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3 years ago

The chemical equation of magnesium and nitric acid with balancing process

Chemistry

1 answer:

Otrada [13]3 years ago

4 0

Mg + HNO3 -> Mg(NO3)2 + H2

On reactants side:

Mg=1

H=1

N=1

O=3

On products side:

Mg=1

H=2

N=2

O=6 (3×2)

Therefore to balance the equation, we must add 2 to HNO3 on the reactants side...

Mg + 2HNO3 -> Mg(NO3)2 + H2

Reactants:

Mg=1

H=2

N=2

O=6

Therefore, it's balanced.

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3 years ago

What volume, in milliliters, of a 0.997 M KOH solution is needed to neutralize 30.0 mL of 0.0400 M HCl?

deff fn [24]

Answer:

1.2 mL

Explanation:

We'll begin by writing the balanced equation for the reaction. This is illustrated:

HCl + KOH —> KCl + H₂O

From the balanced equation,

Mole ratio of the acid, HCl (nₐ) = 1

Mole ratio of base, KOH (n₆) = 1

Finally, we shall determine the volume of the base, KOH needed to neutralize the acid, HCl as follow:

Molarity of base, KOH (M₆) = 0.997 M

Volume of acid, HCl (Vₐ) = 30 mL

Molarity of acid, HCl (Mₐ) = 0.0400 M

Volume of base, KOH (V₆) =?

MₐVₐ / M₆V₆ = nₐ/n₆

0.04 × 30 / 0.997 × V₆ = 1/1

1.2 / 0.997 × V₆ = 1

Cross multiply

0.997 × V₆ = 1.2

Divide both side by 0.997

V₆ = 1.2 / 0.997

V₆ = 1.2 mL

Thus, the volume of the base, KOH needed to neutralize the acid is 1.2 mL.

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3 years ago

What is the mass in grams of 358 mL of ethylene glycol?

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The density of ethylene glycol is: D = 1.11 g/mL
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4 years ago

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3 years ago

Calculate the standard reaction Gibbs free energy for the following cell reactions: (a) 2 Ce41(aq) 1 3 I2(aq) S 2 Ce31(aq) 1 I32

Law Incorporation [45]

Answer:

For a: The standard Gibbs free energy of the reaction is -347.4 kJ

For b: The standard Gibbs free energy of the reaction is 746.91 kJ

Explanation:

Relationship between standard Gibbs free energy and standard electrode potential follows:

$\Delta G^o=-nFE^o_{cell}$ ............(1)

For a:

The given chemical equation follows:

$2Ce^{4+}(aq.)+3I^{-}(aq.)\rightarrow 2Ce^{3+}(aq.)+I_3^-(aq.)$

Oxidation half reaction: $Ce^{4+}(aq.)\rightarrow Ce^{3+}(aq.)+e^-$ ( × 2)

Reduction half reaction: $3I^_(aq.)+2e^-\rightarrow I_3^-(aq.)$

We are given:

$n=2\\E^o_{cell}=+1.08V\\F=96500$

Putting values in equation 1, we get:

$\Delta G^o=-2\times 96500\times (+1.80)=-347,400J=-347.4kJ$

Hence, the standard Gibbs free energy of the reaction is -347.4 kJ

For b:

The given chemical equation follows:

$6Fe^{3+}(aq.)+2Cr^{3+}+7H_2O(l)(aq.)\rightarrow 6Fe^{2+}(aq.)+Cr_2O_7^{2-}(aq.)+14H^+(aq.)$

Oxidation half reaction: $Fe^{3+}(aq.)\rightarrow Fe^{2+}(aq.)+e^-$ ( × 6)

Reduction half reaction: $2Cr^{2+}(aq.)+7H_2O(l)+6e^-\rightarrow Cr_2O_7^{2-}(aq.)+14H^+(aq.)$

We are given:

$n=6\\E^o_{cell}=-1.29V\\F=96500$

Putting values in equation 1, we get:

$\Delta G^o=-6\times 96500\times (-1.29)=746,910J=746.91kJ$

Hence, the standard Gibbs free energy of the reaction is 746.91 kJ

7 0

3 years ago

The chemical equation of magnesium and nitric acid with balancing process​

The chemical equation of magnesium and nitric acid with balancing process