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3 years ago

Is there a relationship between atomie radius and electronegativity? Why or why not?

Chemistry

2 answers:

LenaWriter [7]3 years ago

8 0

Answer:

Yes

Explanation:

Electronegativity is the tendency for atoms to keep electrons closer to them. So, if the atoms are smaller there, there is a stronger pull on the electrons to keep them closer. So, the larger the atom get, due to electron shielding, it become less electro negative. I hope this help. You might get a better explanation from Khan Academy's video on Electronegativity

attashe74 [19]3 years ago

3 0

Answer:

A relationship is intuitively expected

Does electronegativity relate to atomic radius? A relationship is intuitively expected between electronegativity and radius: the size of an atom is determined by the distribution of electrons around its nucleus. The closer the electrons are to the nucleus, the more tightly they are bound, thus increasing the electronegativity of the atom.

Explanation:

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In a neutral atom, the number of protons _______ the number of electrons.

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Answer:

equal the number of electrons.

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Breathing equipment used by rescue workers needs to capture the CO2 the humans breath out and produce O2 for them to breath in,

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Answer: $4KO_2+2CO_2\rightarrow 2K_2CO_3+3O_2$

Explanation:

According to the law of conservation of mass, mass can neither be created nor be destroyed. Thus the mass of products has to be equal to the mass of reactants. The number of atoms of each element has to be same on reactant and product side. Thus chemical equations are balanced.

The balanced chemical equation for reaction of potassium superoxide with carbon dioxide to produce oxygen and potassium carbonate will be:

$4KO_2+2CO_2\rightarrow 2K_2CO_3+3O_2$

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2 years ago

Which of the following is not balanced?

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Is it asking which is or isn’t balanced

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3 years ago

Find the molecular mass of.

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Answer:

sodium oxide Nao
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3 years ago

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onsider the following reaction: CaCN2 + 3 H2O → CaCO3 + 2 NH3 105.0 g CaCN2 and 78.0 g H2O are reacted. Assuming 100% efficiency

mestny [16]

Answer : The excess reactant is, $H_2O$

The leftover amount of excess reagent is, 7.2 grams.

Solution : Given,

Mass of $CaCN_2$ = 105.0 g

Mass of $H_2O$ = 78.0 g

Molar mass of $CaCN_2$ = 80.11 g/mole

Molar mass of $H_2O$ = 18 g/mole

Molar mass of $CaCO_3$ = 100.09 g/mole

First we have to calculate the moles of $CaCN_2$ and $H_2O$ .

$\text{ Moles of }CaCN_2=\frac{\text{ Mass of }CaCN_2}{\text{ Molar mass of }CaCN_2}=\frac{105.0g}{80.11g/mole}=1.31moles$

$\text{ Moles of }H_2O=\frac{\text{ Mass of }H_2O}{\text{ Molar mass of }H_2O}=\frac{78.0g}{18g/mole}=4.33moles$

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

$CaCN_2+3H_2O\rightarrow CaCO_3+2NH_3$

From the balanced reaction we conclude that

As, 1 mole of $CaCN_2$ react with 3 mole of $H_2O$

So, 1.31 moles of $CaCN_2$ react with $1.31\times 3=3.93$ moles of $H_2O$

From this we conclude that, $H_2O$ is an excess reagent because the given moles are greater than the required moles and $CaCN_2$ is a limiting reagent and it limits the formation of product.

Left moles of excess reactant = 4.33 - 3.93 = 0.4 moles

Now we have to calculate the mass of excess reactant.

$\text{ Mass of excess reactant}=\text{ Moles of excess reactant}\times \text{ Molar mass of excess reactant}(H_2O)$

$\text{ Mass of excess reactant}=(0.4moles)\times (18g/mole)=7.2g$

Thus, the leftover amount of excess reagent is, 7.2 grams.

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3 years ago