All categories

3 years ago

Do oxygen atoms become more stable or less stable when oxygen forms compounds!0?

1 answer:

6 0

<span>remember that all elements want to have no valence electrons, they want to be like the noble gases.
Oxygen has six valence electrons, which means that it needs to gain two to become stable.
it takes much more energy for oxygen to lose its 6 valence electrons than it would be to gain 2.
>>When atoms form compounds, they become stable

hope this helps

remember if you ever need anything or a question answered feel free to post it on my profile or you can message me
:)</span>

You might be interested in

Acidity is a measure of the amount of dissolved hydroxide ions in a solution.

Gekata [30.6K]

The concentration of hydrogen ions in a solution is a measure of its acidity. So the correct option is (b) false.

When an Arrhenius acid is dissolved in water, hydrogen ions are produced:

H+(aq) + A- = HA + H2O (aq)

Here, H+ is the hydrogen cation, A- is the solvated anion, also known as the conjugate base, and HA is the non-dissociated acid. When an Arrhenius base is dissolved in water, hydroxide ions are produced:

BOH + H2O → B+(aq) + OH-(aq)

Is a material with at least one hydrogen atom that has the ability to split apart in an aqueous solution to produce an anion and an H + ion (a proton), creating an acidic solution. Bases are substances that, when dissolved in water, create hydroxide ions (OH) and a cation, resulting in a basic solution.

Learn more about hydrogen here:

brainly.com/question/16979348

#SPJ4

4 0

2 years ago

How many are molecules ( or formula) in each sample?

andre [41]

Answer:

$4.010 \times 10^{25} \text { molecules of } \mathrm{NaHCO}_{3} \text { present in } 55.93 \mathrm{kg} \text { of } \mathrm{NaHCO}_{3}$
$16.86 \times 10^{26} \text { molecules of } \mathrm{Na}_{3} \mathrm{PO}_{4} \text { present in } 459 \mathrm{kg}\left(4.59 \times 10^{5} \mathrm{gm}\right) \text { of } \mathrm{Na}_{3} \mathrm{PO}_{4}$

<u>Explanation</u>:

<u>Number of molecules for $55.93 \mathrm{kg} \text { of } \mathrm{NaHCO}_{3}$ </u>

$\text { Firstly molar mass is calculated of } \mathrm{NaHCO}_{3}:$

Atomic mass of Na + H + C + 3(O) = 22.99 + 1.008 + 12.01 + 3 × 16.00 = 84.00 g/mol

$\text { Number of molecules of } \mathrm{NaHCO}_{3} \text { in } 55.93 \text { kg are as follows: }$

$55.93 \times\left(10^{3} \mathrm{gm}\right) \times \frac{1 \mathrm{mol} \mathrm{NaHCO}_{3}}{84.00 \mathrm{gm} \mathrm{NaHCO}_{3}} \times\left(6.022 \times 10^{23} \mathrm{molecules} \text { i.e Avogadro number }\right)$

$=4.010 \times 10^{26} \text { molecules of } \mathrm{NaHCO}_{3} \text { present in } 55.93 \mathrm{kg} \text { of } \mathrm{NaHCO}_{3}$

<u>Number of molecules for for $\left(4.59 \times 10^{5} \mathrm{gm}\right) \text { of } \mathrm{Na}_{3} \mathrm{PO}_{4}$ </u>

$\text { Firstly molar mass is calculated of } \mathrm{Na}_{3} \mathrm{PO}_{4}$

= Atomic mass of 3(Na) + P + 4(O)

= 3(22.99) + 30.97 + 4(16.00) = 163.94 g/mol

$459 \times\left(10^{3} \mathrm{gm}\right) \times \frac{1 \mathrm{mol} N a_{3} P O_{4}}{163.94 \mathrm{gm} N a_{3} P O_{4}} \times\left(6.022 \times 10^{23} \mathrm{molecules} \text { i.e Avogadro number) } / 1 \mathrm{mol}\right.$

$=16.86 \times 10^{26} \text { molecules of } \mathrm{Na}_{3} \mathrm{PO}_{4} \text { present in } 459 \mathrm{kg}\left(4.59 \times 10^{5} \mathrm{gm}\right) \text { of } \mathrm{Na}_{3} \mathrm{PO}_{4}$

8 0

3 years ago

A sample of nitrogen occupies 10.0 liters at 25°C and 98.7 kPa. What would be the volume at 20°C and 102.7 kPa?

krek1111 [17]

To solve the problem, we assume the sample to be ideal. Then, we use the ideal gas equation which is expressed as PV = nRT. From the first condition of the nitrogen gas sample, we calculate the number of moles.

n = PV / RT
n = (98.7x 10^3 Pa x 0.01 m^3) / (8.314 Pa m^3/ mol K) x 298.15 K
n = 0.40 mol N2

At the second condition, the number of moles stays the same however pressure and temperature was changed. So, the new volume is calculated as follows:

V = nRT / P
V = 0.40 x 8.314 x 293.15 / 102.7 x 10^3
V = 9.49 x 10^-3 m^3 or 9.49 L

7 0

3 years ago

Which of the Atoms shown has an atomic number four

dimulka [17.4K]

Answer:

Explanation:

Atomic # = Protons

it says 4 p in the inside of the orbital

4 0

3 years ago

What information does the third quantum number of an electron give? A. The direction the electron in spinning B. The energy leve

Juli2301 [7.4K]

The information that the third quantum number of an electron gives is the direction the electron in spinning. That is option A.

<h3>What are quantum numbers?</h3>

Quantum number are those numbers that are used to specify the properties of the atomic orbitals and the electrons in those orbitals.

The types of quantum numbers include the following:

principal,

azimuthal,

magnetic, and spin.

The third quantum number is shows the direction of the electron while spinning through specifying its angular momentum.

Learn more about orbitals here:

brainly.com/question/14361631

#SPJ1

4 0

2 years ago