The metalloids are mostly concentrated in groups 14, 15, and 16. (Some simpler charts will show them as 4A, 5A, and 6A - take a look at the top of the periodic table your class uses to double-check).
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a) 56g
<h3>Calculation:</h3>
At STP,
22.4 L of N₂ = 1 mol
We have given 44.8 L of N₂, therefore,
44.8 L of N₂ = 
=
mol
We know that,
1 mol of N₂ = 28 g
Hence,
2 mol of N₂ = 28 × 2
= 56g
Hence, there are 56 g of N₂ in 44.8 L of nitrogen gas.
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Answer:
A- Speed = distance/time
Explanation:
Would make sense and that's what I remember
Answer: This description refers to: "hydrogen bonding" .
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<u> Hydrogen bonding </u> is a specific type of "dipole-dipole force" that exists <u /><em><u>between</u> </em> [molecules with hydrogen atoms] <em><u>and</u> </em> [molecules with nitrogen, oxygen, or fluorine atoms.] .
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Answer:
Developing the problem
Explanation:
In order to have an experiment or to make the experiment's data useful, you must first find a problem to solve.