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UNO [17]
3 years ago
10

The element that make up a ---------------- are chemically combined

Chemistry
2 answers:
Bas_tet [7]3 years ago
5 0
Compound is your answer
Nimfa-mama [501]3 years ago
5 0
The elements that make up a compound are chemically combined.

Hope this helps :)
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Suppose you have a dozen carbon atoms, a dozen gold atoms, and a dozen iron atoms. Even though you have the same number of each,
mixer [17]

Answer:

By weight they have the same mass, but the number of atoms is different

Explanation:

3 0
4 years ago
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According to the law of conservation of matter, what cannot change during a chemical reaction?.
Hunter-Best [27]

Answer:

the number of atoms

Explanation:

atoms are never created not destroyed

8 0
3 years ago
If 20.5 g of chlorine is reacted with 20.5 g of sodium, which reactant is in excess? 1. cl 2. neither is in excess. 3. na
zlopas [31]
2. neither is in excess is the answer.
5 0
3 years ago
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A solution of 2-propanol and 1-octanol behaves ideally. Calculate the chemical potential of 2-propanol in solution relative to t
andrew-mc [135]

Answer:

The chemical potential of 2-propanol in solution relative to that of pure 2-propanol is lower by 2.63x10⁻³.    

Explanation:

The chemical potential of 2-propanol in solution relative to that of pure 2-propanol can be calculated using the following equation:

\mu (l) = \mu ^{\circ} (l) + R*T*ln(x)

<u>Where:</u>

<em>μ (l): is the chemical potential of 2-propanol in solution    </em>

<em>μ° (l): is the chemical potential of pure 2-propanol   </em>

<em>R: is the gas constant = 8.314 J K⁻¹ mol⁻¹ </em>

<em>T: is the temperature = 82.3 °C = 355.3 K </em>

<em>x: is the mole fraction of 2-propanol = 0.41 </em>

\mu (l) = \mu ^{\circ} (l) + 8.314 \frac{J}{K*mol}*355.3 K*ln(0.41)

\mu (l) = \mu ^{\circ} (l) - 2.63 \cdot 10^{3} J*mol^{-1}

\mu (l) - \mu ^{\circ} (l) = - 2.63 \cdot 10^{3} J*mol^{-1}  

Therefore, the chemical potential of 2-propanol in solution relative to that of pure 2-propanol is lower by 2.63x10⁻³.    

I hope it helps you!    

8 0
4 years ago
How many seconds are required to produce 4.00 g of aluminum metal from the electrolysis of molten alcl3 with an electrical curre
Brilliant_brown [7]
The correct answer is e. 3.57×10³
Al³+(aq) + 3e→AL(s)
4.00g of AL=4g/26.98 g/mol= 0.1483 mol
t=znF/1 where t is time in seceonds.
Z= valency number of ions of the substance or electrons which are transferred per ion
F= Faraday's constant
I = electric current in'A'CA  C/s
t=(3×0.1483 mol ×96485 C/mol) /12(C15)
t=3577 second = 3.5 ×10³s
8 0
3 years ago
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