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Likurg_2 [28]
4 years ago
13

Write a balance equation for the following.

Chemistry
1 answer:
fomenos4 years ago
6 0
First you must determing the formulas of both of the compounds

Tin (II) Chloride

the (II) means that Tin is 2+ in this compounds; and we know that Cl is -
So the formula for Tin (II) Chloride has to be

SnCl₂

For Sodium Sulfide we know that Sodium is + and Sulfide Ions are 2- so the formula for Sodium Sulfide has to be

Na₂S

As of right now our equation is

SnCl₂ + Na₂S →

When these two compounds react they will form Sodium Chloride and Tin Sulfide

The equations for the two products will be

NaCl      and SnS

now our equation looks like this

SnCl₂ (aq) + Na₂S (aq) → NaCl + SnS

for this to be complete however, we need to first add phases to each of the products and balance the equation

We know that all Group one elements are soluble in water so this means NaCl is soluble


SnCl₂ (aq) + Na₂S (aq) → NaCl (aq) + SnS

we also know that all sulfides are insoluble which means SnS is insoluble

SnCl₂ (aq) + Na₂S (aq) → NaCl (aq) + SnS (s)

Now that we have the phases written in we must balance the equation
There is 1 Sn, 2Cl, 2 Na and 1 S on the left/reactants side
 and 1 Na, 1 Cl, 1Sn, and 1S on the right/products side

to balance it we need to have 2 Cl on the right hand or products side

SnCl₂ (aq) + Na₂S (aq) → 2 NaCl (aq) + SnS

Now everything is balanced!

SnCl₂ (aq) + Na₂S (aq) → 2 NaCl (aq) + SnS is our final reaction equation


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addition of sulfurous acid (a weak acid) to barium hydroxide ( a strong base) results in the formation of a precipitate. The net
Trava [24]

net ionic equation:

SO₄²⁻ (aq) + Ba²⁺ (aq) → BaSO₄ (s)

Explanation:

We have the following chemical reaction:

H₂SO₄ (aq) + Ba(OH)₂ (aq) → BaSO₄ (s) + 2 H₂O (l)

Now we write the ionic equation:

2 H⁺ (aq) + SO₄²⁻ (aq) + Ba²⁺ (aq) + 2 OH⁻ (aq) → BaSO₄ (s) + 2 H⁺ + 2 OH⁻ (aq)

We remove the spectator ions to obtain the net ionic equation:

SO₄²⁻ (aq) + Ba²⁺ (aq) → BaSO₄ (s)

where we have:

(aq) - aqueous

(s) - solid

(l) - liquid

Learn more about:

net ionic equation

brainly.com/question/7018960

#learnwithBrainly

7 0
3 years ago
Balance:<br> Fe2(SO4)3 +<br> _K[OH] —&gt; _K2[SO4] +<br> K[SO] + Fe[OH]3
LenKa [72]

Balance the reaction of Fe2(SO4)3 + KOH = K2SO4 + Fe(OH)3 using this chemical equation balancer! ... Fe2(SO4)3 + 6KOH → 3K2SO4 + 2Fe(OH)3 ...

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6 0
3 years ago
Read 2 more answers
How many moles of cesium (Cs) atoms are in 675 g Cs?
snow_tiger [21]
Just divide 675 by the atomic number of cesium and you will end up with

5.078798426627975
3 0
3 years ago
When 8.0 g H₂ react with 8.0 g O₂ in the reaction 2H₂ + O₂ → 2H₂O, what are the theoretical yield and the limiting reactant?
True [87]

Answer:

Now, we have to determine the limiting reagent.

Now, we have to determine the limiting reagent.4 g of H₂ reacts with 32 g of O₂ 1 g of H₂ reacts with 32/4 g of O₂ 3 g of H₂ reacts with 32/4 x 3 = 24 g of

Now, we have to determine the limiting reagent.4 g of H₂ reacts with 32 g of O₂ 1 g of H₂ reacts with 32/4 g of O₂ 3 g of H₂ reacts with 32/4 x 3 = 24 g ofBut according to the question, 29 g of O₂ is present. 2

Now, we have to determine the limiting reagent.4 g of H₂ reacts with 32 g of O₂ 1 g of H₂ reacts with 32/4 g of O₂ 3 g of H₂ reacts with 32/4 x 3 = 24 g ofBut according to the question, 29 g of O₂ is present. 2So, the limiting reactant is hydrogen.

Now, we have to determine the limiting reagent.4 g of H₂ reacts with 32 g of O₂ 1 g of H₂ reacts with 32/4 g of O₂ 3 g of H₂ reacts with 32/4 x 3 = 24 g ofBut according to the question, 29 g of O₂ is present. 2So, the limiting reactant is hydrogen.Now, 4 g of H₂ forms 36 g of H₂O

Now, we have to determine the limiting reagent.4 g of H₂ reacts with 32 g of O₂ 1 g of H₂ reacts with 32/4 g of O₂ 3 g of H₂ reacts with 32/4 x 3 = 24 g ofBut according to the question, 29 g of O₂ is present. 2So, the limiting reactant is hydrogen.Now, 4 g of H₂ forms 36 g of H₂O1 g of H₂ forms 36/4 g of H₂O. 3 g of H₂ forms 36/4 x 3 = 27 g of H₂O

Now, we have to determine the limiting reagent.4 g of H₂ reacts with 32 g of O₂ 1 g of H₂ reacts with 32/4 g of O₂ 3 g of H₂ reacts with 32/4 x 3 = 24 g ofBut according to the question, 29 g of O₂ is present. 2So, the limiting reactant is hydrogen.Now, 4 g of H₂ forms 36 g of H₂O1 g of H₂ forms 36/4 g of H₂O. 3 g of H₂ forms 36/4 x 3 = 27 g of H₂OMaximum amount of water that can be formed is 27 g.

Now, we have to determine the limiting reagent.4 g of H₂ reacts with 32 g of O₂ 1 g of H₂ reacts with 32/4 g of O₂ 3 g of H₂ reacts with 32/4 x 3 = 24 g ofBut according to the question, 29 g of O₂ is present. 2So, the limiting reactant is hydrogen.Now, 4 g of H₂ forms 36 g of H₂O1 g of H₂ forms 36/4 g of H₂O. 3 g of H₂ forms 36/4 x 3 = 27 g of H₂OMaximum amount of water that can be formed is 27 g.For, amount of oxygen left of unreacted, Only 24 g of oxygen will react.

Now, we have to determine the limiting reagent.4 g of H₂ reacts with 32 g of O₂ 1 g of H₂ reacts with 32/4 g of O₂ 3 g of H₂ reacts with 32/4 x 3 = 24 g ofBut according to the question, 29 g of O₂ is present. 2So, the limiting reactant is hydrogen.Now, 4 g of H₂ forms 36 g of H₂O1 g of H₂ forms 36/4 g of H₂O. 3 g of H₂ forms 36/4 x 3 = 27 g of H₂OMaximum amount of water that can be formed is 27 g.For, amount of oxygen left of unreacted, Only 24 g of oxygen will react.But 29 g is the given amount. Amount of oxygen unreacted = 29 - 24 = 5 g

7 0
3 years ago
What is formed when sediment is blown against an obstacle and settles behind it
Naddik [55]
The answer should be "Dunes"
4 0
3 years ago
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