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3 years ago

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For an alloy containing 50% cu and 50% ni, calculate the amounts of solid s and liquid l presents at 1300c. show that the total

nickel content in the solid and liquid phases adds up to 50%. (5')

1 answer:

iris [78.8K]3 years ago

3 0

100 %cuni calculate the amounts of solid s and liquid

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For the gas phase decomposition of 1-bromopropane, CH3CH2CH2BrCH3CH=CH2 + HBr the rate constant at 622 K is 6.43×10-4 /s and the

Radda [10]

Answer is: activation energy of this reaction is 212,01975 kJ/mol.<span>
Arrhenius equation: ln(k</span>₁/k₂) = Ea/R (1/T₂ - 1/T₁).<span>
k</span>₁ = 0,000643 1/s.<span>
k</span>₂ = 0,00828 1/s.

T₁ = 622 K.

T₂ = 666 K.

R = 8,3145 J/Kmol.

<span> 1/T</span>₁ = 1/622 K = 0,0016 1/K.<span>
1/T</span>₂ = 1/666 K = 0,0015 1/K.<span>
ln(0,000643/0,00828) = Ea/8,3145 J/Kmol · (-0,0001 1/K).
-2,55 = Ea/8,3145 J/Kmol </span>· (-0,0001 1/K).<span>
Ea = 212019,75 J/mol = 212,01975 kJ/mol.</span>

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3 years ago

Which of the following is not an example of a polymer?

drek231 [11]

Sucrose <span>is not an example of a polymer. The answer is letter C. Polystyrene, Nylon and PVC or poly vinyl chloride are examples of polymer</span>

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3 years ago

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Which ancient African Kingdom is located in western Africa?

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3 years ago

g 2BrO3- + 5SnO22-+ H2O5SnO32- + Br2+ 2OH- In the above reaction, the oxidation state of tin changes from to . How many electron

Archy [21]

Answer:

In the above reaction, the oxidation state of tin changes from 2+ to 4+.

10 moles of electrons are transferred in the reaction

Explanation:

Redox reaction is:

2BrO₃⁻ + 5SnO₂²⁻ + H2O ⇄ 5SnO₃²⁻ + Br₂ + 2OH⁻

SnO₂²⁻ → SnO₃²⁻

Tin changes the oxidation state from +2 to +4. It has increased it so this is the oxidation from the redox (it released 2 e⁻). We are in basic medium, so we add water in the side of the reaction where we have the highest amount of oxygen. We have 2 O on left side and 3 O on right side so we add 1 water on the right and we complete with OH⁻ in the opposite side to balance the H.

SnO₂²⁻ + 2OH⁻ → SnO₃²⁻ + 2e⁻ + H₂O <u>Oxidation</u>

BrO₃⁻ → Br₂

First of all, we have unbalance the bromine, so we add 2 on the BrO₃⁻. We have 6 O in left side and there are no O on the right, so we add 6 H₂O on the left. To balance the H, we must complete with 12OH⁻. Bromate reduces to bromine at ground state, so it gained 5e⁻. We have 2 atoms of Br, so finally it gaines 10 e⁻.

6H₂O + 10 e⁻ + 2BrO₃⁻ → Br₂ + 12OH⁻ <u>Reduction</u>

In order to balance the main reaction and balance the electrons we multiply (x5) the oxidation and (x1) the reduciton

(SnO₂²⁻ + 2OH⁻ → SnO₃²⁻ + 2e⁻ + H₂O) . 5

(6H₂O + 10 e⁻ + 2BrO₃⁻ → Br₂ + 12OH⁻) . 1

5SnO₂²⁻ + 10OH⁻ + 6H₂O + 10 e⁻ + 2BrO₃⁻ → Br₂ + 12OH⁻ + 5SnO₃²⁻ + 10e⁻ + 5H₂O

We can cancel the e⁻ and we substract:

12OH⁻ - 10OH⁻ = 2OH⁻ (on the right side)

6H₂O - 5H₂O = H₂O (on the left side)

2BrO₃⁻ + 5SnO₂²⁻ + H2O ⇄ 5SnO₃²⁻ + Br₂ + 2OH⁻

6 0

4 years ago

Five kilograms of liquid carbon tetrachloride undergo a mechanically reversible, isobaric change of state at 1 bar during which

Drupady [299]

Answer:

Explanation:

From the information given:

Mass of carbon tetrachloride = 5 kg

Pressure = 1 bar

The given density for carbon tetrachloride = 1590 kg/m³

The specific heat of carbon tetrachloride = 0.84 kJ/kg K

From the composition, the initial volume of carbon tetrachloride will be: $= \dfrac{5 \ kg }{1590 \ kg/m^3}$

= 0.0031 m³

Suppose $\beta$ is independent of temperature while pressure is constant;

Then:

The change in volume can be expressed as:

$\int ^{V_2}_{V_1} \dfrac{dV}{V} =\int ^{T_2}_{T_1} \beta dT$

$In ( \dfrac{V_2}{V_1}) = \beta (T_2-T_1)$

$V_2 = V_1 \times exp (\beta (T_2-T_1))$

$V_2 = 0.0031 \ m^3 \times exp (1.2 \times 10^{-3} \times 20)$

$V_2 = 0.003175 \ m^3$

However; the workdone = -PdV

$W = -1.01 \times 10^5 \ Pa \times ( 0.003175 m^3 - 0.0031 \ m^3)$

W = - 7.6 J

The heat energy Q = Δ h

$Q = mC_p(T_2-T_1)$

$Q = 5 kg \times 0.84 \ kJ/kg^0 C \times 20$

Q = 84 kJ

The internal energy is calculated by using the 1st law of thermodynamics; which can be expressed as;

ΔU = ΔQ + W

ΔU = 84 kJ + ( -7.6 × 10⁻³ kJ)

ΔU = 83.992 kJ

3 0

3 years ago