Answers
Explanation:
..(1)
Manganese (II) carbonate and oxygen reacts to give manganses (III) oxide with carbon-dioxide gas.
..(2)
Manganese (III) oxide reacts with aluminum metal to give aluminum oxide and manganese metal
On adding (1) and 2(2) we will get the net net chemical equation for the production of manganese. So,
The net balanced equation we got:
Final answer:
The production of manganese from manganese (II) carbonate, oxygen, and aluminum involves two chemical reactions. Heating manganese (II) carbonate with oxygen yields manganese(III) oxide and carbon dioxide. This manganese (III) oxide then reacts with aluminum to produce manganese and aluminum oxide.
Explanation:
The chemical reaction can be described in terms of a net chemical equation after properly balancing the chemicals involved on both sides of the equation. The first step is to understand and correctly write down the formula of reactants and products for the reaction. Here: Manganese (II) Carbonate (MnCO3), Oxygen (O2), and Aluminum (Al).
In the process of manganese production, Manganese (II) Carbonate is first heated in the presence of oxygen to yield Manganese (III) Oxide (Mn2O3) and Carbon Dioxide (CO2):
MnCO3 + 1/2 O2 -> Mn2O3 + CO2.
The resulting Manganese (III) Oxide reacts with Aluminum (which acts as a reducing agent in this case) to generate Manganese and Aluminum Oxide:
3 Mn2O3 + 4 Al -> 6 Mn + 2 Al2O3.
These two equations combined represent the net chemical equation for the production of manganese from manganese (ii) carbonate, oxygen, and aluminum.
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Related Questions
an element has 8 protons and 8 neutrons in its nucleus and 8 electrons revolving around the nucleus. What is the atomic number and atomic mass of this element?
6.579 rounded to nearest hundreth
Calculate the moles of benzoic acid (C6H5COOH) in a 55.66 g sample of benzoic acid
One cubic meter of an ideal gas at 600 K and 1000 KPa expand to five times its initial volume as follows:(a) By a mechanically reversible, isothermal process [T 2 = 600K; P 2 = 200kPa; W= -1609 kJ](b) By a mechanically reversible, adiabatic process [T 2 =208.96K; P 2 = 67.65 kPa; W= -994.4 kJ] For each case calculate the final temperature, pressure and the work done by the gas. C p =21 Jmol- 1K-1.
Answers
Answer:
1.62 g
Explanation:
Given that:
Concentration of HCN = 0.119 M
Assuming the ka 4.00 × 10⁻¹⁰
The pKa of HCN (hydrocyanic acid) = -log (Ka)
= - log ( 4.00 × 10⁻¹⁰)
= 9.398
pH of buffer = 8.809
Using Henderson Hasselbach equation:
[CN^-] = 0.2576[HCN]
[CN^-] = 0.2756 (0.119) L
[CN^-] = 0.033 M
∴
The amount of NaCN (sodium cyanide) is calculated as follows:
= 1.62 g
Round your answer to 3 significant digits.
Answers
A chemist determined by measurements that 0.0250 moles of potassium bromide participate in a chemical reaction is 0.002 moles
What are moles?
The moles are the smallest unit of an atom ion molecule or substance which is used to count the number which is taking part in a chemical reaction and is equal to 2.303 ×10²³ moles of that.
To calculate the participant in a chemical reaction number of moles is 0.0250 moles so the mass will be,
number of moles = mass/ molar mass
substituting the value,
0.0250 moles = mass / 119.002
mass = 119.002 × 0.0250 moles
mass = 0.002 moles
Therefore, 0.002 moles determined by measurements that 0.0250 moles of potassium bromide participate in a chemical reaction.
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My solution
39.0983+126.90447=166.00277
x/166.002277=.06
x=9.96g
Explanation:
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As the temperature of a liquid or stable will increase its vapor strain additionally will increase. Conversely, vapor strain decreases because the temperature decreases.
The better the vapor strain of a substance, the extra the awareness of the compound withinside the gaseous section and the extra the quantity of vaporization
. Liquids range substantially of their vapor pressures. substance with a excessive vapor strain at everyday temperatures is regularly called volatile. The strain exhibited through vapor gift above a liquid floor is referred to as vapor strain. As the temperature of a liquid will increase, the kinetic strength of its molecules additionally will increase.
