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ryzh [129]
3 years ago
10

What type of bonding is found in the molecule HBr?

Chemistry
1 answer:
Liula [17]3 years ago
4 0
D)polar covalent because ionic and metallics include metals and nonpolar covalent wouldn’t work because it’s lewis for structure and en doesn’t match its requirements.
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How many moles are in sample containing 2.71 x 10^24 atoms of iron?
Deffense [45]

Answer:

4.5 moles

Explanation:

One mole is equal to 6.022 x 10^23 atoms

2.71 x 10^24 atoms * 1 mol/ 6.022 x 10^23 atoms = 4.5 moles

5 0
2 years ago
Which of the following is NOT true about one mole?
Lilit [14]
The option in “the mass of 1me of carbon equals the mass of 1 mole of boron atoms” is false since

Mass=molesx molar mass

And the molar masses of these two elements are different so their masses aren’t equal

7 0
3 years ago
What were the limitations of Dobereiner's classifcation?​
yawa3891 [41]

Answer:

Dobereiner could find only three triads; . i.e total of 9 elements only. However the total number of elements were more than that of those encompassed in Dobereiner's Triad

Explanation:

5 0
3 years ago
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If the molecular weight of a semiconductor is 27.9 grams/mole and the diamond lattice constant is 0.503 nm, what is the density
adelina 88 [10]

Explanation:

The given data is as follows.

      Mass = 27.9 g/mol

As we know that according to Avogadro's number there are 6.023 \times 10^{26} atom present in 1 mole. Therefore, weight of 1 atom will be as follows.

            1 atoms weight = \frac{38}{6.023 \times 10^{26}}    

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.

            = \frac{8 \times 27.9 g/mol}{6.023 \times 10^{26}}

            = 37.06 \times 10^{-26}

Now, we will calculate the volume of a lattice with lattice constant 'a' (cubic diamond) as follows.

                   = a^{3}

                   = (0.503 \times 10^{-9})^{3}

                   = 0.127 \times 10^{-27} m^{3}

Formula to calculate density of diamond cell is as follows.

               Density = \frac{mass}{volume}

                             = \frac{37.06 \times 10^{-26}}{0.127 \times 10^{-27} m^{3}}

                            = 2918.1 g/m^{3}

or,                         = 0.0029 g/cc       (as 1 m^{3} = 10^{6} cm^{3})

Thus, we can conclude that density of given semiconductor in grams/cc is 0.0029 g/cc.

4 0
3 years ago
A decrease in temperature reduces the volume of gas
Ksenya-84 [330]

Answer:

<em>Chemical</em><em> </em><em>change</em><em>.</em>

<em> </em><em> </em>

<em>Hope</em><em> </em><em>this</em><em> </em><em>helps</em><em> </em><em>you</em><em>.</em>

7 0
2 years ago
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