Carbon in the lithosphere is held in soil in the form of both organic and inorganic carbon which often as calcium carbonate. Carbon can leave the soil through soil respiration – which releases CO2, or by erosion – which can carry it into rivers or the ocean, where it then enters the hydrosphere.
Answer:
B) FALSE.
Explanation:
First off, its important to understand the following concepts;
A half reaction is either the oxidation or reduction reaction component of a redox (Oxidation - Reduction) reaction. A half reaction is obtained by considering the change in oxidation states of individual substances involved in the redox reaction.
Often, the concept of half-reactions is used to describe what occurs in an electrochemical cell, such as a Galvanic cell battery. Half-reactions can be written to describe both the metal undergoing oxidation (known as the anode) and the metal undergoing reduction (known as the cathode).
Half equations simply much just breaks the reaction into oxidation and reduction steps (irrespective of the order, whether left or right).
An example of half equation is given below;
Mg → Mg2+ + 2e− (Oxidation)
Cu2+ + 2e− → Cu (Reduction)
The answer is false because; oxidation-reduction reactions are NOT represented by equations that group the reaction participants into reactants and products. Rather they are
The correct answer is:
neutrality
on the pH scale 7 is neutral above 7 is basic and below 7 is acidic.
Explanation:
The pH is an example of how acidic/basic water is. The scale goes from 0 - 14, with 7 being neutral, pH of less than 7 designates acidity, whereas a pH of greater than 7 intimates a base. pH is really a pattern of the relative amount of free hydrogen and hydroxyl ions in the water.pH is a measure of how acidic/basic water is.
It is neutral because, at this pH, the molar concentrations of Hydroxide and Hydronium are same, therefore the substance is not a base nor an acid.
6 Na + 1 Fe₂O₃ → 3 Na₂O + 6 Fe
<h3>Explanation</h3>
Method One: Refer to electron transfers.
Oxidation states:
- Na: from 0 to +1; loses one electron.
- Fe: from +3 to 0; gains three electrons.
Each mole of Fe₂O₃ contains two Fe atoms and will gain 2 × 3 = 6 electrons during the reaction. It takes 6 moles of Na to supply all those electrons.
6 Na + 1 Fe₂O₃ → ? Na₂O + ? Fe
- There are two moles of Na atoms in each mole of Na₂O. 6 moles of Na will make 3 moles of Na₂O.
- There are two moles of Fe atoms in each mole of Fe₂O₃. 1 mole of Fe₂O₃ will make 2 moles of Fe.
6 Na + 1 Fe₂O₃ → 3 Na₂O + 2 Fe
Method Two: Atoms conserve.
Fe₂O₃ has the largest number of atoms among one mole of all four species in this reaction. Assume <em>one</em> as its coefficient.
? Na + <em>1</em> Fe₂O₃ → ? Na₂O + ? Fe
There are two moles of Fe atoms and three moles of O atoms in each mol of Fe₂O₃. One mole of Fe₂O₃ contains two moles of Fe and three moles of O. There are one mole of O atom in every mole of Na₂O. Three moles of O will go to three moles of Na₂O.
? Na + <em>1</em> Fe₂O₃ → <em>3</em> Na₂O + <em>2</em> Fe
Each mole of Na₂O contains two moles of Na. Three moles of Na₂O will contain six moles of Na.
<em>6</em> Na + <em>1</em> Fe₂O₃ → <em>3</em> Na₂O + <em>2</em> Fe
Simplify the coefficients. All coefficients in this equation are now full number and relatively prime. Hence the equation is balanced.
6 Na + 1 Fe₂O₃ → 3 Na₂O + 2 Fe
Answer:
Option A:
Zn(s) + Cu^(2+) (aq) → Cu(s) + Zn^(2+)(aq)
Explanation:
The half reactions given are:
Zn(s) → Zn^(2+)(aq) + 2e^(-)
Cu^(2+) (aq) + 2e^(-) → Cu(s)
From the given half reactions, we can see that in the first one, Zn undergoes oxidation to produce Zn^(2+).
While in the second half reaction, Cu^(2+) is reduced to Cu.
Thus, for the overall reaction, we will add both half reactions to get;
Zn(s) + Cu^(2+) (aq) + 2e^(-) → Cu(s) + Zn^(2+)(aq) + 2e^(-)
2e^(-) will cancel out to give us;
Zn(s) + Cu^(2+) (aq) → Cu(s) + Zn^(2+)(aq)
