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Alla [95]
3 years ago
10

Does a precipitate form when a solution of calcium chloride and a solution of mercury(I) nitrate are mixed together? Write the n

et ionic equation.
Chemistry
1 answer:
olganol [36]3 years ago
7 0

Answer : Yes, a precipitate form when a solution of calcium chloride and a solution of mercury(I) nitrate are mixed together.

The net ionic equation will be,

2Hg^{+}(aq)+2Cl^{-}(aq)\rightarrow Hg_2Cl_2(s)

Explanation :

In the net ionic equations, we are not include the spectator ions in the equations.

Spectator ions : The ions present on reactant and product side which do not participate in a reactions. The same ions present on both the sides.

The given balanced ionic equation will be,

CaCl_2(aq)+2HgNO_3(aq)\rightarrow Ca(NO_3)_2(aq)+Hg_2Cl_2(s)

The ionic equation in separated aqueous solution will be,

Ca^{2+}(aq)+2Cl^{-}(aq)+2Hg^{+}(aq)+2NO_3^{-}(aq)\rightarrow Hg_2Cl_2(s)+Ca^{2+}(aq)+2NO_3^{-}(aq)

In this equation, Ca^{2+}\text{ and }NO_3^- are the spectator ions.

By removing the spectator ions from the balanced ionic equation, we get the net ionic equation.

The net ionic equation will be,

2Hg^{+}(aq)+2Cl^{-}(aq)\rightarrow Hg_2Cl_2(s)

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What is the percentage yield of O2 if 12.3 g of KClO3 (molar mass 123 g) is decomposed to produce 3.2 g of O2 (molar mass 32 g)
My name is Ann [436]

Answer:

The percentage yield of O2 is 66.7%

Explanation:

Reaction for decomposition of potassium chlorate is:

2KClO₃ →  2KCl  +  3O₂

The products are potassium chloride and oxygen.

Let's find out the moles of chlorate.

Mass / Molar mass = Moles

12.3 g / 123 g/mol = 0.1 mol

So ratio is 2:3, 2 moles of chlorate produce 3 mol of oxygen.

Then, 0.1 mol of chlorate may produce (0.1  .3)/ 2 = 0.15 moles

Let's convert the moles of produced oxygen, as to find out the theoretical yield.

0.15 mol . 32 g/ 1mol = 4.8 g

To calculate the percentage yield, the formula is

(Produced Yield / Theoretical yield) . 100 =

(3.2g / 4.8g) . 100 = 66.7 %

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2 years ago
Read 2 more answers
30. The density of an unknown gas at 27°C and 2 atm pressure is equal with density of N2 gas at
Zanzabum

Answer:

Molar mass of the unknown gas is 64.6 g/mol

Explanation:

Let's think this excersise with the Ideal Gases Law.

We start from the N₂. At STP conditions we know that 1 mol of anything occupies 22.4L.

We apply: P . V = n . R . T

5 atm . V = 1 mol . 0.082 . 325K

V = (1 mol . 0.082 . 325K) / 5 atm = 5.33 L

It is reasonable to say that, if we have more pressure, we may have less volume.

As this is the volume for 1 mol of N₂, our mass is 28 g. Then, the density of the nitrogen and the unknown gas is 28 g/5.33L = 5.25 g/L

Our unknown gas has, this density at 27°C and 2 atm.

If we star from this, again: 1 mol of any gas occupy 22.4L at STP, we can calculate the volume for 1 mol at those conditions:

P₁ . V₁ / T₁ = P₂ . V₂ / T₂

1 atm . 22,4L / 273K = 2 atm . V₂ / 300K

Remember that the value for T° is Absolute (T°C + 273)

[ (1 atm . 22.4L / 273K) . 300K] / 2 atm = V₂ → 12.3L

This is the volume for 1 mol of the unknown gas at 2 atm and 27°C

We use density to determine the mass: 12.3 L . 5.25 g/L = 64.6 g

That's the molar mass: 64.6 g/mol

6 0
2 years ago
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