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STatiana [176]
3 years ago
11

Equations can be balanced by using the half-reaction method. Which step should be completed immediately after finding the oxidat

ion states of atoms?
A;inserting the coefficients
B;balancing the half reactions
C;identifying the half reactions
D;inspecting the number of atoms
Chemistry
2 answers:
Dovator [93]3 years ago
8 0
After finding the oxidation states of atoms, you identify the half reactions (option c).

The half reactions are given by the change of the oxidation states of the atoms.

For example if Cu is in the left side with oxidation state 0 and in the other side with oxidation state 2+, then there you have a half reaction (oxidation reaction). And if you have O with oxidation state 0 in the left side  and with oxidation state 2- in the right side, there you have other half reaction (reducing reaction).
Marat540 [252]3 years ago
5 0

Answer:

c

Explanation:

You might be interested in
I want to know which ones are molecular equation, complete ionic equation and net ionic equation
NNADVOKAT [17]

Answer:

The molecular equations are:

1. CuSO₄ (aq) + 2 KOH (aq) ----> Cu(OH)₂ (s) + K₂SO₄ (aq)

2. Ba(NO₃)₂ (aq) + K₂SO₄ (aq) + BaSO₄ (s) + 2 KNO₃ (aq)

The complete ionic equations are:

1. Ag + (aq) + NO₃- (aq) + I- (aq) + Na (aq) ---> AgI (s) + No₃- (aq) + Na+ (aq)

2. Cu²+ + SO₄²- (aq) + 2 K+ (aq) + 2 OH- (aq) ---> Cu(OH)₂ (s) + 2K+ (aq) + SO₄²- (aq)

The net ionic equations are:

1. Ca²+ (aq) + SO₄²- (aq) ---> CaSO₄ (s)

2. Ba²+ (aq) +SO₄²- (aq) ---> BaSO₄ (s)

Explanation:

A molecular equation is a balanced chemical equation which shows the reacting species as molecules rather than as componenet ions in their compounds with subscripts written beside the molecules to indicate the state in which they occur in the chemical reaction.

An ionic equation expresses the reacting species as components ions in a chemical reation. All the ions and molecules reacting are shown.

In a net ionic equation, the ions which remain in the ionic state also known as spectator ions are not written as part of the equation.

From the given attachment;

The molecular equations are:

1. CuSO₄ (aq) + 2 KOH (aq) ----> Cu(OH)₂ (s) + K₂SO₄ (aq)

2. Ba(NO₃)₂ (aq) + K₂SO₄ (aq) + BaSO₄ (s) + 2 KNO₃ (aq)

The complete ionic equations are:

1. Ag + (aq) + NO₃- (aq) + I- (aq) + Na (aq) ---> AgI (s) + No₃- (aq) + Na+ (aq)

2. Cu²+ + SO₄²- (aq) + 2 K+ (aq) + 2 OH- (aq) ---> Cu(OH)₂ (s) + 2K+ (aq) + SO₄²- (aq)

The net ionic equations are:

1. Ca²+ (aq) + SO₄²- (aq) ---> CaSO₄ (s)

2. Ba²+ (aq) +SO₄²- (aq) ---> BaSO₄ (s)

8 0
2 years ago
Object A at 40ºC and object B at 80ºC are placed in contact with each other. Which statement describes the heat flow between the
Arada [10]
By the second law of thermodynamics:
Heat can not spontaneously flow from cold regions to hot regions without external work being performed on a system.
Heat transfer is the passage of thermal energy from a hot ( t B = 80° C ) to a colder body ( t A = 40° C ). 
Answer:  B ) Heat flows from object B to object A.
6 0
3 years ago
Read 2 more answers
An initial mixture of nitrogen gas and hydrogen gas is reacted in a rigid container at a certain temperature by the reaction At
anygoal [31]

Answer: [N2]₀ = 10M and [H2]₀ = 11M

Explanation: To calculate the initial concentration, you would have to set up an ICE table, which is an organized way of tracking known quantities or the ones you want to find. ICE stands for:

I is initial amount;

C is change in concentration;

E is for equilibrium concentration;

For the mixture,

       N2                       3H2                2NH3

I      [N2]₀                     [H2]₀                  0

C     - x                          -3x                 +2x

E     [N2]₀ - x =8      [H2]₀ - 3x =5       2x =4

With the product, we can find "x":

2x=4

x=2M

With x=2, find the concentrations:

[N2]₀ - x = 8

[N2]₀ = 10M

[H2]₀ - 3x = 5

[H2]₀ = 11M

The initial concentrations of nitrogen gas [N2] is 10.0 M and of hydrogen gas [H2] is 11.0 M.

8 0
2 years ago
Read 2 more answers
Suppose 0.981 g of iron (II) iodide is dissolved in 150. mL of a 35.0 m M aqueous solution of silver nitrate. Calculate the fina
yaroslaw [1]

Answer:

Final molarity of iodide ion C(I-) = 0.0143M

Explanation:

n = (m(FeI(2)))/(M(FeI(2))

Molar mass of FeI(3) = 55.85+(127 x 2) = 309.85g/mol

So n = 0.981/309.85 = 0.0031 mol

V(solution) = 150mL = 0.15L

C(AgNO3) = 35mM = 0.035M = 0.035m/L

n(AgNO3) = C(AgNO3) x V(solution)

= 0.035 x 0.15 = 0.00525 mol

(AgNO3) + FeI(3) = AgI(3) + FeNO3

So, n(FeI(3)) excess = 0.00525 - 0.0031 = 0.00215mol

C(I-) = C(FeI(3)) = [n(FeI(3)) excess]/ [V(solution)] = 0.00215/0.15 = 0.0143mol/L or 0.0143M

8 0
3 years ago
A 100.0 mL sample of 0.300 M NaOH is mixed with a 100.0 mL sample of 0.300 M HNO3 in a coffee cup calorimeter. If both solutions
Mars2501 [29]

Answer:

Qm  = -55.8Kj/mole

Explanation:

NaOH(aq) + HNO₃(aq) => NaNO₃(aq) + H₂O(l)

Qm = (mc∆T)water /moles acid

Given => 100ml(0.300M) NaOH(aq) + 100ml(0.300M)HNO₃(aq)

=> 0.03mole NaOH(aq) + 0.03mole HNO₃(aq)

=> 0.03mole NaNO₃(aq) + 0.03mole H₂O(l)

ΔH⁰rxn = [(200ml)(1.00cal/g∙°C)(37 – 35)°C]water / 0.03mole HNO₃

= 13,333 cal/mole x 4.184J/cal = 55,787J/mol = 55.8Kj/mole (exothermic)*

Heat of reactions comes from formation of H-Oxy bonds on formation of water of reaction and heats the 200ml of solvent water from 35⁰C to 37⁰C.

4 0
3 years ago
Read 2 more answers
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