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Arturiano [62]
3 years ago
6

A neutral atom of chlorine (Cl) has an average mass of 35 amu and 17 electrons. How many neutrons does it have?

Chemistry
2 answers:
uysha [10]3 years ago
7 0
It has 18 neutrons
hope this helps
nataly862011 [7]3 years ago
6 0
It would have 18 neutrons, as a neutral atom has the same number of electrons as protons, therefore the number of protons is 17. 35-17 leaves you with 18 neutrons
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Explanation:

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Should existing structures build from CCA-treated wood be removed?
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3 0
3 years ago
Which is the correct molar mass for the compound h2SO4
AlladinOne [14]

Explanation: just add the atomic masses on the periodic table together for each atom.

4 0
3 years ago
Teniendo en cuenta los siguientes fenómenos: ebullición del agua- movimiento de un cuerpo- disolución de sal en agua- combustión
miv72 [106K]

Answer:

Las siguientes son reacciones químicas;

combustión de leña

oxidación del hierro

descomposición del agua en hidrógeno y oxígeno

Explanation:

Una reacción química da como resultado la formación de una (s) sustancia (s) nueva (s), mientras que un cambio físico no conduce a la formación de una sustancia nueva.

Las siguientes son reacciones químicas;

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oxidación del hierro: La oxidación del hierro conduce a la formación de óxidos de hierro. Como; 2Fe (s) + O2 (g) ----> 2FeO (s)

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Todos estos procesos enumerados anteriormente conducen a la formación de nuevas sustancias, por lo tanto, son reacciones químicas.

3 0
3 years ago
Consider the reaction: CaCO3(s) à CaO(s) + CO2(g)
Vsevolod [243]

Answer:

131.5 kJ

Explanation:

Let's consider the following reaction.

CaCO₃(s) → CaO(s) + CO₂(g)

First, we will calculate the standard enthalpy of the reaction (ΔH°).

ΔH° = 1 mol × ΔH°f(CaO(s)) + 1 mol × ΔH°f(CO₂(g) ) - 1 mol × ΔH°f(CaCO₃(s) )

ΔH° = 1 mol × (-634.9 kJ/mol) + 1 mol × (-393.5 kJ/mol) - 1 mol × (-1207.6 kJ/mol)

ΔH° = 179.2 kJ

Then, we calculate the standard entropy of the reaction (ΔS°).

ΔS° = 1 mol × S°(CaO(s)) + 1 mol × S°(CO₂(g) ) - 1 mol × S°(CaCO₃(s) )

ΔS° = 1 mol × (38.1 J/mol.K) + 1 mol × (213.8 J/mol.K) - 1 mol × (91.7 J/mol.K)

ΔS° = 160.2 J/K = 0.1602 kJ/K

Finally, we calculate the standard Gibbs free energy of the reaction at T = 25°C = 298 K.

ΔG° = ΔH° - T × ΔS°

ΔG° = 179.2 kJ - 298 K × 0.1602 kJ/K

ΔG° = 131.5 kJ

6 0
3 years ago
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