Si tengo 3 gramos de hidrógeno y 6 de oxígeno.¿ cuánta agua conseguiré?

sveticcg [70]

Answer: The last electron will be filled in first orbital of 3p sub-shell.

Explanation: Filling of electrons in orbitals is done by using Hund's Rule.

Hund's rule states that the electron will be singly occupied in the orbital of the sub-shell before any orbital is doubly occupied.

For filling up of the electrons in Sulfur atom having 16 electrons. First 10 electrons will completely fill according to Aufbau's Rule in 1s, 2s and 2p sub-shells and last 6 electrons are the valence electrons which will be filled in the order of 3s and then 3p.

3s sub-shell will be fully filled and the orbitals of 3p sub-shell will be first singly occupied and then pairing will take place. Hence, the last electron will be filled in the first orbital of 3p-sub-shell.

5 0

2 years ago

Which of the following reactions have a positive ΔSrxn? Check all that apply.

PolarNik [594]

Answer:

The reactions that have a positive ΔS rxn are the first and the fourth choices:

2A(g) + B(s) → 3C(g)

2A(g) + 2B(g) → 5C(g)

Explanation:

ΔS rxn is the change of entropy of the chemical reaction.

ΔS rxn = S after reaction - S before reaction.

Therefore, a positive ΔS rxn means that the entropy after the reaction is greater than the entropy before the reaction.

You may use some assumptions to predict whether a reaction will lead an increase or decrease of the entropy.

First, assume that all the non-shown conditions, such as temperature and pressure, are constant.

Under that assumption, and from the meaning of entropy as a measure of the disorder or randomness of a system you can predict the sign of the change of entropy.

2A(g) + B(s) → 3C(g)

1) The solid compounds, B(s) in this case, are very ordered and so they have low entropy.

2) Gas molecules are highly disordered (scattered), and the greater the number of molecules of the gas the larger the entropy, S).

Hence, since the product side shows 3 gas molecules and the reactant side shows 2 gas molecules and 1 solid molecule, you predict that the products have a larger entropy than the reactants, meaning an increase in entropy: ΔS rxn is positive.

2A(g) + B(g) → C(g)

Using the same reasoning, 3 gas molecules in the reactant side have more entropy than 1 molecule in the product side, and so the reaction leads to a decrease in the entropy: ΔS rxn is negative

A(g) + B(g) → C(g)

Again, 2 gas molecules in the reactant side have more entropy than 1 molecule in the product side, and so the reaction leads to a decrease in the entropy: ΔS rxn is negative

2A(g) + 2B(g) → 5C(g)

With the same reasoing, 5 molecules in the product side, lets you predict that will have more entropy than 4 molecules in the reactant side, and, the entropy will increase: ΔS rxn is positive.

6 0

3 years ago

Consider the balanced equation below.

Bumek [7]

Answer : The correct option is, (2) 8 mole of hydrogen will react with 1 mole of sulfur.

Explanation :

The balanced chemical equation is,

$8H_2+S_8\rightarrow 8H_2S$

By the stoichiometry we conclude that, 8 moles of hydrogen react with 1 mole of sulfur to give 8 moles of hydrogen sulfide.

Hence, the correct options is, (2) 8 mole of hydrogen will react with 1 mole of sulfur.

6 0

2 years ago

Read 2 more answers

What volume of product is produced when 20.0 Mg of liquid hydrogen reacts with excess liquid oxygen in a rocket propellant react

Zinaida [17]

Good luck g hope you get it right

4 0

2 years ago

Calculate the percent activity of the radioactive isotope iodine-131 remaining after 3 half-lives.

vovangra [49]

Half life of a radio-active material is refereed to the time required for reducing its activity to 50 %.

Now, when I-131 passes 1st half life, it's activity will reduce to half i.e. 50%

When, I-131 passes 2nd half life, activity of I-131 will be reduced to 25%

On passing, 3rd half-life, percent activity of the I-131 remaining will be 12.5 %.

3 0

3 years ago