Explanation:
The given data is as follows.
Mass = 27.9 g/mol
As we know that according to Avogadro's number there are
atom present in 1 mole. Therefore, weight of 1 atom will be as follows.
1 atoms weight =
In a diamond cubic cell, the number of atoms are 8. So, n = 8 for diamond cubic cell.
Therefore, total weight of atoms in a unit cell will be as follows.
= 
= 
Now, we will calculate the volume of a lattice with lattice constant 'a' (cubic diamond) as follows.
= 
= 
=
Formula to calculate density of diamond cell is as follows.
Density = 
= 
= 2918.1 
or, = 0.0029 g/cc (as 1
)
Thus, we can conclude that density of given semiconductor in grams/cc is 0.0029 g/cc.
Hello!
Given the density of mercury being 13.57 g/mL, and the mass of 52 kilograms, we need to find the volume.
To find the volume, we need to divide mass by density (V = m/d).
Notice that you are given 52 kilograms, but not grams. To convert kilograms to grams, you need to multiply it by 1000.
52 x 1000 = 52000 grams
With the correct measurements, we can find the volume.
V = 52000 grams / 13.56 grams/milliliter
V ≈ 3834.80826
Therefore, the volume of the mercury is about 3,834.81 mL.
Answer:
Are barium and chlorine a metal and nonmetal?
Explanation:
I had the same question, and my class reviewed the test and this was the answer to this question.
Answer:
you need to have a question or a problem first
63.1 g of NaN3 is formed when 50.0 g of sodium is reacted with 40.5 g of nitrogen according to the balanced chemical reaction
The balanced reaction equation is;
2 Na + 3 N2--> 2 NaN3
Number of moles of Na = 50.0 g/23 g/mol = 2.17 moles of Na
Number of moles of Nitrogen = 40.5 g/28 g/mol = 1.45 moles of N2
We have to obtain the limiting reactant, this is the reactant that yields the least number of moles of product.
For Na
2 moles of Na yields 2 moles of NaN3
2.17 moles of Na yields 2.17 moles of NaN3 (reaction is 1:1).
For N2
3 moles of N2 yields 2 moles of NaN3
1.45 moles of N2 yields 1.45 * 2/3 = 0.97 moles of NaN3
So, N2 is the limiting reactant. Mass of product formed depends on the limiting reactant.
Mass of NaN3 = 0.97 moles of NaN3 * 65 g/mol = 63.1 g of NaN3
Therefore, 63.1 g of NaN3 is formed when 50.0 g of sodium is reacted with 40.5 g of nitrogen according to the balanced chemical reaction
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