Ozone is made of oxygen. What is the formula of ozone if 36g occupies 16.8L at STP?

While hiking in the woods, you see a brown, rotting apple lying on the ground. What changes is the physical properties of the ap

mojhsa [17]

Answer:

The change in color.

Explanation:

The apple turn brown in color because of the oxidation process. When the oxygen and water molecules in air react with it, oxidation take place. The oxidation process is very efficient in ambient temperature.

For example, if the peal off apple is placed into the refrigerator it take a time to got oxidize and turn brown, but if it is placed in room temperature it quickly turn brown.

when oxygen is react with peel off apple , it trigger the polyphenol oxidase enzyme to oxidize the phenolic compound and quinones are formed which then react with amino acids and produced brown color.

3 0

2 years ago

What is the oxidation state of selenium in SeO3?

ra1l [238]

Answer: The oxidation state of selenium in SeO3 is +6

Explanation:

SeO3 is the chemical formula for selenium trioxide.

- The oxidation state of SeO3 = 0 (since it is stable and with no charge)

- the oxidation number of oxygen (O) IN SeO3 is -2

- the oxidation state of selenium in SeO3 = Z (let unknown value be Z)

Hence, SeO3 = 0

Z + (-2 x 3) = 0

Z + (-6) = 0

Z - 6 = 0

Z = 0 + 6

Z = +6

Thus, the oxidation state of selenium in SeO3 is +6

8 0

2 years ago

Reserpine is a natural product isolated from the roots of the shrub Rauwolfia serpentina. It was first synthesized in 1956 by No

liraira [26]

Answer:

Molality = 0.066 m

Molar mass = 608.36 g/mol

Explanation:

It seems the question is incomplete. However a web search us shows this data:

" Reserpine is a natural product isolated from the roots of the shrub Rauwolfia serpentina. It was first synthesized in 1956 by Nobel Prize winner R. B. Woodward. It is used as a tranquilizer and sedative. When 1.00 g reserpine is dissolved in 25.0 g camphor, the freezing-point depression is 2.63 °C (Kf for camphor is 40 °C·kg/mol). Calculate the molality of the solution and the molar mass of reserpine. "

The freezing-point depression is expressed by:

ΔT=Kf * m

We put the data given by the problem and solve for m:

2.63 °C = 40°C·kg/mol * m

m = 0.06575 m

For the calculation of the molar mass: Molality is defined as moles of solute per kilogram of solvent:

0.06575 m = Moles reserpine / kg camphor

25.0 g camphor ⇒ 25.0/1000 = 0.025 kg camphor

We calculate moles of reserpine:

0.06575 m = Moles reserpine / 0.025 kg camphor

Moles reserpine = 1.64x10⁻³ mol

Finally we use the mass of reserpine and the moles to calculate the molar mass:

1.00 g reserpine / 1.64x10⁻³ mol = 608.36 g/mol

Keep in mind that if the data in your problem is different, the results will be different. But the solving method remains the same.

8 0

2 years ago

A particular reaction has an enthalpy and entropy of reaction of ∆H = +33 kJ/mol and ∆S = +0.15 kJ/mol⋅K. At the three indicated

Thepotemich [5.8K]

Answer:

Explanation: check my paage

4 0

1 year ago

Watch the video to determine which of the following relationships are correct according to Boyle’s law.

scoray [572]

Answer:

Part A

Boyle's Law is given mathematically as

P ∝(1/V) or V ∝(1/P)

Options 4 and 5, if they are properly written.

Part B

At constant temperature, and according to the Boyle's law for an ideal gas,

A. What can cause a Volume increase is a corresponding decrease in pressure.

B. What can cause a Volume decrease is a corresponding increase in pressure.

C. The Volume is unchanged if the pressure of the gas is unchanged too.

Part C

The pressure when the gas occupies a volume of 5.0 L = 40 atm

Part D

The pressure when the gas occupies a volume of 4.5 L = 36 atm

Explanation:

Part A

Boyle's Law states that at constant temperature, the pressure of an ideal gas is inversely proportional to the volume occupied by the gas.

So, mathematically, Boyle's Law is given as

P ∝(1/V) or V ∝(1/P)

Part B

Inverse relationship between two quantities means that the higher the value of one of the quantities go, the lower the value of the other quantity goes and vice versa.

So, at constant temperature, and according to the Boyle's law for an ideal gas.

A. What can cause a Volume increase is a corresponding decrease in pressure.

B. What can cause a Volume decrease is a corresponding increase in pressure.

C. The Volume is unchanged if the pressure of the gas is unchanged too.

Part C

A certain gas occupies a volume of 20 L when the applied pressure is 10 atm, find the pressure when the gas occupies a volume of 5.0 L.

According to Boyle's Law for an ideal gas,

P ∝(1/V)

P = (k/V)

where k is the constant of proportionality

PV = k

Therefore,

P₁V₁ = P₂V₂ = k

P₁ = 10 atm

V₁ = 20 L

P₂ = ?

V₂ = 5.0 L

10 × 20 = P₂ × 5

P₂ = 40 atm

Part D

If a certain gas occupies a volume of 18 L when the applied pressure is 9.0 atm , find the pressure when the gas occupies a volume of 4.5 L

P₁V₁ = P₂V₂ = k

P₁ = 9.0 atm

V₁ = 18 L

P₂ = ?

V₂ = 4.5 L

9 × 18 = P₂ × 4.5

P₂ = 36 atm

Hope this Helps!!!

3 0

2 years ago