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NNADVOKAT [17]
2 years ago
8

What does thermal energy refer to?

Chemistry
2 answers:
Trava [24]2 years ago
7 0

Answer:

c

Explanation:

ludmilkaskok [199]2 years ago
5 0

my answer is D sorry if I'm wrong

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The process of an atom releasing energy when it moves to a lower energy state is called
Inga [223]
The process of an atom releasing energy when it moves to a lower energy state is called emission.
4 0
3 years ago
: calculate the mass of the solute in 1.500 l of 0.30 m glucose, c6h12o6, used for intravenous injection
vfiekz [6]
Answer is: <span>the mass of the glucose is 81,07 grams.
</span>c(C₆H₁₂O₆) = 0,3 M = 0,3 mol/L.
V(C₆H₁₂O₆) = 1,500 L.
n(C₆H₁₂O₆) = c(C₆H₁₂O₆) · V(C₆H₁₂O₆).
n(C₆H₁₂O₆) = 0,3 mol/L · 1,5 L.
n(C₆H₁₂O₆) = 0,45 mol.
m(C₆H₁₂O₆) = n(C₆H₁₂O₆) · M(C₆H₁₂O₆).
m(C₆H₁₂O₆) = 0,45 mol · 180,156 g/mol.
m(C₆H₁₂O₆) = 81,07 g.
8 0
3 years ago
Can someone help please!
trasher [3.6K]

Answer:

Mixtures

Explanation:

7 0
3 years ago
Read 2 more answers
2 . what is the formula weight of (nh4)2so4? 118 amu 116 amu 100 amu 132 amu
Nataly [62]
Hey there!

(NH₄)₂SO₄ = 14 * 2 + 1 * 8 + 32+ 16 * 4 => 132 amu


6 0
3 years ago
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
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