An amphoteric species is neither an acid nor a base. True or False

Given the chemical equation: Fe2O3 + 3CO --> 2Fe + 3CO2

lapo4ka [179]

Answer:

The answer to your question is 160 g of Fe₂O₃

Explanation:

Data

mass of Fe = 112 g

mass of CO = in excess

mass of Fe₂O₃ = ?

Balanced chemical reaction

Fe₂O₃ + 3CO ⇒ 2Fe + 3CO₂

Process

1.- Calculate the molar mass of Fe₂O₃ and Fe

Molar mass Fe₂O₃ = (56 x 2) + (16 x 3) = 112 + 48 = 160 g

atomic mass of Fe = 56

2.- Use proportions to calculate the mass of Fe₂O₃ needed

160 g of Fe₂O₃ ------------------- 2(56) g of Fe

x g of Fe₂O₃ ------------------ 112 g of Fe

x = (112 x 160) / 2(56)

x = 17920/112

x = 160 g of Fe₂O₃

6 0

3 years ago

What is the concentration of h+ ions in a 2.20 m solution of hno3?

kvv77 [185]

The question is improperly formatted.

What is the concentration of H+ ions in a 2.2 M solution of HNO3.

Answer:-

2.2 moles of H+ per litre

Explanation:-

M stands for molarity. 2.2 M means 2.2 moles of HNO3 is present per litre of the solution.

Now HNO3 has just 1 H in it's formula. HNO3 would give H+. So 2.2 moles of HNO3 would mean 2.2 moles of H+ per litre.

7 0

3 years ago

A sample of vinegar was found to have an acetic acid concentration of 0.8846 m. What is the acetic acid % by mass? Assume the de

jenyasd209 [6]

Answer:

5.3%

Explanation:

Let the volume be 1 L

volume , V = 1 L

use:

number of mol,

n = Molarity * Volume

= 0.8846*1

= 0.8846 mol

Molar mass of CH3COOH,

MM = 2*MM(C) + 4*MM(H) + 2*MM(O)

= 2*12.01 + 4*1.008 + 2*16.0

= 60.052 g/mol

use:

mass of CH3COOH,

m = number of mol * molar mass

= 0.8846 mol * 60.05 g/mol

= 53.12 g

volume of solution = 1 L = 1000 mL

density of solution = 1.00 g/mL

Use:

mass of solution = density * volume

= 1.00 g/mL * 1000 mL

= 1000 g

Now use:

mass % of acetic acid = mass of acetic acid * 100 / mass of solution

= 53.12 * 100 / 1000

= 5.312 %

≅ 5.3%

3 0

3 years ago

Solve for x 5 to the power 2 x + 4 - 25 to the power x - 1 is equals to

Dafna11 [192]

Answer:

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4 0

3 years ago

Read 2 more answers

Consider the reaction of metallic copper with iron(!!) to give copper(ll) and ironin 0.77V Fe* (aq) + e-Fe (aq) Cup (aq) + 2e --

frosja888 [35]

Answer :

(a) The anode and cathode will be $E^o_{(Cu^{2+}/Cu)}$

and $E^o_{(Fe^{3+}/Fe^{2+})}$ respectively.

(b) The emf of cell potential is 1.022 V

Explanation :

(a) The standard reduction potentials for iron and copper are:

$E^o_{(Fe^{3+}/Fe^{2+})}=0.77V\\E^o_{(Cu^{2+}/Cu)}=0.34V$

In the voltaic cell, the oxidation occurs at an anode which is a negative electrode and the reduction occurs at the cathode which is a positive electrode.

From the standard reduction potentials we conclude that, the substance having highest positive $E^o$ potential will always get reduced and will undergo reduction reaction.