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Elenna [48]
3 years ago
5

Complete and balance the following acid-base equations:

Chemistry
1 answer:
Alchen [17]3 years ago
7 0

Answer:

a) HClO_4 (aq) + LiOH \rightarrow LiClO_4 + H_2O

b) H_2SO_4(aq) + 2NaOH(aq) \rightarrow Na_2SO_4 (aq) + 2H_2O (l)

c) Ba(OH)2 + 2HF\rightarrow BaF_2 + 2H_2O

Explanation:

a)

when HClO_4 is added to LiOH, lithium chlorate and water is formed.

HClO_4 (aq) + LiOH \rightarrow LiClO_4 + H_2O

Balancing of above reaction,

It can be seen that all the atoms in both the sides are balanced. so it is a balanced reaction.

(b) When aqueous H_2SO_4 is added to NaOH, Na_2SO_4 and H_2O is formed. It is a neutrilization reaction.

H_2SO_4(aq) + NaOH(aq) \rightarrow Na_2SO_4 (aq) + H_2O (l)

Balancing of above reaction,

First balance all the atoms except O and H

S atom is already balanced in either side

No. of Na atom in left hand side = 1

No. of Na atom in right hand side = 2

So multiply NaOH by 2, now the reaction becomes:

H_2SO_4(aq) + 2NaOH(aq) \rightarrow Na_2SO_4 (aq) + H_2O (l)

No. of O atoms in left hand side = 6

No. of O atoms in right hand side = 5

So multiply H2O by 2, now the reaction becomes:

H_2SO_4(aq) + 2NaOH(aq) \rightarrow Na_2SO_4 (aq) + 2H_2O (l)

Now, it can been seen that all the atoms are balanced.

c) When Ba(OH)2 reacts with HF, baroum fluoride and water is formed.

Ba(OH)2 + HF\rightarrow BaF_2 + H_2O

Balancing of above reaction,

Barium atoms are already balanced on either side.

No. of F atoms on left hand side = 1

No. of F atoms on right hand side = 2

So, multiply HF by 2, now reaction becomes

Ba(OH)2 + 2HF\rightarrow BaF_2 + H_2O

In order to balance H and O on either side, multiply H2O by 2.

Ba(OH)2 + 2HF\rightarrow BaF_2 + 2H_2O

You might be interested in
How many dimes could you trade for 360 pesos? $1 = 1500 pesos.
sveta [45]

Answer:

The correct option is (d).

Explanation:

It is given that,

1$ = 1500 pesos

We need to convert 360 pesos into dimes

We can convert 360 pesos to dollars as follows:

360\ \text{pesos}=\$\dfrac{1}{1500}\times 360\\\\=$0.24

360 pesos is equal to $0.24

Also, 1 dollar = 10 dimes

We can covert 0.24 dollar to dimes as follows :

0.24 dollar = 10 × 0.24 dimes

0.24 dollar = 2.4 dimes

or

360 pesos = 2.4 dimes

7 0
3 years ago
What is the total number of atoms in 0.20 mol of propanone, CH3COCH3?
Nutka1998 [239]

Answer:

1.2×10²³ atoms.

Explanation:

Data obtained from the question include:

Mole of propanone = 0.20 mole

Number of atoms of propanone =.?

From Avogadro's hypothesis, we understood that 1 mole of any substance contains 6.022×10²³ atoms.

This implies that 1 mole of propanone also contains 6.022×10²³ atoms.

Thus, we can obtain the number of atoms in 0.20 mole of propanone as illustrated below:

1 mole of propanone contains 6.022×10²³ atoms.

Therefore, 0.20 mole of propanone will contain = 0.2 × 6.022×10²³ = 1.2×10²³ atoms.

Thus, 0.20 mole of propanone contain

1.2×10²³ atoms.

6 0
3 years ago
¿Cuáles son las características de los materiales de laboratorio? Por ejemplo: exactitud, resistencia a la temperatura, etc.
boyakko [2]

Answer:

lo siento, no sé punto libre. :p

7 0
3 years ago
0.65743 In scientific notation
Goryan [66]
202827.0000 is the answer I think idrk
6 0
3 years ago
For the reaction o(g) + o2(g) → o3(g) δh o = −107.2 kj/mol given that the bond enthalpy in o2(g) is 498.7 kj/mol, calculate the
Aneli [31]
The reaction;
O(g) +O2(g)→O3(g), ΔH = sum of bond enthalpy of reactants-sum of food enthalpy of products.
ΔH = ( bond enthalpy of O(g)+bond enthalpy of O2 (g) - bond enthalpy of O3(g)
-107.2 kJ/mol = O+487.7kJ/mol =O+487.7 kJ/mol +487.7kJ/mol =594.9 kJ/mol
Bond enthalpy (BE) of O3(g) is equals to 2× bond enthalpy of O3(g) because, O3(g) has two types of bonds from its lewis structure (0-0=0).
∴2BE of O3(g) = 594.9kJ/mol
Average bond enthalpy = 594.9kJ/mol/2
=297.45kJ/mol
∴ Averange bond enthalpy of O3(g) is 297.45kJ/mol.
8 0
3 years ago
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