Composition of the air breathed in and the air breathed out changes

Answers

Answer 1

Answer:

breathed in air contains oxygen while breathed out air contains higher percentage of CO2


Related Questions

What scientifically goes on with a plasma ball?

Answers

Answer:

the electrode at the center of a plasma ball emits a high-frequency,high-voltage alternating electric current. This current flows through the plasma filaments to create colorful tendrils of light.

Explanation:

what colors depend upon the gases used inside the plasma ball? common gases include neon, argon, xenon, and krypton.

what are possible source of error for rusting of a nail​

Answers

Answer:

A nail can rust when exposed to oxygen. the molecules of iron on the surface of the nail exchange atoms with the oxygen in the air and produce a new substance, the reddish brown ferrous oxide i.e rust.

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

How does the entropy change in the reaction 2C3H6(g) + 9O2(g) → 6C02(g) + 6H2O (g)?

I will mark brainliest!! Thank you so much!!

Answers

Answer:

The entropy increases!!!

Explanation:

a pex

The entropy increases in the reaction.

What is entropy?Entropy is defined as the measure of the disorder of a system.Entropy is an extensive property of a thermodynamic system, to put it in simple words, its value changes depending on the amount of matter that is present.Entropy is denoted by the letter S and has units of joules per kelvin    (JK−1)

The entropy increases in the reaction if the total number of product molecules are greater than the total number of reactant molecules.

2C3H6(g) + 9O2(g) → 6C02(g) + 6H2O (g)

In the above reaction, the product molecules are greater than the reactant molecules. Hence, entropy increases.

Hence, we can conclude that option A is the answer.

To learn more about entropy here

https://brainly.com/question/22861773

#SPJ2

Nitric acid and nitrogen monoxide react to form nitrogen dioxide and water, like this: (aq)(g)(g)(l) At a certain temperature, a chemist finds that a reaction vessel containing a mixture of nitric acid, nitrogen monoxide, nitrogen dioxide, and water at equilibrium has the following composition: compoundamount Calculate the value of the equilibrium constant for this reaction. Round your answer to significant digits.

Answers

Answer:

The reaction given is  

Chemistry homework question answer, step 1, image 1

Since in equilibrium constant expression, the concentration of liquid components is not written.

Hence the equilibrium constant expression for the above reaction can be written as  

Chemistry homework question answer, step 1, image 2

where [NO] = equilibrium concentration of NO

[NO2 ] = equilibrium concentration of NO2

[HNO3 ] = equilibrium concentration of HNO3

Step 2

Given : mass of NO at equilibrium = 18.6 g

mass of NO2 at equilibrium = 15.1 g

and mass of HNO3 at equilibrium = 13.1 g

Since molar mass of HNO3 = 63 g/mol, molar mass of NO = 30 g/mol and molar mass of NO2 = 46 g/mol

Hence the moles of gases at equilibrium are

Chemistry homework question answer, step 2, image 1

Step 3

Since mass of solution = mass of HNO3 + mass of water = 13.1 + 234.8 = 247.9 g

Assuming density of solution = density of water = 1.00 g/mL

=> mass of solution = volume of solution X density of solution  

=> 247.9 = 1.00 X volume of solution  

=> volume of solution = 247.9 mL = 0.2479 L      

Explanation:

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

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

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.

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

O True

O False

Answers

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.
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