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3 years ago

Is it likely that an atom of Te would substitute for an atom of O in a solid compound? yes no indeterminate

Chemistry

1 answer:

PolarNik [594]3 years ago

3 0

Answer:

The correct answer is no.

Explanation:

Tellurium is a chemical element denoted by Te and having atomic number 52. It is mildly toxic, brittle, silver-white, and rare metalloid. The element is chemically related to sulfur and selenium, all three of which are chalcogens.

Oxygen is a chemical element, that is, a substance, which comprises only one kind of atom. Its official chemical symbol is O and exhibits an atomic number 8, this signifies that an atom of oxygen possesses eight protons in its nucleus. In the given question, it is not likely that tellurium would replace for oxygen, as the two elements are highly unlike.

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It has been suggested that the surface melting of ice plays a role in enabling speed skaters to achieve peak performance. Carry

vesna_86 [32]

Explanation:

Relation between pressure, latent heat of fusion, and change in volume is as follows.

$\frac{dP}{dT} = \frac{L}{T \times \Delta V}$

Also, $\frac{L}{T} = \Delta S^{fusion}_{m}$

where, $\Delta V^{fusion}_{m}$ is the difference in specific volumes.

Hence, $\frac{dP}{dT} = \frac{\Delta S^{fusion}_{m}}{\Delta V^{fusion}_{m}}$

As, $\Delta S^{fusion}_{m} = \frac{L}{T} = \frac{6010}{273.15}$ = 22.0 J/mol K

And, $\Delta V^{fusion}_{m} = \frac{M}{d_{H_{2}O}} - \frac{M}{d_{ice}}$ ...... (1)

where, $d_{H_{2}O}$ = density of water

$d_{ice}$ = density of ice

M = molar mass of water = $18.02 \times 10^{-3} kg$

Therefore, using formula in equation (1) we will calculate the volume of fusion as follows.

$\Delta V^{fusion}_{m} = \frac{M}{d_{H_{2}O}} - \frac{M}{d_{ice}}$

= $\frac{18.02 \times 10^{-3}}{997} - \frac{18.02 \times 10^{-3}}{920}$

= $-1.51 \times 10^{-6}$

Therefore, calculate the required pressure as follows.

$\frac{dP}{dT} = \frac{22}{-1.51 \times 10^{-6}}$

= $1.45 \times 10^{7} Pa/K$

or, = 145 bar/K

Hence, for change of 1 degree pressure the decrease is 145 bar and for 4.7 degree change dP = $145 \times 4.7 bar$

= 681.5 bar

Thus, we can conclude that pressure should be increased by 681.5 bar to cause 4.7 degree change in melting point.

5 0

2 years ago

Which is the best definition of nonpolar covalent bond?

kondor19780726 [428]

Answer:

It is one of the covalent bonds in which the electrons are shared equally; therefore, dipole moment exists between the atoms in a molecule and there is no charge separation between the atoms in a molecule.

4 0

2 years ago

A 2.50 L solution contains 5.00 g of NaOH, which has a molecular weight of 40.00

Fantom [35]

Answer:

0.313mol/L

Explanation:

$\frac{5.00}{40.00} \times 2.50 = 0.3125$

3 0

2 years ago

The answer to question 2

Bad White [126]

Answer:

Explanation:

You will need a chemical equation with masses and molar masses, so let’s gather all the information in one place.

$M_{r}$ : 258.21 18.02

KAl(SO₄)₂·xH₂O ⟶ KAl(SO₄)₂ + xH₂O

Mass/g: 4.74 2.16

Step 1. Calculate the mass of the KAl(SO₄)₂.

Mass = 4.74 g – 2.16 g = 2.58 g.

Step 2. Calculate the moles of each product.

$\text{Moles of KAl(SO}_{4})_{2} = \text{2.58 g} \times \frac{\text{1 mol} }{\text{258.21 g}} = 9.992 \times 10^{-3} \text{ mol}$

$\text{Moles of H}_{2}\text{O} = \text{2.16 g} \times \frac{\text{1 mol} }{\text{18.02 g}} = \text{ 0.1200 mol}$

Step 3. Calculate the molar ratio of the two products.

$\frac{\text{Moles of KAl(SO}_{4})_{2}}{\text{Moles of H}_{2}\text{O}} = \frac{ 9.992 \times 10^{-3} \text{ mol}}{\text{ 0.1200 mol} } = \frac{1 }{12.01} \approx \frac{ 1}{ 12}$

1 mol of KAl(SO₄)₂ combines with 12 mol H₂O, so x = 12.

3 0

3 years ago

Which of the following best illustrates a natural process acting as a constructive force

Alina [70]

Answer:

Wind depositing sand to build up sand dunes.

Explanation:

3 0

2 years ago