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3 years ago

At this pressure, how many molecules of air would there be in a 20 ∘C experimental chamber with a volume of 0.020 m3 ?

Chemistry

1 answer:

jonny [76]3 years ago

8 0

Answer:

Explanation: The lowest pressure in a laboratory is 4.0×10^-11Pa

Using Ideal gas equation

PV = nRT

P= 4.0×10^-11Pa

V= 0.020m^3

T= 20+273= 293k

n=number of moles = m/A

Where m is the number of molecules and A is the Avogradro's number=6.02×10²³/mol

R=8.314J/(mol × K)

PV= m/A(RT)

4.0×10^-11 ×0.020 = m/6.02×10²³(8.314×293)

m = 4.0×10^-11×0.020×6.02×10^23 / (8.314×293)

m = 1.98×10^8 molecules

Therefore,the number of molecules is 1.98×10^8

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3 years ago

32 g of sulfur will react with 48 g of oxygen to produce 80 g of sulfur trioxide. If 32 g of sulfur and 100 g of oxygen are plac

Lina20 [59]

Answer:

Since the container is consealed, and O2 will no be completely consumed, the total mass of material in the container will be 80 grams SO3+ 52 grams O2 = 132 grams (option B)

Explanation:

Step 1: Data given

Mass of sulfur = 32.00 grams

Mass of oxygen = 48.00 grams

Molar mass of sulfur = 32.07 g/mol

Molar mass of oxygen = 32 g/mol

Molar mass of SO3 = 80.07 g/mol

Step 2: The balanced equation

2S + 3O2 → 2SO3

Step 3: Calculate moles S

Moles S = Mass S / molar mass S

Moles S = 32.0 grams / 32.07 g/mol

Moles S = 0.998 moles

Step 4: Calculate moles O2

Moles O2 = 100.0 grams / 32.0 g/mol

Moles O2 = 3.125 moles

Step 5: Calculate the limiting reactant

For 2 moles S we need 3 moles O2 to produce 2 moles SO3

S is the limiting reactant. It will completely be consumed (0.998 moles)

O2 is in excess, there will be consumed 3/2 * 0.998 = 1.497 moles

There will remain 3.125- 1.497 = 1.628 moles O2

This is 1.628 moles * 32 g/mol = 52.1 grams

Step 6: Calculate moles SO3

For 2 moles S we need 3 moles O2 to produce 2 moles SO3

For 0.998 moles S there will react 0.998 moles SO3

Step 6: Calculate mass SO3

Mass SO3 = moles SO3 * molar mass SO3

Mass SO3 = 0.998 moles * 80.07 g/mol

Mass SO3 = 79.9 grams ≈ 80 grams

There will be produced 80 grams of SO3

Since the container is consealed, and O2 will no be completely consumed, the total mass of material in the container will be 80 grams SO3+ 52 grams O2 = 132 grams (option B)

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3 years ago

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lys-0071 [83]

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6 0

3 years ago

Oxygen and Oxygen Ionic or Covalent? Defense how do you know

Marat540 [252]

Answer:

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8 0

2 years ago

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olga nikolaevna [1]

Answer:

Acid: HCl(aq), conjugate base: Cl⁻(aq)

Base: CO₃⁻²(aq), conjugate acid: HCO₃⁻(aq)

The rewrite reaction is shown below.

Explanation:

The acid compound is the one that loses an H⁺, and the compound formed when it happens is its conjugate base. The base compound is the acceptor of H⁺, and its conjugate acid is the compound formed (Brosted-Lowry theory).

So, the acid-base pairs are:

Acid: HCl(aq), conjugate base: Cl⁻(aq)

Base: CO₃⁻²(aq), conjugate acid: HCO₃⁻(aq)

The TUMS® is an antacid, so it intends to reduce the concentration of the strong acid HCl. So, the forward reaction is favored. It can be represented with the forward arrow larger than the reversible arrow, as shown in the image below.

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