Answer:
When we look at an arbitrary point in the sky, away from the sun, we see only the light that was redirected by the atmosphere into our line of sight. Because that occurs much more often for blue light than for red, the sky appears blue. Violet light is actually scattered even a bit more strongly than blue.
Explanation:
Moles of H₂ are needed to produce 9.33 moles of NH₃ : 13.995
<h3>Further explanation</h3>
A reaction coefficient is a number in the chemical formula of a substance involved in the reaction equation. The reaction coefficient is useful for equalizing reagents and products.
The reaction coefficient in a chemical equation shows the mole ratio of the reactants and products
Reaction for the synthesis of ammonia :
N₂+3H₂⇒2NH₃
moles of NH₃ = 9.33
From equation, mol ratio of H₂ : NH₃ = 3 : 2, so mol H₂ :
Stoichiometry time! Remember to look at the equation for your molar ratios in other problems.
31.75 g Cu | 1 mol Cu | 2 mol Ag | 107.9 g Ag 6851.65
⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻ → ⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻ = 107.9 g Ag
∅ | 63.5 g Cu | 1 mol Cu | 1 mol Ag 63.5
There's also a shorter way to do this: Notice the molar ratio from Cu to Ag, which is 1:2. When you plug in 31.75 into your molar mass for Cu, it equals 1/2 mol. That also means that you have 1 mol Ag because of the ratio, qhich you can then plug into your molar mass, getting 107.9 as well.
The statement which is true is
Fluorine is more reactive than nitrogen because fluorine needs only one electron to fill its outermost shell.
<u><em>Explanation</em></u>
Fluorine has electron configuration of 1S²2S²2P⁵ while nitrogen has 1S²2S²2P³ electron configuration.
The 2P sub shell for nitrogen is half filled therefore it is sable than fluorine.
since p orbital can hold a maximum of 6 electrons ,Fluorine requires 1 electron to completely fill it's 2P sub shell which make it more reactive than nitrogen.
Answer:
See solution.
Explanation:
Hello there!
In this case, according to the given information, it turns out possible for us to set up the formula for the calculation of the by-mass percentage of the metal:
Thus, we solve for the molar mass of the metal to obtain:
For the subsequent problems, we proceed as follows:
a.
b.
c.
Regards!
