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SSSSS [86.1K]
4 years ago
9

Balance the following half reaction in basic conditions. Then, indicate the coefficients for H2O and OH– for the balanced half r

eaction, and which side they appear on. Si (s) + Mg(OH)2 (s) → Mg (s) + SiO32- (aq)
Chemistry
1 answer:
Ugo [173]4 years ago
7 0

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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A vapor composed of 0.60 mole fraction ethanol and 0.40 mole fraction acetic acid at 120.0 mmHg (absolute), is in equilibrium wi
Kazeer [188]

Answer:

0.00833

Explanation:

Log p1sat(mm Hg)= 8.20417-1642.89/(t+230.3), t in deg,c

Log p2sat(mm Hg)= 7.5596-1644.05/(t+233.524), t in deg,c

Let y1= mole fraction of ethanol in the vapor phase= 0.6, y2= mole fraction of acetic acid in the vapor phase= 0.4

Let x1=mole fraction of ethanol in the liquid phase, x2= mole fraction of acetic acid in the liquid phase.

From Raoult’s law, y1P= x1P1sat   (1) , P1sat= saturation pressure of ethanol , P2sat= saturation pressure of acetic acid, P= total pressure, y2P=x2p2sat (2)

Eq.1 can be written as x1= y1P/P1sat and x2= y2P//P2sat

Addition of x1 and x2 gives x1+x2= y1P/P1sat + y2P/P2sat

Hence y1/P1sat + y2/P2sat= 1/P

0.6/P1sat +0.4/P2sat= 1/120 = 0.00833 (1)

this proves by trial and error procedure. we can assume some temperature, Calculate p1sat ( for ethanol and P2sat for acetic acid) and calculate the LHS of Eq.3 and check whether it relates with RHS of the equation. The calculations are done in excel.

The iterative procedure gives temperature as 53.16 deg.c        

T(deg.c)                                                         53.16                                                                                                                                    

P1sat ( mm Hg)                                            256.0505949                                                                                              

P2sat ( mm Hg)                                             66.81725201                                                                                                                                      0.6/P1sat                                                         0.002343287                                                                                                

0.4/P2sat                                                        0.005986478                                                                                                    

Calc                                                                 0.008329765                                                                                                

Answer                                                            0.00833        

Hence y1P=x1P1sat, x1=y1P/P1sat =0.6*120/256 =0.28, x2= 1-0.28=0.72                                                                                                

8 0
3 years ago
The general form for a double replacement reaction is
yawa3891 [41]

Answer:

A is the right answer

Explanation:

In double replacement reaction two compounds combines to give other two compounds

As example:

CaCl2 + 2AgNO3 = Ca(NO3)2+ 2AgCl

7 0
3 years ago
How many moles of calcium chloride, cacl2, can be added to 1.5 l of 0.020 m potassium sulfate, k2so4, before a precipitate is ex
Brut [27]
When CaSO4 → Ca2+ + SO4
So when we have Ksp = [Ca2+][SO4]

when Ksp = 4.93 x 10^-5
and [SO4] = 0.02 M 
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4.93x10^-5 = [Ca2+] [0.02]
∴ [Ca2+] = 0.0025 mol/L

∴ the moles of calcium chloride = 0.0025 mol / L * 1.5 L
                                                      = 0.00167 mol

4 0
4 years ago
Entropy change in an internally reversible isothermal process is _______ to the temperature of the system.
alukav5142 [94]

Answer:

Internally reversible is the answer.

Explanation:

3 0
3 years ago
An ideal gas is contained in a cylinder with a volume of 5.0x102 mL at a temperature of 30°C and a pressure of 710. Torr. The ga
Scorpion4ik [409]

Answer:

51207 torr is the new pressure of the gas

Explanation:

We can solve this question using combined gas law that states:

P1V1T2 = P2V2T1

<em>Where P is pressure, V volume and T absolute temperature of 1, initial state and 2, final state of the gas</em>

<em> </em>

Computing the values of the problem:

P1 = 710torr

V1 = 5.0x10²mL

T1 = 273.15 + 30°C = 303.15K

P2 = ?

V2 = 25mL

T2 = 273.15 + 820°C = 1093.15K

Replacing:

710torr*5.0x10²mL*1093.15K = P2*25mL*303.15K

3.881x10⁸torr*mL*K = P2 * 7.579x10³mL*K

P2 = 51207 torr is the new pressure of the gas

4 0
3 years ago
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