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3 years ago

H

Chemistry

1 answer:

Irina18 [472]3 years ago

3 0

Answer:

Hehehe :)

Interesting.

Phase 1.

Zinc heated in Air

2Zn + O₂ = 2ZnO

Zinc Oxide is formed.

SOLID A = (ZnO)ZINC OXIDE

Phase 2.

Zinc Oxide is heated with Powdered Magnesium.

magnesium would displace Zinc since its higher in the series.

ZnO + Mg = MgO + Zn

SOLID B + SOLID C = MgO(Magnesium Oxide) + Zn(Zinc metal)

At Phase 3...( 1st Aspect)

Both MgO and Zn can react with Dilute HCl.

Now the products required from the question are a "SOLUTION D" AND "GAS E".

When MgO reacts with Dilute HCl

The products are

MgO + 2HCl = MgCl₂ + H₂O

Now H₂O could be steam(gas) or Water :(

This is a little Uncertain.

But if we use Zn as the one reacting with Dilute HCl

We have

Zn + 2HCl = ZnCl₂ + H₂

Now I'll go with this cos H₂ is simply Hydrogen Gas.

SOLUTION D + GAS E = ZnCl₂(Zinc Chloride) + H₂(Hydrogen Gas)

2nd aspect of Phase 3 ( or denominator)

ELECTROLYSIS OF MOLTEN B

B = MgO

MOLTEN B = MOLTEN MAGNESIUM CHLORIDE.

From Electrolysis of MgO

Mg²⁺ + 2e- = Mg

2O²⁻ = O₂ + 2e⁻

Combining Both Equations.

Mg²⁺ + 2O²⁻ = 2Mg(ₛ) + O₂(g)

Magnesium is a Silvery Metal :)

and Oxygen is a Gas.

Wow!!!

So this automatically makes "SILVERY METAL F" AND "GAS G" TO BE MAGNESIUM AND OXYGEN RESPECTIVELY.

SILVERY METAL F = MAGNESIUM

SILVERY METAL F = MAGNESIUMGAS G = OXYGEN

All the way down to phase 5 now.

Hope you're with me?

Now CuSO₄ is added to Mixture of Mg and O₂. Oxygen would prolly escape into the atmosphere

Leaving Mg to react with CuSO₄

Mg + CuSO₄ = MgSO₄(aq) + Cu(s)

Copper when precipitated from solution sometimes has Brown Pink Color.

MgSO₄ is Solid.

Niceeeee

That's exactly what they need as the answer.

Automatically...

BROWN PINK SOLID H = COPPER(CU)

SOLUTION I = MgSO₄(MAGNESIUM SULPHATE)

Hope this Helps :)

Have a great day.

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Calculate the Equilibrium constant Kc at 25 C from the free - energy change for the following reaction . Zn(s) + 2Ag+(aq)<---

FinnZ [79.3K]

Answer: $Kc=1.24*10^5^2$

Explanation: For the given reaction:

$Kc=\frac{[Zn^+^2]}{[Ag^+]^2}$

Concentrations of the ions are not given so we need to think about another way to calculate Kc.

We can calculate the free energy change using the standard cell potential as:

$\Delta G^0=-nFE^0_c_e_l_l$

$E^0_c_e_l_l$ can be calculated using standard reduction potentials.

Standard reduction potential for zinc is -0.76 V and for silver, it is +0.78 V.

$E^0_c_e_l_l$ = $E^0_c_a_t_h_o_d_e+E^0_a_n_o_d_e$

Reduction takes place at anode and oxidation at cathode. As silver is reduced, it is cathode. Zinc is oxidized and so it is anode.

$E^0_c_e_l_l$ = 0.78 V - (-0.76 V)

$E^0_c_e_l_l$ = 0.78 V + 0.76 V

$E^0_c_e_l_l$ = 1.54 V

Value of n is two as two moles of electrons are transferred in the cell reaction F is Faraday constant and its value is 96485 C/mol of electron .

$\Delta G^0=-(2*96485*1.54)$

$\Delta G^0$ = -297173.8 J

Now we can calculate Kc using the formula:

$\Delta G^0=-RTlnKc$

T = 25+273 = 298 K

R = $8.314JK^-^1mol^-^1$

--297173.8 = -(8.314*298)lnKc

297173.8 = 2477.572*lnKc

$lnKc=\frac{297173.8}{2477.572}$

lnKc = 119.946

$Kc=e^1^1^9^.^9^4^6$

$Kc=1.24*10^5^2$

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4 years ago

The balanced combustion reaction for C6H6 is 2C6H6(l)+15O2(g)⟶12CO2(g)+6H2O(l)+6542 kJ If 8.600 g C6H6 is burned and the heat pr

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Answer : The final temperature of the water is, $22.5166^oC$

Explanation : Given,

Mass of benzene = 8.600 g

Molar mass of benzene = 78 g/mole

First we have to calculate the moles of benzene.

$\text{Moles of benzene}=\frac{\text{Mass of benzene}}{\text{Molar mass of benzene}}=\frac{8.600g}{78g/mol}=0.1103mole$

Now we have to calculate the energy of combustion.

The given balanced chemical reaction is:

$2C_6H_6(l)+15O_2(g)\rightarrow 12CO_2(g)+6H_2O(l)+6542 kJ$

According to reaction,

As, 2 moles of benzene gives 6542 kJ of energy on combustion.

So, 0.1103 mole of benzene gives $\frac{6542 kJ}{2}\times 0.1103=360.7913kJ$ of energy on combustion.

Now we have to calculate the final temperature of the water.

Formula used : $q_w=m_w\times c_w\times \Delta T=m_w\times c_w\times (T_{final}-T_{initial})$

where,

$q_w$ = heat released = 360.7913 kJ = 36079.13 J

$m_w$ = mass of water = 5691 g

$c_w$ = specific heat of water= $4.18J/g^oC$

$T_{final}$ = final temperature = ?

$T_{initial}$ = initial temperature = $21^oC$

Now put all the given values in the above formula, we get:

$36079.13J=5691g\times 4.18J/g^oC\times (T_{final}-21^oC)$

$T_{final}=22.5166^oC$

Therefore, the final temperature of the water is, $22.5166^oC$

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10. B.
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