTP</em>

Replacing:

PV / RT = n

1atm*900L / 0.082atmL/molK*273.15 = n

n = 40.18mol methane

<em>Mass methane:</em>

40.18 moles * (16g/mol) =

<h3>643g of methane will there be in the room</h3>

5 0

2 years ago