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4 years ago

AlCl3 + NaOH —> Al(OH)3 + NaCl

Chemistry

1 answer:

victus00 [196]4 years ago

6 0

Hey there!

AlCl₃ + NaOH → Al(OH)₃ + NaCl

Balance OH.

1 on the left, 3 on the right. Add a coefficient of 3 in front of NaOH.

AlCl₃ + 3NaOH → Al(OH)₃ + NaCl

Balance Cl.

3 on the left, 1 on the right. Add a coefficient of 3 in front of NaCl.

AlCl₃ + 3NaOH → Al(OH)₃ + 3NaCl

Balance Na.

3 on the left, 3 on the right. Already balanced.

Balance Al.

1 on the left, 1 on the right. Already balanced.

Our final balanced equation: AlCl₃ + 3NaOH → Al(OH)₃ + 3NaCl

Hope this helps!

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If you took a spoonful of tea out of the cup, which would be more painful if it spilled on you? *

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Calculate the energy (in kj/mol) required to remove the electron in the ground state for each of the following one-electron spec

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Explanation:

$E_n=-13.6\times \frac{Z^2}{n^2}ev$

where,

$E_n$ = energy of $n^{th}$ orbit

n = number of orbit

Z = atomic number

a) Energy change due to transition from n = 1 to n = ∞ ,hydrogen atom .

Z = 1

Energy of n = 1 in an hydrogen like atom:

$E_1=-13.6\times \frac{1^2}{1^2}eV=-13.6 eV$

Energy of n = ∞ in an hydrogen like atom:

$E_{\infty}=-13.6\times \frac{1^2}{(\infty)^2}eV=0$

Let energy change be E for 1 atom.

$E=E_{\infty}-E_1=0-(-13.6 eV)=13.6 eV$

$1 mole = 6.022\times 10^{-23}$

Energy for 1 mole = E'

$E'=6.022\times 10^{-23} mol^{-1}\times 13.6 eV$

$1 eV=1.60218\times 10^{-22} kJ$

$E'=6.022\times 10^{23}\times 13.6 \times 1.60218\times 10^{-22} kJ/mol$

$E'=1,312.17 kJ/mol$

The energy required to remove the electron in the ground state is 1,312.17 kJ/mol.

b) Energy change due to transition from n = 1 to n = ∞ , $B^{4+}$ atom .

Z = 5

Energy of n = 1 in an hydrogen like atom:

$E_1=-13.6\times \frac{5^2}{1^2}eV=-340 eV$

Energy of n = ∞ in an hydrogen like atom:

$E_{\infty}=-13.6\times \frac{5^2}{(\infty)^2}eV=0$

Let energy change be E.

$E=E_{\infty}-E_1=0-(-340eV)=340 eV$

$1 mole = 6.022\times 10^{-23}$

Energy for 1 mole = E'

$E'=6.022\times 10^{-23} mol^{-1}\times 340eV$

$1 eV=1.60218\times 10^{-22} kJ$

$E'=6.022\times 10^{23}\times 340\times 1.60218\times 10^{-22} kJ/mol$

$E'=32,804.31 kJ/mol$

The energy required to remove the electron in the ground state is 32,804.31 kJ/mol.

c) Energy change due to transition from n = 1 to n = ∞ , $Li^{2+}$ atom .

Z = 3

Energy of n = 1 in an hydrogen like atom:

$E_1=-13.6\times \frac{3^2}{1^2}eV=-122.4 eV$

Energy of n = ∞ in an hydrogen like atom:

$E_{\infty}=-13.6\times \frac{3^2}{(\infty)^2}eV=0$

Let energy change be E.

$E=E_{\infty}-E_1=0-(-122.4 eV)=122.4 eV$

$1 mole = 6.022\times 10^{-23}$

Energy for 1 mole = E'

$E'=6.022\times 10^{-23} mol^{-1}\times 122.4 eV$

$1 eV=1.60218\times 10^{-22} kJ$

$E'=6.022\times 10^{23}\times 122.4\times 1.60218\times 10^{-22} kJ/mol$

$E'=11,809.55 kJ/mol$

The energy required to remove the electron in the ground state is 11,809.55 kJ/mol.

d) Energy change due to transition from n = 1 to n = ∞ , $Mn^{24+}$ atom .

Z = 25

Energy of n = 1 in an hydrogen like atom:

$E_1=-13.6\times \frac{25^2}{1^2}eV=-8,500 eV$

Energy of n = ∞ in an hydrogen like atom:

$E_{\infty}=-13.6\times \frac{25^2}{(\infty)^2}eV=0$

Let energy change be E.

$E=E_{\infty}-E_1=0-(-8,500 eV)=8,500 eV$

$1 mole = 6.022\times 10^{-23}$

Energy for 1 mole = E'

$E'=6.022\times 10^{-23} mol^{-1}\times 8,500eV$

$1 eV=1.60218\times 10^{-22} kJ$

$E'=6.022\times 10^{23}\times 8,500 \times 1.60218\times 10^{-22} kJ/mol$

$E'=820,107.88 kJ/mol$

The energy required to remove the electron in the ground state is 820,107.88 kJ/mol.

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Answer:

a. Al(s) ⇄ Al⁺³(aq) + 3e⁻ (oxidation)

Mg²⁺(aq) + 2e⁻ ⇄ Mg(s) (reduction)

b. ΔE° = + 0.715 V

c. It's an electrolytic cell, because it's a nonspontaneous reaction.

d. 2Al(s) + 3Mg²⁺(aq) ⇄ 2Al⁺³(aq) + 3Mg(s)

Explanation:

a. By the notation given, first is represented the oxidation reaction and then the reduction reaction, so they are:

Al(s) ⇄ Al⁺³(aq) + 3e⁻ (oxidation)

Mg²⁺(aq) + 2e⁻ ⇄ Mg(s) (reduction)

b. The standard potential of the cell (ΔE°) is the reduction potential of the oxidation less the reduction potential of the reduction. The reduction potentials are:

Al(s) = -1.66 V

Mg(s) = -2.375 V

ΔE° = -1.66 - (-2.375)

ΔE° = + 0.715 V

c. It's an electrolytic cell.

A galvanic cell is spontaneous, so the cathode (reduction) has a higher E° than the cathode (oxidation). In this case, the oxidation reaction has a higher E°, so the reaction is nonspontaneous and it's necessary an external force to it happen, so it's an electrolytic cell.

d. 2Al(s) + 3Mg²⁺(aq) ⇄ 2Al⁺³(aq) + 3Mg(s)

The number of electrons must be the same, so the oxidation reaction is multiplied by 2, and the reduction reaction by 3.

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