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BaLLatris [955]
3 years ago
13

A. Use the following steps to balance the redox reaction below. Al(s) Ni2 (aq) Ni(s) Al3 (aq)

Chemistry
1 answer:
Alja [10]3 years ago
8 0

Answer:

2Al(s) +3Ni²⁺(aq) ⟶ 2Al³⁺(aq) + 3Ni(s)  

Explanation:

The unbalanced equation is

Al(s) + Ni²⁺(aq) ⟶ Ni(s) + Al³⁺(aq)

(i) Half-reactions

Al(s) ⟶ Al³⁺(aq) + 3e⁻

Ni²⁺(aq) + 2e⁻ ⟶ Ni(s)

(ii) Balance charges

2 × [Al(s) ⟶ Al³⁺(aq) + 3e⁻]

3 × [Ni²⁺(aq) + 2e⁻ ⟶ Ni(s)]

gives

2Al(s) ⟶ 2Al³⁺(aq) + 6e⁻

3Ni²⁺(aq) + 6e⁻ ⟶ 3Ni(s)

(iii) Add equations

2Al(s) ⟶ 2Al³⁺(aq) + 6e⁻

<u>3Ni²⁺(aq) + 6e⁻ ⟶ 3Ni(s)                                       </u>  

2Al(s) +3Ni²⁺(aq) + <em>6e</em>⁻ ⟶ 2Al³⁺(aq) + 3Ni(s) + <em>6e⁻ </em>

Simplify (cancel electrons)

2Al(s) +3Ni²⁺(aq) ⟶ 2Al³⁺(aq) + 3Ni(s)

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The atoms of which element will typically form ions
Musya8 [376]

Ions are electrically charged particles, meaning, that whenever any type of atom loses or gains electrons it is forming a new type of structure, or ion.

Metal atoms form postive ions, Non-Metals form negative ions.

The strongest type of bond that can occur between two ions, when they are oppositely charged, are Ionic Bonds. The most commonly known ionic bond is seen in Sodium and Chlorine, which when combined give us NaCl or Sodium Chlorine, aka Salt.

+1 charge = Group 1, Metals

+2 charge = Group 2, Alkali Metals

-1 charge = Group 17, Halogens

Provided is a periodic table marked with the charges for each group.

7 0
3 years ago
Consider the reaction PCl5(g) ⇌ PCl3(g) + Cl2(g). If 0.02 moles of PCl5, 0.04 moles of PCl3, and 0.08 moles of Cl2 are combined
Furkat [3]

Answer:

The reaction quotient (Q) before the reaction is 0.32

Explanation:

Being the reaction:

aA + bB ⇔ cC + dD

Q=\frac{[C]^{c} *[D]^{d} }{[A]^{a}*[B]^{b}  }

where Q is the so-called reaction quotient and the concentrations expressed in it are not those of the equilibrium but those of the different reagents and products at a certain instant of the reaction.

The concentration will be calculated by:

Concentration=\frac{number of moles of solute}{Volume}

You know  the reaction:

PCl₅ (g) ⇌ PCl₃(g) + Cl₂(g).

So:

Q=\frac{[PCl_{3} ] *[Cl_{2} ] }{[PCl_{5} ]}

The concentrations are:

  • [PCl₃]=\frac{0.04 moles}{0.5 L} =0.08 \frac{moles}{L}
  • [Cl₂]=\frac{0.08 moles}{0.5 L} =0.16 \frac{moles}{L}
  • [PCl₅]=\frac{0.02 moles}{0.5 L} =0.04 \frac{moles}{L}

Replacing:

Q=\frac{0.08*0.16}{0.04}

Solving:

Q= 0.32

<u><em>The reaction quotient (Q) before the reaction is 0.32</em></u>

4 0
3 years ago
If 4.0 g of helium gas occupies a volume of 22.4 L at 0 o C and a pressure of 1.0 atm, what volume does 3.0 g of He occupy under
WINSTONCH [101]

Answer:

the volume occupied by 3.0 g of the gas is 16.8 L.

Explanation:

Given;

initial reacting mass of the helium gas, m₁ = 4.0 g

volume occupied by the helium gas, V = 22.4 L

pressure of the gas, P = 1 .0 atm

temperature of the gas, T = 0⁰C = 273 K

atomic mass of helium gas, M = 4.0 g/mol

initial number of moles of the gas is calculated as follows;

n_1 = \frac{m_1}{M} \\\\n_1 = \frac{4}{4} = 1

The number of moles of the gas when the reacting mass is 3.0 g;

m₂ = 3.0 g

n_2 = \frac{m_2}{M} \\\\n_2 = \frac{3}{4} \\\\n_2 = 0.75 \ mol

The volume of the gas at 0.75 mol is determined using ideal gas law;

PV = nRT

PV = nRT\\\\\frac{V}{n} = \frac{RT}{P} \\\\since, \ \frac{RT}{P} \ is \ constant,\  then;\\\frac{V_1}{n_1} = \frac{V_2}{n_2} \\\\V_2 = \frac{V_1n_2}{n_1} \\\\V_2 = \frac{22.4 \times 0.75}{1} \\\\V_2 = 16.8 \ L

Therefore, the volume occupied by 3.0 g of the gas is 16.8 L.

4 0
3 years ago
What does cellular respiration do?
Anton [14]
I believe the answer is A the 1st one
7 0
3 years ago
I need to find the reactants and products for each of these 7 problems. I need it in 3/5 hours help please!!!
Nezavi [6.7K]

Answer:

The arrow points from the reactants to the products, so just follow the arrows.

Explanation:

some have the reactants on the left and the products on the right, and others are the opposite... just know that

reactants---------> products

or

products<-----------reactants

4 0
4 years ago
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