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marissa [1.9K]
3 years ago
5

Hamad stated that Nitrogen gas is noncombustible so is water vapour. If Both nitrogen and water are in gaseous state then why th

e term gas and vapour were used by Hamad?
Chemistry
2 answers:
Rudiy273 years ago
5 0

Answer:

Explanation:

Unlike clouds, fog, or mist which are simply suspended particles of liquid water in the air, water vapour itself cannot be seen because it is in gaseous form. Water vapour in the atmosphere is often below its boiling point. When water is boiled the water evaporates much faster and makes steam.

Read more on Brainly.com - brainly.com/question/12860575#readmore

taurus [48]3 years ago
3 0
<h3>Answer:</h3>

Unlike clouds, fog, or mist which are simply suspended particles of liquid water in the air, water vapour itself cannot be seen because it is in gaseous form. Water vapour in the atmosphere is often below its boiling point. When water is boiled the water evaporates much faster and makes steam.

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We need (i) the stoichiometric equation, and (ii) the equivalent mass of dihydrogen.
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Whatever this molar quantity is, it is clear from the stoichiometry of the reaction that 3/2 equiv of dihydrogen gas were required. How much dinitrogen gas was required?
4 0
3 years ago
What is the molar mass of a compound?
Masteriza [31]

Answer:

The molar mass of a compound is The mass in grams of 1 mole of the compound (Option A)

Explanation:

Let's take ammonia as an example (NH3)

Mass of N = 14 g

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Molar mass of ammonia is Mass of N + (Mass of H).3

14 + 3 = 17 g/m

Ammonia is a compound that has 1 mol of N, plus 3 moles of H (see the formula)

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7 0
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A mixture consists of 28% oxygen, 14% hydrogen, and 58% nitrogen by volume. A sample of this mixture has a pressure of 4.0 atm i
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Answer:

C) 1.3 mol

Explanation:

Using gas law we can find the initial moles of the sample of the mixture, as follows:

PV = nRT

PV / RT = n

<em>Where P is pressure: 4.0atm</em>

<em>V is volume: 9.6L</em>

<em>R is gas constant: 0.082atmL/molK</em>

<em>T is absolute temperature: 300K</em>

<em>And n are moles of the gas</em>

<em />

PV / RT = n

4.0atm*9.6L / 0.082atmL/molK300K = n

n = 1.56moles of the mixture of the gas are present into the 9.6L container

Now, 14% of this gas is hydrogen that was removed of the system, that is:

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1.56mol - 0.22mol = 1.34mol.

Right answer is:

<h3>C) 1.3 mol</h3>

6 0
3 years ago
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