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3 years ago

Use the information below to explain why the atomic radius decreases across a period.

Chemistry

1 answer:

pishuonlain [190]3 years ago

7 0

Answer:

Detail is given below.

Explanation:

Atomic radius trend along period:

As we move from left to right across the periodic table the number of valance electrons increases. The atomic size tend to decrease in same period of periodic table because the electrons are added with in the same shell. When the electron are added, at the same time protons are also added in the nucleus. The positive charge is going to increase and this charge is greater in effect than the charge of electrons. This effect lead to the greater nuclear attraction. The electrons are pull towards the nucleus and valance shell get closer to the nucleus. As a result of this greater nuclear attraction atomic radius decreases and ionization energy increases because it is very difficult to remove the electron from atom and more energy is required.

we can see in picture A there is one positive charge and force of attraction is 2.30×10⁻⁸ N

In picture D there are 2 positive charge and force of attraction is 4.60×10⁻⁸ N

So by increasing the positive charge force of attraction also increases.

In F we can see there is more positive charge thus force of attraction is also greater.

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Magnesium (used in the manufacture of light alloys) reacts with iron(III) chloride to form magnesium chloride and iron. A mixtur

yuradex [85]

Answer: The limiting reactant is magnesium and mass of excess reactant present in the vessel is 96.35 grams.

Explanation:

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ .....(1)

For magnesium:

Given mass of magnesium = 41.0 g

Molar mass of magnesium = 24 g/mol

Putting values in equation 1, we get:

$\text{Moles of magnesium}=\frac{41.0g}{24g/mol}=1.708mol$

For iron(III) chloride:

Given mass of iron(III) chloride = 175.0 g

Molar mass of iron(III) chloride = 162.2 g/mol

Putting values in equation 1, we get:

$\text{Moles of iron(III) chloride}=\frac{175g}{162.2g/mol}=1.708mol$

The chemical equation for the reaction of magnesium and iron(III) chloride follows:

$3Mg+2FeCl_3\rightarrow 3MgCl_2+2Fe$

By Stoichiometry of the reaction:

3 moles of magnesium reacts with 2 moles of iron(III) chloride

So, 1.708 moles of magnesium will react with = $\frac{2}{3}\times 1.708=1.114mol$ of iron(III) chloride

As, given amount of iron(III) chloride is more than the required amount. So, it is considered as an excess reagent.

Thus, magnesium is considered as a limiting reagent because it limits the formation of product.

Moles of excess reactant left (iron(III) chloride) = [1.708 - 1.114] = 0.594 moles

Now, calculating the mass of iron(III) chloride from equation 1, we get:

Molar mass of iron(III) chloride = 162.2 g/mol

Moles of iron(III) chloride = 0.594 moles

Putting values in equation 1, we get:

$0.594mol=\frac{\text{Mass of iron(III) chloride}}{162.2g/mol}\\\\\text{Mass of iron(III) chloride}=(0.594mol\times 162.2g/mol)=96.35g$

Hence, the limiting reactant is magnesium and mass of excess reactant present in the vessel is 96.35 grams.

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