Where do the nitrogen bases that are floating go

I need the answers to all three questions pls:

Klio2033 [76]

Answer:

Explanation:

8) As a sulfur atom gains electrons, its radius

A) decreases

B) increases

C) remains the same

Answer:

Increases

Explanation:

when an atom form ion its atomic radius changed from neutral atomic radius.

if the atom form cation its atomic radius is smaller than neutral atomic radius.

If atom form anion its anionic radius is larger than the neutral atomic radius.

The reason is that when electron is remove energy shell deduced so cation get smaller ionic radii while in case of anion electron is added and size increase.

when sulfur gain electrons it form anion that's why its radius increases.

9) Which element forms an ion larger than its atom?

A) Na B) Ne C) Ba D) Br

Answer;

Br

Explanation:

when an atom form ion its atomic radius changed from neutral atomic radius.

if the atom form cation its atomic radius is smaller than neutral atomic radius.

If atom form anion its anionic radius is larger than the neutral atomic radius.

The reason is that when electron is remove energy shell deduced so cation get smaller ionic radii while in case of anion electron is added and size increase.

In given list of elements sodium and barium form cation while bromine form anion.

That's why anion of bromine is larger than atom.

while neon is inert it cannot form ion.

10. Which of the following particles has the smallest

radius?

A) Na B) K° C) Na+ D) K+

Answer:

Na⁺

Explanation:

Atomic radii trend along group:

As we move down the group atomic radii increased with increase of atomic number. The addition of electron in next level cause the atomic radii to increased. The hold of nucleus on valance shell become weaker because of shielding of electrons thus size of atom increased .

Sodium and potassium are present in group one. Sodium is present in period three while potassium is present in period four. So atomic size of sodium is smaller and when it form the cation its cation is smaller than its atom because of losing of electron.The reason is that when electron is remove energy shell deduced so cation get smaller ionic radii

8 0

2 years ago

Explain why the boiling point of H2S with relative molecular mass 34 is lower than that of H2O with relative molecular mass 18

Levart [38]

Answer and Explanation:

In H2O molecules, due to formation of intermolecular hydrogen bonds, there is molecular association. Large amount of energy is required to break these intermolecular hydrogen bonds. Intermolecular hydrogen bonding is not possible in H2S. Hence, its boiling point is lower and is a gas.

4 0

3 years ago

How many molecules of O2 are in 153.9 g O2?

MrRissso [65]

Answer:

2.895*10^24

Explanation:

mass of Oxygen give = 153.9g

molar mass of O2 molecule = 16*2=32g/mol

n= mole

To find the mole

n= mass/ molar mass

n= 153.9/32

n=4.81mol.

To find the number of molecules of o

Nm= number of molecule

Nn = Number of mole

NA = number of Avogados

Nm= Nn * NA

Nm= 4.81 *6.02*10^23

Nm= 2.895*10^24

3 0

2 years ago

Suppose the reaction between nitrogen and hydrogen was run according to the amounts presented in Part A, and the temperature and

andrew11 [14]

Explanation:

Assuming that moles of nitrogen present are 0.227 and moles of hydrogen are 0.681. And, initially there are 0.908 moles of gas particles.

This means that, for $N_{2}(g) + 3H_{2}(g) \rightarrow 2NH_{3}$

moles of $N_{2}$ + moles of $H_{2}$ = 0.908 mol

Since, 2 moles of $N_{2}$ = $2 \times 0.227$ = 0.454 mol

As it is known that the ideal gas equation is PV = nRT

And, as the temperature and volume were kept constant, so we can write

$\frac{P(_in)}{n_(in)}$ = $\frac{P_(final)}{n_(final)}$

$\frac{10.4}{0.908}$ = $\frac{P_(final)}{0.454
}$

$P_(final)$ = $10.4 \times \frac{0.454}{0.908}$

= 5.2 atm

Therefore, we can conclude that the expected pressure after the reaction was completed is 5.2 atm.

7 0

2 years ago

Write the net cell equation for this electrochemical cell. Phases are optional. Do not include the concentrations. Co ( s ) ∣ ∣

Tatiana [17]

Answer: The overall equation will be $Co(s)+2Ag^+(aq)\rightarrow Co^{2+}(aq)+Ag(s)$

Explanation:

The representation is given by writing the anode on left hand side followed by its ion with its molar concentration. It is followed by a salt bridge. Then the cathodic ion with its molar concentration is written and then the cathode.

Oxidation reaction is defined as the reaction in which a substance looses its electrons. The oxidation state of the substance increases.

Anode : $Co(s)\rightarrow Co^{2+}(aq)+2e^{-}$

Reduction reaction is defined as the reaction in which a substance gains electrons. The oxidation state of the substance gets reduced.

Cathode : $Ag^+(aq)+e^{-1}\rightarrow Ag(s)$ $\times 2$

The number of electrons lost must be equal to the number of electrons gained , thus overall equation will be :

$Co(s)+2Ag^+(aq)\rightarrow Co^{2+}(aq)+Ag(s)$

5 0

2 years ago