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Answers
Answer:
a) H₂SO₄ + Ba(OH)₂ ⇄ BaSO₄(s) + 2 H₂O(l)
b) H₂SO₄, H⁺, HSO₄⁻, SO₄²⁻. H₂O, H⁺, OH⁻.
c) H⁺, HSO₄⁻, SO₄²⁻
d) As the titration takes place, reaction [1] proceeds to the right. The conductivity of the solution decreases because the amount of H⁺, HSO₄⁻, SO₄²⁻ decreases. The formed solid is barium sulfate BaSO₄. Since BaSO₄ is very insoluble, the main responsible for conductivity are still H⁺, HSO₄⁻ and SO₄²⁻,
e) At the equivalence point equivalent amounts of H₂SO₄ and Ba(OH)₂ react. The conducting species are Ba²⁺, SO₄²⁻, H⁺ and OH⁻.
f) After the equivalence point there is an excess of Ba(OH)₂. The ions Ba²⁺ and OH⁻ are responsible for the increase in the conductivity, being the major conducting species.
Explanation:
a) Write an equation (including states of matter) for the reaction between H₂SO₄ and Ba(OH)₂.
The balanced equation is:
H₂SO₄ + Ba(OH)₂ ⇄ BaSO₄(s) + 2 H₂O(l) [1]
b) At the very start of the titration, before any titrant has been added to the beaker, what is present in the solution?
In the beginning there is H₂SO₄ and the ions that come from its dissociation reactions: H⁺, HSO₄⁻, SO₄²⁻. There is also H₂O and a very small amount of H⁺ and OH⁻ coming from its ionization.
H₂SO₄(aq) ⇄ H⁺(aq) + HSO₄⁻(aq)
HSO₄⁻(aq) ⇄ H⁺(aq) + SO₄²⁻(aq)
H₂O(l) ⇄ H⁺(aq) + OH⁻(aq)
c) What is the conducting species in this initial solution?
The main responsible for conductivity are the ions coming from H₂SO₄: H⁺, HSO₄⁻, SO₄²⁻.
d) Describe what happens as titrant is added to the beaker. Why does the conductivity of the solution decrease? What is the identity of the solid formed? What is the conducting species present in the beaker?
As the titration takes place, reaction [1] proceeds to the right. The conductivity of the solution decreases because the amount of H⁺, HSO₄⁻, SO₄²⁻ decreases. The formed solid is barium sulfate BaSO₄. Since BaSO₄ is very insoluble, the main responsible for conductivity are still H⁺, HSO₄⁻ and SO₄²⁻,
e) What happens when the conductivity value reaches its minimum value (which is designated as the equivalence point for this type of titration)? What is the conducting species in the beaker?
At the equivalence point equivalent amounts of H₂SO₄ and Ba(OH)₂ react. Only BaSO₄ and H₂O are present, and since they are weak electrolytes, there is a small amount of ions to conduct electricity. The conducting species are Ba²⁺ and SO₄²⁻ coming from BaSO₄ and H⁺ and OH⁻ coming from H₂O.
f) Describe what happens at additional titrant is added past the equivalence point. Why does the conductivity of the solution increase? What is the conducting species present in the beaker?
After the equivalence point there is an excess of Ba(OH)₂. The ions Ba²⁺ and OH⁻ are responsible for the increase in the conductivity, being the major conducting species.
Final answer:
The chemical reaction between H2SO4 and Ba(OH)2 forms BaSO4 and water, reducing conductivity by reducing the number of free ions. Beyond the equivalence point, the conductivity increases due to the dissociated ions from the excess Ba(OH)2 in the solution.
Explanation:
Chemical Reaction and Metric Titration
Firstly, the equation representing the reaction between sulfuric acid (H2SO4) and barium hydroxide (Ba(OH)2) is:
Ba(OH)2 (aq) + H2SO4 (aq) → BaSO4 (s) + 2H2O (l)
In the beginning, the solution only contains H2SO4 with its dissociated ions serving as the conducting species. As titrant (Ba(OH)2) is added, they react to form BaSO4, a solid precipitate reducing the number of free ions in the solution, thus decreasing conductivity. At the equivalence point, all H2SO4 has reacted, and conductivity reaches its minimum as there are lesser free ions for conduction. If additional titrant is added past the equivalence point, conductivity increases due to excess Ba(OH)2's dissociated ions that increase ion concentration in solution.
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Answers
Answer:
53.1 mL NaOH
Explanation:
Answers
Answer:
See Explanation
Explanation:
Pb2O3 is better formulated as PbO.PbO2. It is actually a mixture of the two oxides of lead, lead II oxide and lead IV oxide.
This implies that this compound Pb2O3 (sometimes called lead sesquioxide) is a mixture of the oxides of lead in its two known oxidation states +II and +IV.
Hence Pb2O3 contains PbO and PbO2 units.