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Brilliant_brown [7]
3 years ago
11

What kind of solid often has the highest melting points?

Chemistry
2 answers:
nevsk [136]3 years ago
7 0

I think it is C for this question

kupik [55]3 years ago
3 0

B solids made of atoms

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What is the difference between the Lewis model and the valence-shell electron pair repulsion (VSEPR) model?
grandymaker [24]

Answer: only the VSEPR mode shows the geometric shape of a formula.

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3 years ago
Consider an Al-4% Si alloy. Determine (a) if the alloy is hypoeutectic or hypereutectic, (b) the composition of the first solid
ozzi

Answer:

(a) Hypoeutectic

(b) Alpha solid, aluminium

(c) 70% α , 30% β

(d) 97.6% α, 2.4% β

(e) 97.6% α, 2.4% β

(f) 97% α, 3% β

Explanation:

(a) The eutectic composition for Al Si alloy is 11.7 wt% silicon, therefore, an Al-4% Si alloy is hypoeutectic

(b) For the hypoeutectic alloy, aluminium, Al, is expected to form first, such that the aluminium content is reduced till the point it gets to the eutectic proportion of 11.7 wt% silicon

(c) At  578°C we have

% α:  Al      (11 - 4)/(11 - 1) = 70% α

% L:  Si      100 - 70 = 30% β

(d) At  576°C we have

α: 99.83% Si    (99.83 - 4)/(99.83- 1.65) = 97.6% α

β: 1.65% Si (4 - 1.65)/(99.83- 1.65) = 2.4% β

(e) Primary α: 1.65% α (99.83 - 4)/(99.83 - 1.65) = 97.6% α

Eutectic 4% Si  = 100 - 97.6 = 2.4% β

(f) At 25°C we have;

α%: (99.83 - 4)/(99.83 - 1) = 97% α

β%:   100 - 97 = 3% β.

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In a first-order decomposition reaction, 50.0% of a compound decomposes in 10.5 min.
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In this reaction 50% of the compound decompose in 10.5 min thus, it is half life of the reaction and denoted by symbol t_{1/2}.

(a) For first order reaction, rate constant and half life time are related to each other as follows:

k=\frac{0.6932}{t_{1/2}}=\frac{0.6932}{10.5 min}=0.066 min^{-1}

Thus, rate constant of the reaction is 0.066 min^{-1}.

(b) Rate equation for first order reaction is as follows:

k=\frac{2.303}{t_{1/2}}log\frac{[A_{0}]}{[A_{t}]}

now, 75% of the compound is decomposed, if initial concentration [A_{0} ] is 100 then concentration at time t [A_{t} ] will be 100-75=25.

Putting the values,

0.066 min^{-1}=\frac{2.303}{t}log\frac{100}{25}=\frac{2.303}{t}(0.6020)

On rearranging,

t=\frac{2.303\times 0.6020}{0.066 min^{-1}}=21 min

Thus, time required for 75% decomposition is 21 min.

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