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spin [16.1K]
2 years ago
15

You are presented with a white solid and told that due to careless labeling it is not clear if the substance is barium chloride,

lead chloride, or zinc chloride.when you transfer the solid to a beaker and add water,the solid dissolves to give a clear solution.next a na2so4(aq) solution is added and a white precipitate forms. what is the identity of the unknown white solid?
Chemistry
1 answer:
Lady bird [3.3K]2 years ago
5 0

The given salts are Barium chloride, lead chloride and zinc chloride.

The balanced chemical reaction between barium chloride and sodium sulfate:

BaCl_{2}(aq)+Na_{2}SO_{4}(aq)-->2NaCl(aq)+BaSO_{4}(s)_{white precipitate}

The balanced chemical reaction between lead chloride and sodium sulfate:

PbCl_{2}(aq)+Na_{2}SO_{4}(aq)-->2NaCl(aq)+PbSO_{4}(s)_{white precipitate}

The balanced chemical reaction between zinc chloride and sodium sulfate:

ZnCl_{2}(aq)+Na_{2}SO_{4}(aq)-->2NaCl(aq)+ZnSO_{4}(aq)

Both Barium chloride and lead chloride produce a white precipitate of their respective sulfates upon reaction with sodium sulfate. But BaCl2 is readily soluble in water where as Lead chloride is insoluble in water.

Therefore, the compound is barium chloride.

You might be interested in
Combustion analysis of a 13.42-g sample of the unknown organic compound (which contains only carbon, hydrogen, and oxygen) produ
kirza4 [7]

<u>Answer:</u> The molecular formula for the given organic compound is C_{18}H_{20}O_2

<u>Explanation:</u>

The chemical equation for the combustion of hydrocarbon having carbon, hydrogen and oxygen follows:

C_xH_yO_z+O_2\rightarrow CO_2+H_2O

where, 'x', 'y' and 'z' are the subscripts of Carbon, hydrogen and oxygen respectively.

We are given:

Mass of CO_2=39.61g

Mass of H_2O=9.01g

We know that:

Molar mass of carbon dioxide = 44 g/mol

Molar mass of water = 18 g/mol

<u>For calculating the mass of carbon:</u>

In 44 g of carbon dioxide, 12 g of carbon is contained.

So, in 39.61 g of carbon dioxide, \frac{12}{44}\times 39.61=10.80g of carbon will be contained.

<u>For calculating the mass of hydrogen:</u>

In 18 g of water, 2 g of hydrogen is contained.

So, in 9.01 g of water, \frac{2}{18}\times 9.01=1.00g of hydrogen will be contained.

Mass of oxygen in the compound = (13.42) - (10.80 + 1.00) = 1.62 g

To formulate the empirical formula, we need to follow some steps:

  • <u>Step 1:</u> Converting the given masses into moles.

Moles of Carbon = \frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{10.80g}{12g/mole}=0.9moles

Moles of Hydrogen = \frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{1g}{1g/mole}=1moles

Moles of Oxygen = \frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{1.62g}{16g/mole}=0.10moles

  • <u>Step 2:</u> Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 0.10 moles.

For Carbon = \frac{0.9}{0.10}=9

For Hydrogen = \frac{1}{0.10}=10

For Oxygen = \frac{0.10}{0.10}=1

  • <u>Step 3:</u> Taking the mole ratio as their subscripts.

The ratio of C : H : O = 9 : 10 : 1

Hence, the empirical formula for the given compound is C_9H_{10}O

For determining the molecular formula, we need to determine the valency which is multiplied by each element to get the molecular formula.

The equation used to calculate the valency is :

n=\frac{\text{Molecular mass}}{\text{Empirical mass}}

We are given:

Mass of molecular formula = 268.34 g/mol

Mass of empirical formula = 134 g/mol

Putting values in above equation, we get:

n=\frac{268.34g/mol}{134g/mol}=2

Multiplying this valency by the subscript of every element of empirical formula, we get:

C_{(9\times 2)}H_{(10\times 2)}O_{(1\times 2)}=C_{18}H_{20}O_2

Thus, the molecular formula for the given organic compound is C_{18}H_{20}O_2.

