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3 years ago

Write a balanced half-reaction describing the reduction of gaseous dioxygen to aqueous oxide anions.

Chemistry

1 answer:

maksim [4K]3 years ago

8 0

Answer: The balanced half reaction is written below.

Explanation:

Reduction reaction is defined as the chemical reaction in which an atom gains electrons. The oxidation number of the atom gets reduced during this reaction.

$X^{n+}+ne^-\rightarrow X$

When oxygen gas is reduced to oxide ions, the number of electron transferred are 2

The chemical equation for the reduction of oxygen gas to oxide ions follows:

$O_2+2e^-\rightarrow 2O^{2-}$

Hence, the balanced half reaction is written above.

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Chlorine can be prepared in the laboratory by the reaction of manganese dioxide with hydrochloric acid, HCl ( aq ) , as describe

Nataly [62]

Answer:

0.9483 grams of manganese dioxide should be added to excess HCl.

Explanation:

Pressure of the chlorine gas = P = 795 Torr = 1.046 atm (1 atm = 760 Torr)

Volume of the chlorine gas = V = 255 ml = 0.255 L

Temperature of the chlorine gas = T = 25°C= 298.15 K

Moles of chlorine gas = n

Using ideal gas equation:

PV = nRT

$n=\frac{PV}{RT}=\frac{1.046 atm\times 0.255 L}{0.0821 atm l/mol k\times 298.15 K}$

n = 0.01090 mol

$MnO_2 ( s ) + 4 HCl ( aq)\rioghtarrow MnCl_2 ( aq ) + 2 H_2O ( l ) + Cl_2 ( g )$

According to reaction , 1 mole of chlorine gas is obtained from 1 mole of manganese dioxide.

Then 0.01090 moles of chlorine gas will be obtained from:

$\frac{1}{1}\times 0.01090 mol=0.01090 mol$ manganese dioxide

Mass of 0.01090 moles of manganese dioxide:

0.01090 mol × 87 g/mol = 0.9483 g

8 0

2 years ago

Combining 0.350 mol Fe 2 O 3 with excess carbon produced 18.9 g Fe . Fe 2 O 3 + 3 C ⟶ 2 Fe + 3 CO

lapo4ka [179]

Answer:

1. 0.338 moles of Fe

2. 0.700 moles of Fe

3. 48.3%

Explanation:

This is the reaction:

Fe₂O₃ + 3C → 2Fe + 3CO

We were told that we produce 18.9 g of Fe. Let's convert the mass to moles:

18.9 g . 1mol/ 55.85 g = 0.338 moles of Fe

Let's make a rule of three; ratio is 1:2.

1 mol of oxide can produce 2 moles of elemental iron

Then, 0.350 moles must produce (0.350 .2) / 1 = 0.700 moles of Fe

Let's determine the percent yield:

(Yield produced /Theoretical Yield) . 100 = 48.3 %

3 0

3 years ago

Read 2 more answers

A 50.00 mL sample of groundwater is titrated with 0.0300 M EDTA . Of 12.40 mL of EDTA is required to titrate the 50.00 mL sample

Rzqust [24]

Answer:

7.44x10⁻³ mol/L and 744 ppm

Explanation:

Let's assume that the hardness of the water is totally from Ca⁺² ions only(the hardness is the measure of Ca⁺² and Mg⁺² ions). The titration with EDTA will form a complex. The EDTA is always in 1:1 proportion, so the number of moles of it will be the number of moles of Ca⁺², which will be the number of moles of CaCO₃.

n = 0.0124 L * 0.0300 mol/L

n = 3.72x10⁻⁴ mol

The molarity is the number of moles divided by the volume (0.05 L)

M = 3.72x10⁻⁴/0.05

M = 7.44x10⁻³ mol/L

1 part per million = 1 mg/L. The molar mass of the CaCO₃ is 100 g/mol, so the mass of it is:

m = 3.72x10⁻⁴ mol * 100 g/mol

m = 0.0372 g = 37.2 mg

Then, the ppm:

37.2/0.05 = 744 ppm

5 0

3 years ago

Question 14 of 25

belka [17]

double displacement

bcoz each of the reactants combines with other reactants to obtain the product

4 0

2 years ago

Find the density of a substance whose volume is 5 cm^3 and whose mass is 25 g.

Ilya [14]

Cm3 is Centimeter to the 3 power or Cubic Centimeter, just like for figuring concrete you need to know the volume of the pour expressed in cubic yards, length x width x height.

cm3 is centimeters instead of yards.

So 25 grams/5 cubic centimeters (CM3) = 5
HOPE THIS HELPED

7 0

2 years ago