Air masses form fronts when B, they collide with each other
The amount, in grams, of N that the sample will contain will be 2.1 grams.
<h3>Stoichiometric mass ratio</h3>
According to the problem. the mass ratio of H and N in ammonia is 4.7:1.
In other words, every 4.7 grams of H in ammonia must have 1 gram of N.
Now, in a particular ammonia sample, 10 grams of H is present.
The amount of N present can be calculated from the standard mass ratio.
4.1 grams H = 1 gram N
10 grams H = x
4.1x = 1 x 10
x = 10/4.1
x = 2.1 grams
Thus, the mass of nitrogen in the ammonia sample will be 2.1 grams.
More on mass ratios can be found here: brainly.com/question/14577772
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Answer:
Explanation:
mass of 1mole of brick = mass of 1 brick x avagadro number
= 4 kg x 6.022 x 10^23 bricks = 2.4 x 10^24 kg
No. of moles of bricks have a mass equal to the mass of the earth
= mass of earth / mass of 1 mole of brick
= [ 6 x 10^27 ] / [ 2.4 x 10^24 ]
= 2.5 x 10^3
= 2500 moles
23.0 g/ (4.50 g/ mL) gives 5.11 mL as the volume.
Answer:
17)c: the reactivity of a metal.
Explanation:
