22.4 L is the volume of any gas regardless of atmospheric conditions.

O True

O False

Answers

Answer 1
This is false. One mole of a gas occupies 22.4 L at STP, which is taken to be 0°C (273 K) and 1 atm. If atmospheric conditions depart from these values, this assumption cannot be used.

Related Questions

100 POINTS!!!!!!!
The following diagram shows the branching tree for four kingdoms and some of their shared derived characteristics.

A slanting line is shown. There are four lines drawn vertically on this line at equal intervals. There is a black circle between each pair of these vertical lines. Protists, Plants, Fungi, Animals are written on top of each vertical line in order from left to right. Common Cell is written on the left most end of the sloping line. The second black circle has Mostly Multicellular written on it. The first black circle has X written on it.

What shared characteristic can be written at point X? Use complete sentences to explain your answer.

Answers

Answer:

The answer could be "Autotrophs that photosynthesis".

Explanation:

At point X, the common characteristic can be written. The common character between protists and plants is autotrophs that photosynthesize which is the characteristic of plant-like protists. Plant-like protists are autotrophs, use carbon dioxide and water to produce glucose by the process of photosynthesis.

How tightly particles are packed determines what state it takes.
True or False​

Answers

Answer:

True

Explanation:

tighter the molecules the slower it moves and the harder it gets

solid has very tight particles packed together

gas has its particles spread out

Please give me the answer please

Answers

Answer:

A. 30cm³

Explanation:

Based on the chemical reaction:

CaCO₃ + 2HCl → CaCl₂ + H₂O + CO₂

1 mol of calcium carbonate reacts with 2 moles of HCl to produce 1 mol of CO₂

To solve this question we must convert the mass of each reactant to moles. With the moles we can find limiting reactant and the moles of CO₂ produced. Using PV = nRT we can find the volume of the gas:

Moles CaCO₃ -Molar mass: 100.09g/mol-

1.00g * (1mol / 100.09g) = 9.991x10⁻³ moles

Moles HCl:

50cm³ = 0.0500dm³ * (0.05 mol / dm³) = 2.5x10⁻³ moles

For a complete reaction of 2.5x10⁻³ moles HCl there are necessaries:

2.5x10⁻³ moles HCl * (1mol CaCO₃ / 2mol HCl) = 1.25x10⁻³ moles CaCO₃. As there are 9.991x10⁻³ moles, HCl is limiting reactant.

The moles produced of CO₂ are:

2.5x10⁻³ moles HCl * (1mol CO₂ / 2mol HCl) = 1.25x10⁻³ moles CO₂

Using PV = nRT

Where P is pressure = 1atm assuming STP

V volume in L

n moles = 1.25x10⁻³ moles CO₂

R gas constant = 0.082atmL/molK

T = 273.15K at STP

V = nRT / P

1.25x10⁻³ moles * 0.082atmL/molK*273.15K / 1atm = V

0.028L = V

28cm³ = V

As 28cm³ ≈ 30cm³

Right option is:

A. 30cm³

What volume of 0.215 M HCl is required to neutralize 50.0 mL of
0.800 M NaOH?

Answers

Answer:

186 mL HCl

Explanation:

M1V1 = M2V2

M1 = 0.215 M HCl

V1 = ?

M2 = 0.800 M NaOH

V2 = 50.0 mL

Solve for V1 --> V1 = M2V2/M1

V1 = (0.800 M)(50.0 mL) / (0.215 M) = 186 mL HCl

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