Which one of the following is not an agent of chemical weathering?

Complete and balance the following redox equation using the set of smallest whole– number coefficients. Now sum the coefficients

Elena L [17]

Answer : The balanced chemical equation in a acidic solution is,

$BrO_3^-(aq)+6H^+(aq)+3Sb^{3+}(aq)\rightarrow Br^-(aq)+3H_2O(l)+3Sb^{5+}(aq)$

The sum of the coefficients is, 17

Explanation :

Redox reaction or Oxidation-reduction reaction : It is defined as the reaction in which the oxidation and reduction reaction takes place simultaneously.

Oxidation reaction : It is defined as the reaction in which a substance looses its electrons. In this, oxidation state of an element increases. Or we can say that in oxidation, the loss of electrons takes place.

Reduction reaction : It is defined as the reaction in which a substance gains electrons. In this, oxidation state of an element decreases. Or we can say that in reduction, the gain of electrons takes place.

Rules for the balanced chemical equation in acidic solution are :

First we have to write into the two half-reactions.

Now balance the main atoms in the reaction.

Now balance the hydrogen and oxygen atoms on both the sides of the reaction.

If the oxygen atoms are not balanced on both the sides then adding water molecules at that side where the less number of oxygen are present.

If the hydrogen atoms are not balanced on both the sides then adding hydrogen ion $(H^+)$ at that side where the less number of hydrogen are present.

Now balance the charge.

The given chemical reaction is,

$BrO_3^-(aq)+Sb^{3+}(aq)\rightarrow Br^-(aq)+Sb^{5+}(aq)$

The oxidation-reduction half reaction will be :

Oxidation : $Sb^{3+}\rightarrow Sb^{5+}$

Reduction : $BrO_3^-\rightarrow Br^-$

First balance the main element in the reaction.

Oxidation : $Sb^{3+}\rightarrow Sb^{5+}$

Reduction : $BrO_3^-\rightarrow Br^-$

Now balance oxygen atom on both side.

Oxidation : $Sb^{3+}\rightarrow Sb^{5+}$

Reduction : $BrO_3^-\rightarrow Br^-+3H_2O$

Now balance hydrogen atom on both side.

Oxidation : $Sb^{3+}\rightarrow Sb^{5+}$

Reduction : $BrO_3^-+6H^+\rightarrow Br^-+3H_2O$

Now balance the charge.

Oxidation : $Sb^{3+}\rightarrow Sb^{5+}+2e^-$

Reduction : $BrO_3^-+6H^++6e^-\rightarrow Br^-+3H_2O$

The charges are not balanced. Now multiplying oxidation reaction by 3 and then adding both equation, we get the balanced redox reaction.

Oxidation : $3Sb^{3+}\rightarrow 3Sb^{5+}+6e^-$

Reduction : $BrO_3^-+6H^++6e^-\rightarrow Br^-+3H_2O$

The balanced chemical equation in acidic medium will be,

$BrO_3^-(aq)+6H^+(aq)+3Sb^{3+}(aq)\rightarrow Br^-(aq)+3H_2O(l)+3Sb^{5+}(aq)$

The sum of the coefficients = 1 + 6 + 3 + 1 + 3 + 3

The sum of the coefficients = 17

7 0

3 years ago

Identify the number and kinds of atoms present in a molecule of each compound. caffeine c8h10n4o2 iron iii sulfate fe2(so4)3

finlep [7]

This problem is very easy to answer. You simply have to look at the subscripts of each element of the compound.

1. For caffeine, which has a molecular formula of C₈H₁₀N₄O₂, it contains 8 atoms of Carbon, 10 atoms of Hydrogen, 4 atoms of Nitrogen and 2 atoms of Oxygen.

2. For Iron(III) Sulfate, which has a molecular formula of Fe₂(SO₄)₃, it contains 2 atoms of Iron, 3 atoms of Sulfur, and 12 atoms of Oxygen.

6 0

3 years ago

A buffer solution is made that is 0.347 M in H2C2O4 and 0.347 M KHC2O4.

irga5000 [103]

Answer:

1. pH = 1.23.

2. $H_2C_2O_4(aq) +OH^-(aq)\rightarrow HC_2O_4^-(aq)+H_2O(l)$

Explanation:

Hello!

1. In this case, for the ionization of H2C2O4, we can write:

$H_2C_2O_4\rightleftharpoons HC_2O_4^-+H^+$

It means, that if it is forming a buffer solution with its conjugate base in the form of KHC2O4, we can compute the pH based on the Henderson-Hasselbach equation:

$pH=pKa+log(\frac{[base]}{[acid]} )$

Whereas the pKa is:

$pKa=-log(Ka)=-log(5.90x10^{-2})=1.23$

The concentration of the base is 0.347 M and the concentration of the acid is 0.347 M as well, as seen on the statement; thus, the pH is:

$pH=1.23+log(\frac{0.347M}{0.347M} )\\\\pH=1.23+0\\\\pH=1.23$

2. Now, since the addition of KOH directly consumes 0.070 moles of acid, we can compute the remaining moles as follows:

$n_{acid}=0.347mol/L*1.00L=0.347mol\\\\n_{acid}^{remaining}=0.347mol-0.070mol=0.277mol$

It means that the acid remains in excess yet more base is yielded due to the effect of the OH ions provided by the KOH; therefore, the undergone chemical reaction is:

$H_2C_2O_4(aq) +OH^-(aq)\rightarrow HC_2O_4^-(aq)+H_2O(l)$

Which is also shown in net ionic notation.

Best regards!

4 0

3 years ago

The flame in a torch used to cut metal is produced by burning acetylene (C2H2) in pure oxygen. Assuming the combustion of 1 mole

Nookie1986 [14]

Answer:

19.8 kg of C₂H₂ is needed

Explanation:

We solve this by a rule of three:

If 1251 kJ of heat are relased in the combustion of 1 mol of acetylene

95.5×10⁴ kJ of heat may be released by the combustion of

(95.5×10⁴ kJ . 1) /1251kJ = 763.4 moles of C₂H₂

Let's convert the moles to mass → 763.4 mol . 26 g/1 mol = 19848 g

If we convert the mass from g to kg → 19848 g . 1kg / 1000g = 19.8 kg

7 0

2 years ago

Read 2 more answers

Which phase change results in atoms with the highest kinetic energy? A. Boiling B. Melting C. Freezing D. Condensing

suter [353]

Answer: Boiling because it makes the molecules in water bounce around.

Hope it helps.

4 0

3 years ago