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4 years ago

Why does sulfur have a higher melting point than phosphorus

Chemistry

1 answer:

tiny-mole [99]4 years ago

5 0

Because phosphorus is more flammable than sulfur!

Have a nice day! :)

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Define the term diffusion

maw [93]

Answer:

Spreading, like dispersing.

Explanation:

7 0

4 years ago

Read 2 more answers

g Phosphorus -32 is a commonly used radioactive nuclide in biochemical research, particularly in studies of nucleic acids. The h

Setler79 [48]

Answer:

6.88 mg

Explanation:

Step 1: Calculate the mass of ³²P in 175 mg of Na₃³²PO₄

The mass ratio of Na₃³²PO₄ to ³²P is 148.91:31.97.

175 mg g Na₃³²PO₄ × 31.97 g ³²P/148.91 g Na₃³²PO₄ = 37.6 mg ³²P

Step 2: Calculate the rate constant for the decay of ³²P

The half-life (t1/2) is 14.3 days. We can calculate k using the following expression.

k = ln2/ t1/2 = ln2 / 14.3 d = 0.0485 d⁻¹

Step 3: Calculate the amount of P, given the initial amount (P₀) is 37.6 mg and the time elapsed (t) is 35.0 days

For first-order kinetics, we will use the following expression.

ln P = ln P₀ - k × t

ln P = ln 37.6 mg - 0.0485 d⁻¹ × 35.0 d

P = 6.88 mg

3 0

3 years ago

What is the correct sequence of coefficients when this equation is balanced? ___cs2(l) + ___o2(g) → ___co2(g) + ___so2(g)?

iris [78.8K]

Answer: the sequence is 1, 3, 1, 2.

Explanation:

1) Given equation: CS₂(l) + O₂(g) → CO₂(g) + SO₂(g)

2) To balance:

i) initially C is balanced: 1 atom in the left side, and 1 atom in the right sides

ii) Add coefficient 2 in front of SO₂ to have 2 atoms in each side.

That leads to: CS₂(l) + O₂(g) → CO₂(g) + 2SO₂(g)

iii) Count the O atoms; there are 2 atoms in the left, and 6 atoms in the right side. So, add a coefficient 3 in front of O₂ (in the left) and you get:

CS₂(l) + 3O₂(g) → CO₂(g) + 2SO₂(g)

4) You can verigy that now the equation is balanced:

C: 1 in the left, and 1 in the right

S: 2 in the left, and 2 in the right

O: 6 in the left, and 6 in the right.

5) So, the sequence is 1, 3, 1, 2.

Remember, when the none coefficient is shown it is because it is 1.

4 0

3 years ago

Read 2 more answers

In which of the following compounds does the carbonyl stretch in the IR spectrum occur at the lowest wavenumber?

klio [65]

Answer:

a. Cyclohexanone

Explanation:

The principle of IR technique is based on the vibration of the bonds by using the energy that is in this region of the electromagnetic spectrum. For each bond, there is a specific energy that generates a specific vibration. In this case, you want to study the vibration that is given in the carbonyl group C=O. Which is located around 1700 cm-1.

Now, we must remember that the lower the wavenumber we will have less energy. So, what we should look for in these molecules, is a carbonyl group in which less energy is needed to vibrate since we look for the molecule with a smaller wavenumber.

If we look at the structure of all the molecules we will find that in the last three we have heteroatoms (atoms different to carbon I hydrogen) on the right side of the carbonyl group. These atoms allow the production of resonance structures which makes the molecule more stable. If the molecule is more stable we will need more energy to make it vibrate and therefore greater wavenumbers.

The molecule that fulfills this condition is the cyclohexanone.

See figure 1

I hope it helps!

4 0

4 years ago

Which variable is tested during a lab investigation

Licemer1 [7]

Actually, it could indeed be the independent variable.

5 0

4 years ago