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3 years ago

What is the trend in electronegativity as elements are considered from left to right in the periodic table?

Chemistry

1 answer:

Readme [11.4K]3 years ago

8 0

Answer:

the answer is

the electronegativity of the elements from L to R increases

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Any substance that when dissolved in water will give off h ions is referred to as a(n) ________.

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3 years ago

The smallest unit which maintains the physical properties of a compound is a(n) _____.

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Molecule
The smallest unit in a compound is its molecule.

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4 years ago

What is the ideal gas constant?

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It is a universal constant used for all gases

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3 years ago

Carbon and silicon are tetravalent but ge sn pb show di valency. Why

Anuta_ua [19.1K]

<u>Answer:</u>

Carbon and silicon both are tetravalent elements as compared to germanium, tin, and lead which are divalent.

That's because Ge, tin, and Pb show inert pair effect and has a greater nuclear effective charge on the 's' electrons due to poor shielding effect. .That's why these elements are not able to share their valence electrons while carbon and silicon does and show "catenation" which is the ability to form long chain molecules.

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4 years ago

If you feed 100 kg of N2 gas and 100 kg of H2 gas into a

torisob [31]

Answer : The mass of ammonia produced can be, 121.429 k

Solution : Given,

Mass of $N_2$ = 100 kg = 100000 g

Mass of $H_2$ = 100 kg = 100000 g

Molar mass of $N_2$ = 28 g/mole

Molar mass of $H_2$ = 2 g/mole

Molar mass of $NH_3$ = 17 g/mole

First we have to calculate the moles of $N_2$ and $H_2$ .

$\text{ Moles of }N_2=\frac{\text{ Mass of }N_2}{\text{ Molar mass of }N_2}=\frac{100000g}{28g/mole}=3571.43moles$

$\text{ Moles of }H_2=\frac{\text{ Mass of }H_2}{\text{ Molar mass of }H_2}=\frac{100000g}{2g/mole}=50000moles$

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

$N_2+3H_2\rightarrow 2NH_3$

From the balanced reaction we conclude that

As, 1 mole of $N_2$ react with 3 mole of $H_2$

So, 3571.43 moles of $N_2$ react with $3571.43\times 3=10714.29$ moles of $H_2$

From this we conclude that, $H_2$ is an excess reagent because the given moles are greater than the required moles and $N_2$ is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of $NH_3$

From the reaction, we conclude that

As, 1 mole of $N_2$ react to give 2 mole of $NH_3$

So, 3571.43 moles of $N_2$ react to give $3571.43\times 2=7142.86$ moles of $NH_3$

Now we have to calculate the mass of $NH_3$

$\text{ Mass of }NH_3=\text{ Moles of }NH_3\times \text{ Molar mass of }NH_3$

$\text{ Mass of }NH_3=(7142.86moles)\times (17g/mole)=121428.62g=121.429kg$

Therefore, the mass of ammonia produced can be, 121.429 kg

6 0

3 years ago