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AURORKA [14]
3 years ago
11

3. What noble gas would be part of the electron configuration notation for Mn?

Chemistry
1 answer:
shusha [124]3 years ago
7 0

Answer:

Argon {Ar}

Explanation:

The noble gas used for a condensed electron configuration is the one before the element which you are configuring.

In this case, the element (Mn) is manganese

The noble gas that is before this element is Argon which is the row above it

so your configuration would be {Ar} 3d^5 4s^2

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Titanium is a transition metal used in many alloys because it is extremely strong and lightweight. Titanium tetrachloride (TiCl4
vova2212 [387]

Answer:

a) 226.6 grams of Cl₂

b) 19.2 grams of C

c) 303.2 grams of TiCl₄ and 70.4 grams of CO₂

Explanation:

The balanced chemical reaction is the following:

TiO₂(s) + C(s) + 2 Cl₂(g) → TiCl₄(s) + CO₂(g)

(a) What mass of Cl₂ gas is needed to react with 1.60 mol TiO₂?

From the chemical equation, 1 mol of TiO₂ reacts with 2 moles of Cl₂. So, the stoichiometric ratio is 2 mol Cl₂/1 mol TiO₂. We multiply this ratio by the moles of TiO₂ we have to calculate the moles of Cl₂ we need:

1.60 mol TiO₂ x 2 mol Cl₂/1 mol TiO₂ = 3.2 mol Cl₂

Now, we convert from moles to mass by using the molecular weight (MW) of Cl₂:

MW(Cl₂) = 35.4 g/mol x 2 = 70.8 g/mol

mass of Cl₂= 3.2 mol x 70.8 g/mol = 226.6 g

<em>Therefore, 226.6 grams of Cl₂ are needed to react with 1.6 mol of TiO₂. </em>

(b) What mass of C is needed to react with 1.60 mol of TiO₂?

From the chemical equation, 1 mol of TiO₂ reacts with 1 moles of C(s). So, the stoichiometric ratio is 1 mol C/1 mol TiO₂. We multiply this ratio by the moles of TiO₂ we have to calculate the moles of C(s) we need:

1.60 mol TiO₂ x 1 mol C(s)/1 mol TiO₂ = 1.60 mol C(s)

So, we convert the moles of C(s) to grams as follows:

MW(C) = 12 g/mol

1.60 mol x 12 g/mol = 19.2 g C(s)

<em>Therefore, a mass of 19.2 grams of C is needed to react with 1.60 mol of TiO₂. </em>

(c) What is the mass of all the products formed by reaction with 1.60 mol of TiO₂?

From the chemical equation, we can notice that 1 mol of TiO₂ produces 1 mol of TiCl₄ and 1 mol of CO₂. So, from 1.60 moles of TiO₂, 1 mol of each product will be produced:

1 mol TiO₂/1 mol TiCl₄ ⇒ 1.60 mol TiO₂/1.60 mol TiCl₄

1 mol TiO₂/1 mol CO₂ ⇒ 1.60 mol TiO₂/1.60 mol CO₂

Finally, we convert the moles to grams by using the molecular weight of each compound:

MW(TiCl₄) = 47.9 g/mol Ti + (35.4 g/mol x 4 Cl) = 189.5 g/mol

1.60 mol x 189.5 g/mol = 303.2 g

MW(CO₂) = 12 g/mol C + (16 g/mol x 2 O) = 44 g/mol

1.60 mol x 44 g/mol = 70.4 g

<em>Therefore, from the reaction of 1.60 mol of TiO₂ are formed 303.2 grams of TiCl₄ and 70.4 grams of CO₂.</em>

3 0
3 years ago
20. What is the area of a piece of metal foil that measures 43.9 cm by 29.21 cm? Express the answer to
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Answer:

The correct answer is b. 1280 cm^2

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Answer: which*

Explanation:

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Copper oxide, CuO, reacts with hydrochloric acid, HCI, to produce copper chloride, CuCL2 and water
spayn [35]

Explanation:

El óxido de cobre (II), también llamado antiguamente óxido cúprico ({\displaystyle {\ce {CuO}}}{\displaystyle {\ce {CuO}}}), es el óxido de cobre con mayor número de oxidación. Como mineral se conoce como tenorita.

{\displaystyle {\ce {2Cu + O2 = 2CuO}}}{\displaystyle {\ce {2Cu + O2 = 2CuO}}}

Aquí, se forma junto con algo de óxido de cobre (I) como un producto lateral, por lo que es mejor prepararlo por calentamiento de nitrato de cobre (II), hidróxido de cobre (II) o carbonato de cobre (II):

{\displaystyle {\ce {2 Cu(NO3)2 = 2 CuO + 4 NO2+ O2}}}{\displaystyle {\ce {2 Cu(NO3)2 = 2 CuO + 4 NO2+ O2}}}

{\displaystyle {\ce {Cu(OH)2 (s) = CuO (s) + H2O (l)}}}{\displaystyle {\ce {Cu(OH)2 (s) = CuO (s) + H2O (l)}}}

{\displaystyle {\ce {CuCO3 = CuO + CO2}}}{\displaystyle {\ce {CuCO3 = CuO + CO2}}}

El óxido de cobre (II) es un óxido básico, así se disuelve en ácidos minerales tales como el ácido clorhídrico, el ácido sulfúrico o el ácido nítrico para dar las correspondientes sales de cobre (II):

{\displaystyle {\ce {CuO + 2 HNO3 = Cu(NO3)2 + H2O}}}{\displaystyle {\ce {CuO + 2 HNO3 = Cu(NO3)2 + H2O}}}

{\displaystyle {\ce {CuO + 2 HCl =CuCl2 + H2O}}}{\displaystyle {\ce {CuO + 2 HCl =CuCl2 + H2O}}}

{\displaystyle {\ce {CuO + H2SO4 = CuSO4 + H2O}}}{\displaystyle {\ce {CuO + H2SO4 = CuSO4 + H2O}}}

Reacciona con álcali concentrado para formar las correspondientes sales cuprato.

{\displaystyle {\ce {3 XOH + CuO + H2O = X3[Cu(OH)6]}}}{\displaystyle {\ce {3 XOH + CuO + H2O = X3[Cu(OH)6]}}}

Puede reducirse a cobre metálico usando hidrógeno o monóxido de carbono:

{\displaystyle {\ce {CuO + H2 = Cu + H2O}}}{\displaystyle {\ce {CuO + H2 = Cu + H2O}}}

{\displaystyle {\ce {CuO + CO = Cu + CO2}}}{\displaystyle {\ce {CuO + CO = Cu + CO2}}}

6 0
3 years ago
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