11.Calculate the molar mass of Fe2(SO3)3

Answers

Answer 1
Answer:

The molar mass of Fe2(SO3)3 is 351.8796 moles.

What is molar mass?

Molar mass is defined as the mass equivalent of an element's or a chemical compound's Avogadro number of atoms or molecules, respectively. A mole is defined as the quantity of atoms, molecules, or ions that are present in a substance. It is additionally referred to as the volume of material that contains the same number of discrete units.

Molar mass of Fe2(SO3)3

= 2 × 55.845 + 3 × 32.065 + 9 × 15.9994

= 111.69 +  96.195 + 143.9946

= 351.8796 moles.

The substance iron(III) sulfite, sometimes known as ferrous sulfite, has the chemical formula Fe2(SO3)3. The family of inorganic compounds known as iron(III) Sulphate, sometimes known as ferric Sulphate, has the formula Fe2(SO4)3(H2O)n.

Thus, the molar mass of Fe2(SO3)3 is 351.8796 moles.

To learn more about molar mass, refer to the link below:

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Answer 2
Answer:

Answer:

The molar mass and molecular weight of Fe2(SO3)3 is 351.8796.


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Determine the concentrations of K2SO4, K+, and SO42− in a solution prepared by dissolving 2.07 × 10−4 g K2SO4 in 2.50 L of water. Express all three concentrations in molarity. Additionally, express the concentrations of the ionic species in parts per million (ppm). Note: Determine the formal concentration of SO42−. Ignore any reactions with water.
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The ratio of oxygen-16 and oxygen-18 isotopes in plankton fossils in deep-sea sediments can be used to determine ________.

Answers

The ratio of oxygen-16 and oxygen-18 isotopes in plankton fossils in deep-sea sediments can be used to determine past temperatures.

Answer;

-past temperatures

The ratio of oxygen-16 and oxygen-18 isotopes in plankton fossils in deep-sea sediments can be used to determine past temperatures.

Explanation;

-O-16 will evaporate more readily than O-18 since it is lighter, therefore; during a warm period, the relative amount of O-18 will increase in the ocean waters since more of the O-16 is evaporating.

-Hence, looking at the ratio of O16 to O18 in the past can give clues about global temperatures.

Animal fats tend to be _______________ with hydrogen atoms. (fifth paragraph of this sub-section

Answers

Animal fats tend to be _______________ with hydrogen atoms. Fats are long chain amino acids, mostly carbon and hydrogen. The answer is "Animal fats tend to be single bonded to or surrounded by or saturated with hydrogen atoms." Saturated means the maximum number of hydrogen atoms are bonded to the carbon atom.

Answer:

Saturated.

Explanation:

Hello,

Animal fats are lipids derived from animals which are commonly solid at room temperature and mainly constituted by triglycerides which are strictly chemically saturated with hydrogen, it means  they do not tend to have double or triple bonded carbon atoms but just single-bonded carbons. This fact suggests that animal fats provide more energy than vegetable fats because they have more C-H bonds that when broken increase the total provided energy.

Best regards.

Polymers are large molecules composed of simple units repeated many times. Thus, they often have relatively simple empirical formulas. Calculate the empirical formulas of the following polymers: (a) Lucite (Plexiglas); 59.9% C, 8.06% H, 32.0% O (b) Saran; 24.8% C, 2.0% H, 73.1% Cl (c) polyethylene; 86% C, 14% H (d) polystyrene; 92.3% C, 7.7% H (e) Orlon; 67.9% C, 5.70% H, 26.4% N

Answers

Answer:

(a) C_5H_8O_2

(b) CHCl

(c) CH_2

(d) CH

(e) C_3H_3N

Explanation:

Hello,

(a) For the lucite, one computes the moles of C, H and O that are present:

n_C=0.599gC*(1molC)/(12gC)=0.05molC\nn_H=0.0806gH*(1molH)/(1gH)=0.0806molH\nn_O=0.32gO*(1molO)/(16gO)=0.02molO\n

Now, dividing each moles by the smallest moles (oxygen's moles), one obtains:

C=(0.05)/(0.02) =2.5;H=(0.0806)/(0.02) =4;O=(0.02)/(0.02) =1

Finally, we look for the smallest whole number subscript by multiplying by 2, so the empirical formula turns out into:

C_5H_8O_2

(b) For the Saran, one computes the moles of C, H and Cl that are present:

n_C=0.248gC*(1molC)/(12gC)=0.021molC\nn_H=0.02gH*(1molH)/(1gH)=0.02molH\nn_(Cl)=0.731gCl*(1molCl)/(35.45gCl)=0.021molCl\n

