What’s Silicon’s melting point?

Chemistry

2 answers:

torisob [31]3 years ago

8 0

Answer:

2,577°F

Also,

Boron: 3,769°F (2,076°C)

Neon: -415.5°F (-248.6°C)

Beryllium: 2,349°F (1,287°C)

Bond [772]3 years ago

4 0

Answer: 2,577°F or 1,414°C

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Why would it be difficult to sift the iron fillings from the sand

slamgirl [31]

Explanation:

Sand and iron fillings have the same size of particles.

It would be difficult to use a mesh to separate them. Also, sometimes , the iron filing clings well to the surface of the sand particles.

The best way to separate them is to use a bar magnet. The magnetic properties of the iron fillings will get it attracted to the bar magnet leaving the sand particles behind.

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Extraction brainly.com/question/5881840

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3 years ago

How much heat energy, in kilojoules, is required to convert 72.0 gg of ice at −−18.0 ∘C∘C to water at 25.0 ∘C∘C ? Express your a

Ghella [55]

Answer: The enthalpy change is 34.3 kJ

Explanation:

The conversions involved in this process are :

$(1):H_2O(s)(-18^0C)\rightarrow H_2O(s)(0^0C)\\\\(2):H_2O(s)(0^0C)\rightarrow H_2O(l)(0^0C)\\\\(3):H_2O(l)(0^0C)\rightarrow H_2O(l)(25^0C)$

Now we have to calculate the enthalpy change.

$\Delta H=[m\times c_{s}\times (T_{final}-T_{initial})]+n\times \Delta H_{fusion}+[m\times c_{l}\times (T_{final}-T_{initial})]$

where,

$\Delta H$ = enthalpy change = ?

m = mass of water = 72.0 g

$c_{s}$ = specific heat of ice = $2.09J/g^0C$

$c_{l}$ = specific heat of liquid water = $4.184J/g^0C$

n = number of moles of water = $\frac{\text{Mass of water}}{\text{Molar mass of water}}=\frac{72.0g}{18g/mole}=4.00moles$

$\Delta H_{fusion}$ = enthalpy change for fusion = 6010 J/mole

Now put all the given values in the above expression, we get

$\Delta H=[72.0g\times 2.09J/g^0C\times (0-(-18)^0C]+4.00mole\times 6010J/mole+[72.0g\times 4.184J/g^)C\times (25-0)^0C]$ $\Delta H=34279.8J=34.3kJ$ (1 KJ = 1000 J)