3 0
2 years ago
How many grams of the excess reagent are left over when 6.00g of CS2 gas react with 10.0g of Cl2 gas in the following reaction:
WARRIOR [948]

Answer:

CS₂ = 2.43 g

Explanation:

Data Given:

CS₂ gas = 6.00g

Cl₂ = 10.0g

Reaction Given:

                   CS₂(g) + 3Cl₂(g) --------> CCl₄(l) + S₂Cl₂(l)

Solution

Limiting Reagent :

The reactant which is less in amount control the amount of product and know as limiting reagent.

Excess Reagent:

The amount of reactant which are in excess and leftover at the end of reaction and product formed.

Now we have to find the reactant that is in excess

For this we will look at the Reaction

                       CS₂   +    3Cl₂  -------->   CCl₄   +  S₂Cl₂

                       1 mol     3 mol               1 mol      1 mol

we come to know from the above reaction that

1 mole of CS₂ react with 3 mole of Cl₂ to produce 1 mole of CCl₄ and 1 mole of  S₂Cl₂

Now to convert mole to mass we required molar masses

molar mass of CS₂ = 12 +  2(32)

molar mass of CS₂ = 76 g/mol

molar mass of Cl₂ = 2(35.5)

molar mass of Cl₂ = 71 g/mol

if we represent mole in grams then

             CS₂              +    3Cl₂             -------->   CCl₄   +  S₂Cl₂

      1 mol (76 g/mol)     3 mol (71 g/mol)

                CS₂   +    3Cl₂   -------->   CCl₄   +  S₂Cl₂

                 76 g       213 g

So,

we come to know that 76 g of CS₂ will combine with 213 g of Cl₂ to form product.

So now look for the ratio of both reactant

                   CS₂    :    Cl₂

                   76 g   :    213 g

                    1  g     :      2.8 g

So, the for every one gram of CS₂ required 2.8g Cl₂

So from this details

we apply unity formula

                 2.8 g of Cl₂ ≅  1 g of CS₂

                 10 g of Cl₂ ≅  ? g of CS₂

by doing cross multiplication

                    g of CS₂ = 10 x 1 / 2.8

                    g of CS₂ = 3.6 g

So,

10.0g of Cl₂ used 3.57 g of CS₂

It showed that 3.57 g of CS₂ used and the remaining amount of CS₂ is

               Remaining amount of CS₂ = 6 - 3.57 = 2.43 g

 

8 0
2 years ago
Which of the following procedures demonstrates repetition? A. Jeremiah measures the mass of five different strips of metal. He p
lara31 [8.8K]

The answer is C bc i did it already.

7 0
2 years ago
Read 2 more answers
How many grams of KClO3 are needed to produce of 4.26 moles of O2? 2 KClO3 2 KCl + 3 O2 a. 348 g b. 136 g c. 174 g d. 522 g e. 7
JulsSmile [24]
Ok so this is what we know :

2KClO3 -> 2KCl + 3O2         (Always check if equation is balanced - in this                                               case it is)
                              4.26moles
So we know that we have 4.26 moles of oxygen (O2). Now lets look at the ratio between KClO3 and O2.
We see that the ratio is 2:3 meaning that we need 2KClO3 in order to produce 3O2.
Therefore divide 4.26 by 3 and then multiply by 2.
4.26/3 = 1.42
1.42 * 2 = 2.84
Now we know that the molarity of KClO3 is 2.84 moles.
Multiply by R.M.M to find how many grams of KClO3 we have.

R.M.M of KClO3
K- 39
Cl- 35.5
3O- 3 * 16 -> 48
---------------------------
                      <span>122.5
</span>2.84 * 122.5 = 347.9 grams therefore the answer is (a)
                       348 grams needed of KClO3 to produce 4.26 moles of O2.
Hope this helps :).

8 0
2 years ago
Read 2 more answers
H2A and BOH are acid and base and they react according to the following balanced equation: H2A(aq) + 2 BOH(aq) → B2A(aq) + 2 H2O
Rudiy27

Answer:

ΔH=15000 J  =  15KJ

Explanation:

In this exercise  you have find the enthalpy of reaction this is the difference between enthalpy of reactans and products,

For the following equation

H2A(aq) + 2 BOH(aq) → B2A(aq) + 2 H2O(l)

We know that 0.20 moles of BOH reacted with excess amount of H2A solution and 1500. J

so,

(2mol/0,2mol)*1500J=15000J

for de reactions exothermics tha enthalpy is negative so:

ΔH=15000 J  =  15KJ

3 0
2 years ago
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