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Anna [14]
4 years ago
9

The elements selenium and oxygen are both poor electrical conductors in which section of the periodic table are they found

Chemistry
2 answers:
Nutka1998 [239]4 years ago
6 0

Answer:

This question appears incomplete

Explanation:

The periodic table is divided into 4 blocks; s-block, p-block, d-block and the f-block.

Elements found in the s-block are metals; alkali metals and alkali earth metals. Elements found in the p-block include metals and non-metals while those found in the d-block are metals (transition metals). The F-block elements are the radioactive elements.

Oxygen (O) and selenium (Se) are non-metals found in group 16 of the periodic table which is found in the p-block of the periodic table. The p-block is located on the right side of the periodic table. Generally, non-metals are poor conductors of electricity.

Strike441 [17]4 years ago
5 0

Answer:

No metals. Right side.

Explanation:

On the right. Non metallic elements

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Interacting with animals has been shown to decrease levels of cortisol (a stress-related hormone) and lower blood pressure. Other studies have found that animals can reduce loneliness, increase feelings of social support, and boost your mood.

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What is the density of marble that has a mass of 4.3 g and a volume of 2.7 cm?
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Explanation:

density(d) = mass(m) × volume(v)

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Nitric oxide reacts with chlorine to form nocl. the data refer to 298 k. 2no(g) + cl2(g) → 2nocl(g) substance: no(g) cl2(g) nocl
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Answer:

- 10.555 kJ/mol.

Explanation:

∵ ∆G°rxn = ∆H°rxn - T∆S°rxn.

Where, ∆G°rxn is the standard free energy change of the reaction (J/mol).

∆H°rxn is the standard enthalpy change of the reaction (J/mol).

T is the temperature of the reaction (K).

∆S°rxn is the standard entorpy change of the reaction (J/mol.K).

  • Calculating ∆H°rxn:

∵ ∆H°rxn = ∑∆H°products - ∑∆H°reactants

<em>∴ ∆H°rxn = (2 x ∆H°f NOCl) - (1 x ∆H°f Cl₂) - (2 x ∆H°f NO) </em>= (2 x 51.71 kJ/mol) - (1 x 0) - (2 x 90.29 kJ/mol) = - 77.16 kJ/mol.

  • Calculating ∆S°rxn:

∵  ∆S°rxn = ∑∆S°products - ∑∆S°reactants

<em>∴ ∆S°rxn = (2 x ∆S° NOCl) - (1 x ∆S° Cl₂) - (2 x ∆S° NO) </em>= (2 x 261.6 J/mol.K) - (1 x 223.0 J/mol.K) - (2 x 210.65 J/mol.K) =<em> - 121.1 J/mol.K. = - 0.1211 kJ/mol.K.</em>

<em></em>

  • Calculating ∆G°rxn:

∵ ∆G°rxn = ∆H°rxn - T∆S°rxn.

<em>∴ ∆G°rxn = ∆H°rxn - T∆S°rxn </em>= (- 77.16 kJ/mol) - (550 K)(- 0.1211 kJ/mol.K) = <em>- 10.555 kJ/mol.</em>

4 0
3 years ago
Walter gets a paper back in lab with “-2.75% error” written on it. He found the mass of an object to be 100.7 grams. What should
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0.0275x100.7= 2.8(approximately). 100.7-2.8=97.9grams
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