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ozzi
1 year ago
11

In general terms, how does each of the following atomic properties influence the metallic character of the main-group elements i

n a period?
(d) Effective nuclear charge
Chemistry
1 answer:
vekshin11 year ago
6 0

Metallic property also rises with increasing atomic radius. Metallic character reduces with an increase in the amount of outer electrons.

<h3>What is atomic radius?</h3>

A chemical element's atomic radius, which is typically the average or typical distance between the nucleus's core and the outermost isolated electron, serves as a gauge for the size of an atom. There are numerous non-equivalent definitions of atomic radius since the border is not a clearly defined physical entity. Van der Waals radius, ionic radius, metallic radius, and covalent radius are the four most frequently used definitions of atomic radius. The atomic radius is often measured in a chemically linked condition because it is difficult to isolate individual atoms to measure their radii separately. However, theoretical computations are easier when considering isolated atoms.

To learn more about atomic radius from the given link:

brainly.com/question/13607061

#SPJ4

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Two adaptations to plants are described below:
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Half-reaction for the reduction of liquid water to gaseous hydrogen in basic aqueous solution
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Answer:

2H₂O (liq) + 2e⁻⇒ H₂ (g) + 2OH⁻ (aq)

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In reduction-oxidation reaction two reactions take place, one is oxidation and the other is reduction reaction. In an oxidation reaction, there is the loss of an electron whereas in the reduction reaction there is gain of electron occus.

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2H₂O (liq) + 2e⁻⇒ H₂ (g) + 2OH⁻ (aq)

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3 years ago
What is the percent yield of CuS for the following reaction given that you start with 15.5 g of Na2S and 12.1 g CuSO4? The actua
notka56 [123]

Answer:

(B) 42.1%

Explanation:

The formula for the calculation of moles is shown below:

moles = \frac{Mass\ taken}{Molar\ mass}

Given: For Na_2S

Given mass = 15.5 g

Molar mass of Na_2S = 78.0452 g/mol

<u>Moles of Na_2S = 15.5 g / 78.0452 g/mol = 0.1986 moles</u>

Given: For CuSO_4

Given mass = 12.1 g

Molar mass of CuSO_4 = 159.609 g/mol

<u>Moles of CuSO_4 = 12.1 g / 159.609 g/mol = 0.0758 moles</u>

According to the given reaction:

Na_2S+CuSO_4\rightarrow Na_2SO_4+CuS

1 mole of Na_2S react with 1 mole of CuSO_4

0.1986 mole of Na_2S react with 0.1986 mole of CuSO_4

Available moles of CuSO_4 = 0.0758 moles

Limiting reagent is the one which is present in small amount. Thus, CuSO_4 is limiting reagent. (0.0758 < 0.1986)

The formation of the product is governed by the limiting reagent. So,

1 mole of CuSO_4 gives 1 mole of CuS

0.0758 mole of CuSO_4 gives 0.0758 mole of CuS

Molar mass of CuS = 95.611 g/mol

Mass of CuS = Moles × Molar mass = 0.0758 × 95.611 g = 7.2473 g

Theoretical yield = 7.2473 g

Given experimental yield = 3.05 g

% yield = (Experimental yield / Theoretical yield) × 100 = (3.05/7.2473) × 100 = 42.1 %

<u>Option B is correct.</u>

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