CHEMISTRY COLLEGE

Balance the following half reaction in basic conditions. Then, indicate the coefficients for H2O and OH– for the balanced half reaction, and which side they appear on. Si (s) + Mg(OH)2 (s) → Mg (s) + SiO32- (aq)

Answers

Answer 1

Answer:

Answer:

The ballance half reactions are:

Mg²⁺ + 2e⁻ → Mg

6OH⁻ + Si → SiO₃²⁻ + 4e⁻ + 3 H₂O

Coefficients for H2O and OH– are 3 for H₂O (in products side) and 6 for OH⁻ (in reactants side)

Explanation:

Si (s) + Mg(OH)₂ (s) → Mg (s) + SiO₃²⁻ (aq)

Let's see the oxidations number.

As any element in ground state, we know that oxidation state is 0, so Si in reactants and Mg in products, have 0.

Mg in reactants, acts with +2, so the oxidation number has decreased.

This is the reduction, so it has gained electrons.

Si in reactants acts with 0 so in products we find it with +4. The oxidation number increased it, so this is oxidation. The element has lost electrons.

Let's take a look to half reactions:

Mg²⁺ + 2e⁻ → Mg

Si → SiO₃²⁻ + 4e⁻

In basic medium, we have to add water, as the same amount of oxygen we have, IN THE SAME SIDE. We have 3 oxygens in products, so we add 3 H₂O and in the opposite site we can add OH⁻, to balance the hydrogen. The half reaciton will be:

6OH⁻ + Si → SiO₃²⁻ + 4e⁻ + 3 H₂O

If we want to ballance the main reaction we have to multiply (x2) the half reaction of oxidation. So the electrons can be ballanced.

2Mg²⁺ + 4e⁻ → 2Mg

Now, that they are ballanced we can sum the half reactions:

2Mg²⁺ + 4e⁻ → 2Mg

6OH⁻ + Si → SiO₃²⁻ + 4e⁻ + 3 H₂O

2Mg²⁺ + 4e⁻ + 6OH⁻ + Si → 2Mg + SiO₃²⁻ + 4e⁻ + 3 H₂O

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What may happen to the human body when exposed to an infectious agent? A. When infectious agents get into the human body, your body responds by functioning normally.

B. When infectious agents get into the human body, the body gets a surge of energy, causing a slight increase in body temperature, and you feel great.

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Answers

Answer:

when infectious agents get into the human body your body responds by raising the core body temperature causing a fever

The answer is C it will raise your body temp and cause a fever

6.579 rounded to nearest hundreth

Answers

Answer:

6.58

Explanation:

Write the equilibrium expression, calculate KEQ and then tell where the equilibrium lies: Fe (s) + O2 (g) ↔ Fe2O3 (s) In a 2.0 L Container At equilibrium: Fe = 1.0 mol O2 = 1.0 E-3 mol Fe2O3 = 2.0 mol

Answers

Answer:

The value of Keq is 4e-9. See the solution below

Explanation:

We need to balanced rhe equation and use the formula of the Keq

A 9.70-g piece of solid CO 2 is allowed to sublime in a balloon. The final volume of the balloon is 1.00 L at 298 K. What is the pressure of the gas inside the balloon?a. 5.39 atmb. 2.37  102 atmc. 2.52 atmd. 0.186 atme. none of these

Answers

Answer:

a. 5.39 atm

Explanation:

Pressure = ?

Volume = 1 L

Temperature = 298 K

Mass = 9.70g

The formular relating these variables is the ideal gas equation;

PV = nRT

where R = gas constant = 0.082057 L.atm.K-1.mol-1

n can be obtained from the formular below;

n = mass / molar mass = 9.70 / 44

n = 0.2205 mol

P = nRT / V

P = 0.2205 * 0.082057 * 298 / 1

P = 5.392 atm

The correct option is option A.

Iron has density of 7.9g/cm3. What is the mass of a cube with the length of one side equal to 1.64x10squared cm3?

Answers

Given data:

Density of iron (Fe) = 7.9 g/cm3

Length of one side of the iron cube = 1.64 * 10^2 cm

Now, the volume (V) of a cube in which the length of the side is 'a' cm is given as:

V = a^3

Volume of iron cube = (1.64 *10^2 cm)^3 = 4.41 * 10^6 cm3

The density (D) of an object of mass (m) and volume (V) is given as:

D = m/V

or, m = D*V

Therefore, mass of iron cube = 7.9 g/cm3 * 4.41 * 10^6 cm3

= 34.84 *10^6 g

Consider the following unbalanced equation for the combustion of hexane: αC6H14(g)+βO2(g)→γCO2(g)+δH2O(g) Part A Balance the equation. Give your answer as an ordered set of numbers α, β, γ, ... Use the least possible integers for the coefficients. α α , β, γ, δ = nothing Request Answer Part B Determine how many moles of O2 are required to react completely with 5.6 moles C6H14. Express your answer using two significant figures. n n = nothing mol Request Answer Provide Feedback