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SVEN [57.7K]
2 years ago
10

Which of the following species contains manganese with the highest oxidation number?

Chemistry
1 answer:
ioda2 years ago
7 0

In NaMnO₄, Mn has the highest oxidation number.

The question is incomplete, the complete question is;

Which of the following species contains manganese with the highest oxidation number?

A) Mn

B) MnF₂

C) Mn₃(PO₄)₂

D) MnCl₄

E) NaMnO₄

In order to ascertain the specie that contains manganese with the highest oxidation number, we must calculate the oxidation number of manganese in each of the species one after the other.

1) For Mn, the oxidation number of Mn is zero because the atom is uncombined.

2) For MnF₂;

Mn has an oxidation number of +2

3) For Mn₃(PO₄)₂

Mn has an oxidation number of +2

4) For MnCl₄

Mn has an oxidation number of +4

5) For NaMnO₄

Mn has an oxidation number of +7

Hence in NaMnO₄, Mn has the highest oxidation number.

Learn more: brainly.com/question/10079361

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You evaporate all of the water from 100 mL of NaCl solution and obtain 11.3 grams of NaCl. What was the molarity of the NaCl sol
kicyunya [14]

Answer:

                      Molarity  =  1.93 mol.L⁻¹

Explanation:

             Molarity is the unit of concentration used to specify the amount of solute in given amount of solution. It is expressed as,

                         Molarity  =  Moles / Volume of Solution    ----- (1)

Data Given;

                  Mass  =  11.3 g

                  Volume  =  100 mL  =  0.10 L

First calculate Moles for given mass as,

                   Moles  =  Mass / M.mass

                   Moles  =  11.3 g / 58.44 g.mol⁻¹

                   Moles  =  0.1933 mol

Now, putting value of Moles and Volume in eq. 1,

                        Molarity  =  0.1933 mol ÷ 0.10 L

                        Molarity  =  1.93 mol.L⁻¹

3 0
2 years ago
What is the molar mass of al (clo2)3
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The molar mass is 229.33
4 0
2 years ago
What is the percent yield of this reaction if 22.o g of Mgl2 is produced by the reaction of 25.0 g of Mg with 25.0 g of l2?
expeople1 [14]

% yield = 80.719

<h3>Further explanation</h3>

Given

22.0 g of Mgl₂

25.0 g of Mg

25.0 g of l₂

Required

The percent yield

Solution

Reaction

Mg + I₂⇒ MgI₂

mol Mg = 25 g : 24.305 g/mol = 1.029

mol I₂ = 25 g : 253.809 g/mol = 0.098

Limiting reactant = I₂

Excess reactant = Mg

mol MgI₂ based on I₂, so mol MgI₂ = 0.098

Mass MgI₂ (theoretical):

= mol x MW

= 0.098 x 278.114

= 27.255 g

% yield = (actual/theoretical) x 100%

% yield = (22 / 27.255) x 100%

% yield = 80.719

8 0
1 year ago
9 The Haber process is a reversible reaction. N2(g) + 3H2(g) 2NH3(g) The reaction has a 30% yield of ammonia. Which volume of am
Illusion [34]

Answer:  3.36 L of ammonia gas

Explanation:

The balanced chemical reaction is:

N_2(g)+3H_2(g)\rightarrow 2NH_3(g)  

According to stoichiometry :

3 moles of H_2 produce = 2 moles of NH_3

Thus 0.75 moles of H_2 will producee=\frac{2}{3}\times 0.75=0.50moles  of NH_3

But as percent yield is 30 %, amount of ammonia produced = \frac{30}{100}\times 0.50moles=0.15moles

According to ideal gas equation:

PV=nRT

P = pressure  = 1 atm

V = Volume  = ?

n = number of moles = 0.15

R = gas constant =0.0821Latm/Kmol

T =temperature =273K

V=\frac{nRT}{P}

V=\frac{0.15\times 0.0820 L atm/K mol\times 273K}{1atm}=3.36L

Thus 3.36 L of ammonia gas is obtained by reacting 0.75 moles of hydrogen with excess nitrogen.

3 0
2 years ago
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