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2 years ago

Why sulphur has two melting point values

1 answer:

8 0

I don't know how well known/accepted this is (it's in my textbook so I'm guessing it's right), but Sulphur has two forms - the alpha and beta forms ,apparently gamma sulphur exists as well.

The alpha form is rhombic, yellow in color and has a MP of 385.8 K. The beta form is colorless and has a MP of 393 K and is formed by melting rhombic sulphur and cooling it till a crust forms on top. Poke a hole and pour out the liquid inside and you get beta sulphur. The transition point is 369K - below it, alpha sulphur is stable and above it, beta sulphur is stable. Both have helped. I had to pull out an old textbook and that's something that I don't usually do.

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For each property listed, identify the type of element it describes.

Juliette [100K]

Answer:

1.Very good electrical conductivity : Metals (Decreacing order of conductivity)

Silver > Copper > Gold > aluminium

2. Amphoteric : Metal elements

Beryllium , Aluminium , Zinc ,

3.Gaseous at room temperature: mostly Nobel gases elements and some non - metal elements.

Helium ,neon , argon , krypton , fluorine , Oxygen , nitrogen

4.Solid at room temperature: Mostly Metals (few non-metals, metalloid elements)

Metals (Sodium , potassium , calcium , gold are solid)

Non- metals(Carbon ,Boron )

Metalloids(antimony)

5. Brittle : non - metals (can't be rolled into wires)

Hydrogen , carbon , sulfur , phosphorus

Explanation:

6 0

2 years ago

Read 2 more answers

For a phase change, H0 = 2 kJ/mol and A S0 = 0.017 kJ/(K•mol). What are

34kurt

Answer:

ΔG = -6.5kJ/mol at 500K

Explanation:

We can find ΔG of a reaction using ΔH, ΔS and absolute temperature with the equation:

ΔG = ΔH - TΔS

Computing the values in the problem:

ΔG = ?

ΔH = 2kJ/mol

T = 500K

And ΔS = 0.017kJ/(K•mol)

Replacing:

ΔG = 2kJ/mol - 500K*0.017kJ/(K•mol)

ΔG = 2kJ/mol - 8.5kJ/mol

8 0

2 years ago

Can someone help me with this please!

pashok25 [27]

NitITE is NO2
nitrATE is NO3

5 0

2 years ago

What is the final temperature of the solution formed when 1.52 g of NaOH is added to 35.5 g of water at 20.1 °C in a calorimeter

Inessa [10]

Answer : The final temperature of the solution in the calorimeter is, $31.0^oC$

Explanation :

First we have to calculate the heat produced.

$\Delta H=\frac{q}{n}$

where,

$\Delta H$ = enthalpy change = -44.5 kJ/mol

q = heat released = ?

m = mass of $NaOH$ = 1.52 g

Molar mass of $NaOH$ = 40 g/mol

$\text{Moles of }NaOH=\frac{\text{Mass of }NaOH}{\text{Molar mass of }NaOH}=\frac{1.52g}{40g/mole}=0.038mole$

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

$44.5kJ/mol=\frac{q}{0.038mol}$

$q=1.691kJ$

Now we have to calculate the final temperature of solution in the calorimeter.

$q=m\times c\times (T_2-T_1)$

where,

q = heat produced = 1.691 kJ = 1691 J

m = mass of solution = 1.52 + 35.5 = 37.02 g

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

$T_1$ = initial temperature = $20.1^oC$

$T_2$ = final temperature = ?

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

$1691J=37.02g\times 4.18J/g^oC\times (T_2-20.1)$

$T_2=31.0^oC$

Thus, the final temperature of the solution in the calorimeter is, $31.0^oC$

4 0

3 years ago

What is the age of mummified primate skin that contains 8.25% of the original quantity of 14C?

rjkz [21]

Answer:

20619.4793 years

Explanation:

The half life of carbon-14 = 5730 years

The formula for the half life for a first order kinetic reaction is:

$t_{1/2}=\frac {ln\ 2}{k}$

Where,

$t_{1/2}$ is the half life

k is the rate constant.

Thus rate constant is:

5730 years=ln(2)/k

k = 1.21×10⁻⁴ years ⁻¹

Using integrated rate law as:

$ln\ \frac {A_t}{A_0}=-k\times t$

Where,

$A_t$ is the concentration at time t

$A_0$ is the initial concentration

Given that the final concentration contains 8.25 % of the original quantity which means that:

$\frac {A_t}{A_0}=0.0825$

So,

ln(.0825)= -1.21×10⁻⁴×t

t = 20619.4793 years

8 0

2 years ago