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3 years ago

A common oxidizing agent is..?

Chemistry

1 answer:

zimovet [89]3 years ago

8 0

Answer:

C. KMnO4/H+

Explanation:

Oxidizing agent is a substance that has the ability to accept electrons from other substances.

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Water expands in volume as it freezes, because it's molecules form an open ___ structure

Grace [21]

Because it is Disordered

6 0

3 years ago

How to balance _h2s+ _o2 = _h2o+ _s

goldfiish [28.3K]

Answer:

2H₂S + 1O₂ → 2H₂O + 2S

Explanation:

SOLUTION :-

Balance it by using 'hit & trial' method , and you'll get the answer :-

2H₂S + 1O₂ → 2H₂O + 2S

VERIFICATION :-

In reactant side of equation :-

Number of atoms in H = 2×2 = 4
Number of atoms in S = 2×1 = 2
Number of atoms in O = 1×2 = 2

In product side of equation :-

Number of atoms in H = 2×2 = 4
Number of atoms in O = 2×1 = 2
Number of atoms in S = 2×1 = 2

Number of atoms of each element is equal in both reactant & product side of equation. Hence , the equation is balanced.

5 0

3 years ago

A galvanic cell at a temperature of 42 degrees Celcius is powered by the following redox reaction:

seropon [69]

Answer: The cell voltage of the given reaction is 1.86 V

Explanation:

The given chemical equation follows:

$3Cu^{2+}(aq.)+2Al\rightarrow 2Al^{3+}(aq.)+2Au(s)$

Oxidation half reaction: $Al(aq.)\rightarrow Al^{3+}(aq.)+3e^-;E^o_{Al^{3+}/Al}=1.66V$ ( × 2)

Reduction half reaction: $Cu^{2+}(aq.)+2e^-\rightarrow Cu(s);E^o_{Cu^{2+}/Cu}=0.16V$ ( × 3)

Oxidation reaction occurs at anode and reduction reaction occurs at cathode.

To calculate the $E^o_{cell}$ of the reaction, we use the equation:

$E^o_{cell}=E^o_{cathode}-E^o_{anode}$

Putting values in above equation, we get:

$E^o_{cell}=0.16-(-1.66)=1.82V$

To calculate the EMF of the cell, we use the Nernst equation, which is:

$E_{cell}=E^o_{cell}-\frac{2.303RT}{nF}\log \frac{[Al^{3+}]^2}{[Cu^{2+}]^3}$

where,

$E_{cell}$ = electrode potential of the cell = ? V

$E^o_{cell}$ = standard electrode potential of the cell = +1.82 V

n = number of electrons exchanged = 2

R = Gas constant = 8.314 J/mol Kl

T = temperature = $42^oC=[42+273]K=315K$

F = Faraday's constant = 96500

$[Al^{3+}]=1.63M$

$[Cu^{2+}]=3.43M$

Putting values in above equation, we get:

$E_{cell}=1.82-\frac{2.303\times 8.314\times 315}{2\times 96500}\times \log(\frac{(1.63)^2}{(3.43)^3})\\\\E_{cell}=1.86V$

Hence, the cell voltage of the given reaction is 1.86 V

4 0

3 years ago

A lab requires 3.50L of a 2.0M solution of HCl. What volume of a 6.0M solution is needed to make this solution?

Alchen [17]

The answer is 1.167L of 6.0M solution.

4 0

3 years ago

Read 2 more answers

Estimate ΔH for the reaction: C2H6(g) + Cl2(g)--> C2H5Cl(g) + HCl(g) given the following average bond energies (in kJ/mol): C

Leno4ka [110]

Explanation:

The reaction equation will be as follows.

$C_{2}H_{6}(g) + Cl_{2}(g) \rightarrow C_{2}H_{5}Cl(g) + HCl(g)$

Using bond energies, expression for calculating the value of $\Delta H$ is as follows.

$\Delta H = \sum B.E_{reactants} - \sum B.E_{products}$

On reactant side, from $C_{2}H_{6}$ number of bonds are as follows.

C-C bonds = 1

C-H bonds = 6

From $Cl_{2}$ ; Cl-Cl bonds = 1

On product side, from $C_{2}H_{5}Cl$ number of bonds are as follows.

C-C bonds = 1

C-H bonds = 5

C-Cl bonds = 1

From HCl; H-Cl bonds = 1

Hence, using the bond energies we will calculate the enthalpy of reaction as follows.

$\Delta H = \sum B.E_{reactants} - \sum B.E_{products}$

= $[(1 \times 348 kJ/mol) + (6 \times 414 kJ/mol) + (1 \times 242 kJ/mol)] - [(1 \times 348 kJ/mol) + (5 \times 414 kJ/mol) + (1 \times 327 kJ/mol) + (1 \times 431 kJ/mol)]$ = -102 kJ/mol

Thus, we can conclude that change in enthalpy for the given reaction is -102 kJ/mol.

5 0

3 years ago