Now, dividing each moles by the smallest moles (hydrogen's moles), one obtains:

C=(0.021)/(0.02) =1;H=(0.02)/(0.02) =1;Cl=(0.021)/(0.02) =1

Finally, as all of the subscripts are whole numbers, the empirical formula turns out into:

CHCl

(c) For the polyethylene, one computes the moles of C and H that are present:

n_C=0.86*(1molC)/(12gC)=0.072molC\nn_H=0.14gH*(1molH)/(1gH)=0.14molH

Now, dividing each moles by the smallest moles (carbon's moles), one obtains:

C=(0.072)/(0.072) =1;H=(0.14)/(0.072) =2

Finally, as all of the subscripts are whole numbers, the empirical formula turns out into:

CH_2

(d) For the polystyrene, one computes the moles of C and H that are present:

n_C=0.923*(1molC)/(12gC)=0.077molC\nn_H=0.077gH*(1molH)/(1gH)=0.077molH

Now, dividing each moles by the smallest moles (either carbon's or hydrogen's moles), one obtains:

C=(0.077)/(0.077) =1;H=(0.077)/(0.077) =1

Finally, as all of the subscripts are whole numbers, the empirical formula turns out into:

CH

(e) For the orlon, one computes the moles of C, H and N that are present:

n_C=0.679*(1molC)/(12gC)=0.057molC\nn_H=0.057gH*(1molH)/(1gH)=0.057molH\nn_N=0.264gN*(1molN)/(14gN)=0.019molN

Now, dividing each moles by the smallest moles (nitrogen's moles), one obtains:

C=(0.057)/(0.019) =3;H=(0.057)/(0.019) =3;N=(0.019)/(0.019) =1

Finally, as all of the subscripts are whole numbers, the empirical formula turns out into:

C_3H_3N

Best regards.

Consider the dissolution of AB(s):AB(s)⇌A+(aq)+B−(aq)Le Châtelier's principle tells us that an increase in either [A+] or [B−] will shift this equilibrium to the left, reducing the solubility of AB. In other words, AB is more soluble in pure water than in a solution that already contains A+ or B− ions. This is an example of the common-ion effect.The generic metal hydroxide M(OH)2 has Ksp = 1.05×10−18. (NOTE: In this particular problem, because of the magnitude of the Ksp and the stoichiometry of the compound, the contribution of OH− from water can be ignored. However, this may not always be the case.)What is the solubility of M(OH)2 in pure water?

Answers

Answer:

S = 6.40 × 10⁻⁷ M

Explanation:

In order to calculate the solubility (S) of M(OH)₂ in pure water we will use an ICE Chart. We recognize 3 stages: Initial, Change and Equilibrium, and we complete each row with the concentration or change in concentration.

            M(OH)₂(s) ⇄ M²⁺(aq) + 2 OH⁻(aq)

I                                   0                  0

C                                 +S               +2S

E                                   S                 2S

The solubility product (Kps) is:

Kps = 1.05 × 10⁻¹⁸ = [M²⁺].[OH⁻]²=S.(2S)²

1.05 × 10⁻¹⁸ = 4S³

S = 6.40 × 10⁻⁷ M

____________ acids are proton donors. ____________ bases are proton acceptors. The ____________ the pKₐ the stronger the acid. ____________ acids are electron pair acceptors.

Answers

Answer:

1. Bronsted—Lowry acid

2. Bronsted—Lowry Base

3. Lower the pka

4. Lewis acids

Explanation:

If the mass percentage composition of a compound is 72.1% Mn and 27.9% O, its empirical formula is

Answers

Answer:

MnO- Manganese Oxide

Explanation:

Empirical formula: This is the formula that shows the ratio of elements

present in a  

compound.

   

How to determine Empirical formula

1. First arrange the symbols of the elements present in the compound

alphabetically to  determine the real empirical formula. Although, there

are exceptions to this rule, E.g H2So4

2. Divide the percentage composition by the mass number.

3. Then divide through by the smallest number.

4. The resulting answer is the ratio attached to the elements present in

a compound.

           

                                                                              Mn                         O    

                         

% composition                                                      72.1                      27.9    

                       

Divide by mass number                                       54.94                     16  

                                 

                                                                               1.31                      1.74    

                       

Divide by the smallest number                         1.31                      1.31                          

                                                                               1                    1.3

                                                 

The resulting ratio is 1:1

 

Hence the Empirical formula is MnO, Manganese oxide