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Mazyrski [523]
3 years ago
8

What is the result when atoms combine in fixed ratios?

Chemistry
1 answer:
Liono4ka [1.6K]3 years ago
4 0

Answer:

Chemical reaction

Explanation:

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spayn [35]
The balanced chemical reaction would be expressed as follows:

2Na + Cl2 = 2NaCl

We are given the amount of sodium metal to be used reaction. This would be the starting value for the calculations. We do as follows:

6.25 g ( 1 mol / 22.99 g ) ( 1 mol Cl2 / 2 mol Na ) ( 22.4 L / 1 mol ) = 3.045 L Cl2 needed

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Which is NOT true of a mixture?
Reika [66]

Answer:

OPTUON C,CHEMICALLY BONDED

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For a ternary solution at constant T and P, the composition dependence of molar property M is given by: M = x1M1 + x2M2 + x3M3 +
AveGali [126]

Answer:

M_{i} = M_{i} + C_{xjxk} (1-2x_{i}) ...1

M^{\alpha } = M_{i} + CX_{xjxk}          ...2

Explanation:

The ternary constant is given by the following equation:

The symbol XiXi, where XX is an extensive property of a homogeneous mixture and the subscript ii identifies a constituent species of the mixture, denotes the partial molar quantity of species ii defined by

M_{i}  = [\frac{d(nM)}{dn_{i} }]_{P,t,n,j}

This is the rate at which property  X  changes with the amount of species  i  added to the mixture as the temperature, the pressure, and the amounts of all other species are kept constant.  A partial molar quantity is an intensive state function.  Its value depends on the temperature, pressure, and composition of the mixture.

In a multi phase system (in this case, a ternary system), the components resolved give:

M_{i} = M_{i} + C_{xjxk} (1-2x_{i})

and M^{\alpha } = M_{i} + CX_{xjxk}

5 0
3 years ago
The following equation for the reaction of alum with barium chloride is not balanced. KAl(SO4)2•12H2O(aq) + BaCl2(s) → KCl(aq) +
Pepsi [2]

Answer:

A. 1.04g of BaCl2.

B. Percentage yield of BaSO4 is 87.6%

Explanation:

A. The balanced equation for the reaction. This given below:

KAl(SO4)2•12H2O(aq) + 2BaCl2(s) → KCl(aq) + AlCl3(aq) + 2BaSO4(s) + 12H2O(l)

Next, we shall determine the number of mole in 25 mL of a 0.10 M alum. This is illustrated below:

Volume = 25mL = 25/1000 = 0.025L

Molarity = 0.1M

Mole =..?

Mole = Molarity x Volume

Mole of alum = 0.1 x 0.025 = 2.5x10¯³ mol.

Next, we shall convert 2.5x10¯³ mol of alum to grams.

Number of mole alum, KAl(SO4)2•12H2O = 2.5x10¯³ mol

Molar Mass of alum, KAl(SO4)2•12H2O = 39 + 27 + 2[32+(16x4)] + 12[(2x1) + 16]

= 39 + 27 + 2[32 + 64] + 12[2 + 16]

= 39 + 27 + 2[96] + 12[18]

= 474g/mol

Mass of alum, KAl(SO4)2•12H2O =..?

Mass = mole x molar mass

Mass of alum, KAl(SO4)2•12H2O = 2.5x10¯³ x 474 = 1.185g

Next, we shall determine the mass of alum and BaCl2 that reacted and the mass of BaSO4 produced from the balanced equation. This is illustrated below:

Molar mass of alum, KAl(SO4)2•12H2O = 474g

Mass of alum, KAl(SO4)2•12H2O from the balanced equation = 1 x 474 = 474g

Molar mass of BaCl2 = 137 + (35.5x2) = 208g/mol

Mass of BaCl2 from the balanced equation = 2 x 208 = 416g

Molar mass of BaSO4 = 137 + 32 + (16x4) = 233g/mol

Mass of BaSO4 from the balanced equation = 2 x 233 = 466g

Summary:

From the balanced equation above,

474g of alum reacted with 416g of BaCl2 to produce 466g of BaSO4.

Finally, we can calculate the mass of BaCl2 needed for the reaction as follow:

From the balanced equation above,

474g of alum reacted with 416g of BaCl2.

Therefore, 1.185g of alum will react with = (1.185 x 416)/474 = 1.04g of BaCl2.

Therefore, 1.04g of BaCl2 is needed for the reaction.

B. Determination of the percentage yield of BaSO4(s).

We'll begin by calculating the theoretical yield of BaSO4. This is illustrated below:

From the balanced equation above,

474g of alum reacted to produce 466g of BaSO4.

Therefore, 1.185g of alum will react to produce = (1.185 x 466)/474 = 1.165g of BaSO4.

Therefore, the theoretical yield of BaSO4 is 1.165g.

Finally, we shall determine the percentage of BaSO4 as follow:

Actual yield of BaSO4 = 1.02g.

Theoretical yield of BaSO4 = 1.165g.

Percentage yield of BaSO4 =..?

Percentage yield = Actual yield /Theoretical yield x 100

Percentage yield of BaSO4 = 1.02/1.165 x 100

Percentage yield of BaSO4 = 87.6%

4 0
3 years ago
Read 2 more answers
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