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3 years ago

The number of proton, neutrons and electrons in oxygen atom

Chemistry

1 answer:

Ksenya-84 [330]3 years ago

4 0

Answer:

in an oxygen atom there are:

protons:8

electrons:8

neutrons:8

Explanation:

this is because the atomic number of oxygen is 8 and that is the proton number and the electron number is the same as the atomic number

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An aluminum can weighing 10 g absorbs 106.8 J of heat and warms by 12 degrees C. What is the specific heat of the aluminum can?

Sladkaya [172]

Answer:

$c_{Al} = 0.89\,\frac{J}{kg\cdot ^{\circ}C}$

Explanation:

The heating process is modelled after the First Law of Thermodynamics:

$Q = m \cdot c_{Al}\cdot \Delta T$

The specific heat of the aluminium can is:

$c_{Al} = \frac{Q}{m\cdot \Delta T}$

$c_{Al} = \frac{106.8\,J}{(10\,g)\cdot (12^{\circ}C)}$

$c_{Al} = 0.89\,\frac{J}{kg\cdot ^{\circ}C}$

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3 years ago

A reaction has activation energy of 85kjper mol. What is the effect on the rate of raising the temperature from 20degree to 30 d

DENIUS [597]

Answer: The rate increases 3 times on raising the temperature from 20degree to 30 degree

Explanation:

According to Arrhenius equation with change in temperature, the formula is as follows.

$ln \frac{k_{2}}{k_{1}} = \frac{-E_{a}}{R}[\frac{1}{T_{2}} - \frac{1}{T_{1}}]$

where $k_2$ = rate constant at temp $T_2$

$k_1$ = rate constant at temp $T_1$

$E_a$ = activation energy

R= gas constant

$T_1$ = temperature = $20^0C=(20+273)K=293K$

$T_2$ = temperature = $30^0C=(30+273)K=303K$

$ln \frac{k_{2}}{k_{1}} = \frac{-85\times 1000J/mol}{8.314J/Kmol}[\frac{1}{303} - \frac{1}{293}]$

$ln \frac{k_{2}}{k_{1}}=1.15$

$\frac{k_{2}}{k_{1}}=3$

Thus rate increases 3 times on raising the temperature from 20degree to 30 degree

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Answer:

A group of cells ready to form roots a stem and the first leaves

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