All categories

3 years ago

What is the molar mass, in grams, of a mole of an element equal to?

Chemistry

2 answers:

guajiro [1.7K]3 years ago

5 0

Answer:

The molar mass of a mole of an element equal to the atomic weight of the element in grams.

Explanation:

The mass number is the sum of the numbers of protons and neutrons present in the nucleus of an atom. The mass number is an integer that is approximately equal to the numerical value of the atomic mass. Although the mass number is unitless, it is assigned units called atomic mass units (amu).

The standard atomic weight is the average of the relative atomic masses of all the isotopes of a chemical element weighted by their respective abundance on Earth. It reflects the prevalence of each natural isotope of an element.

A mole is defined as the amount of a substance that contains the number of carbon atoms in exactly 12 g of isotopically pure carbon-12. According to the most recent experimental measurements, this mass of carbon-12 contains 6.022142×10²³ atoms.

Hence, the molar mass, in grams, of a mole of an element equal to the atomic weight of the element.

Paha777 [63]3 years ago

3 0

The molar mass, in grams, of a mole for an element is equal to the atomic weight if the element.

You might be interested in

The images show two unrelated events occurring at separate times.

madreJ [45]

Event 1 is an example of a chemical reaction.

<u>Explanation:</u>

Whenever there is a chemical reaction, we can find that by means of a color change, formation of any gas, vapors, bubbles or any color or colorless precipitation, or by means of heat generation.

In the event 1 there is only a clear liquid in the beaker again it is added to a clear liquid in another beaker, forming an orange colored liquid , which shows that there is an occurrence of some chemical reaction.

So Event 1 is most likely an example of a chemical reaction.

5 0

3 years ago

Read 2 more answers

In the equation below

yulyashka [42]

Answer:

6.25 moles of N₂ is produced, and 18.8 moles of Cu and H₂O is produced.

Explanation:

We are given the chemical equation:

$\displaystyle 2\text{NH$_3$}_\text{(g)} + 3\text{CuO}_\text{(s)} \longrightarrow \text{N$_2$}_\text{(g)} + 3\text{Cu}_\text{(s)}+3\text{H$_2$O}_\text{(g)}$

And we want to determine the amount of products produced when 12.5 moles of NH₃ is reacted with excess CuO.

Compute using stoichiometry. From the equation, we can see the following stoichiometric ratios:

The ratio between NH₃ and N₂ is 2:1. (i.e. One mole of N₂ is produced from every two moles of NH₃.)

The ratio between NH₃ and Cu is 2:3.
The ratio between NH₃ and H₂O is 2:3. (i.e. Three moles of H₂O or Cu is produced frome every two moles of NH₃.)

Dimensional Analysis:

The amount of N₂ produced:

$\displaystyle 12.5\text{ mol NH$_3$} \cdot \frac{1\text{ mol N$_2$}}{2\text{ mol NH$_3$}} = 6.25\text{ mol N$_2$}$

The amount of Cu produced:

$\displaystyle 12.5\text{ mol NH$_3$} \cdot \frac{3\text{ mol Cu}}{2\text{ mol NH$_3$}} = 18.8\text{ mol Cu}$

And the amount of H₂O produced:

$\displaystyle 12.5\text{ mol NH$_3$} \cdot \frac{3\text{ mol H$_2$O}}{2\text{ mol NH$_3$}} = 18.8\text{ mol H$_2$O}$

In conclusion, 6.25 moles of N₂ is produced, and 18.8 moles of Cu and H₂O is produced.

8 0

2 years ago

Show how acid and base give below dissociates in water <br>HNO3

lara31 [8.8K]

The answer is b thank me later :)

3 0

2 years ago

What is the entropy change of the surroundings

KiRa [710]

Answer: The entropy change of the surroundings will be -17.7 J/K mol.

Explanation: The enthalpy of vapourization for 1 mole of acetone is 31.3 kJ/mol

Amount of Acetone given = 10.8 g

Number of moles is calculated by using the formula:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

Molar mass of acetone = 58 g/mol

Number of moles = $\frac{10.8}{58}=0.1862moles$

If 1 mole of acetone has 32.3 kJ/mol of enthalpy, then

0.1862 moles will have = $\frac{32.3}{1}\times 0.1862=5.828kJ/mol$

To calculate the entropy change for the system, we use the formula:

$\Delta S_{sys}=\frac{\Delta H_{vap}}{T(\text{ in K)}}$

Temperature = 56.2°C = (273 + 56.2)K = 329.2K

Putting values in above equation, we get

$\Delta S_{sys}=\frac{5.828}{329.2}=0.0177kJ/Kmol=17.7J/Kmol$ (Conversion Factor: 1 kJ = 1000J)

At Boiling point, the liquid phase and gaseous phase of acetone are in equilibrium. Hence,

$\Delta S_{system}+\Delta S_{surrounding}=0$

$\Delta S_{surrouding}=-\Delta S_{system}=-17.7J/Kmol$

5 0

3 years ago

5. __NH3 + __O2 >>>>>__ NO +__ H20

inn [45]

2 NH3+ 2 O2 —> 2 NO+ 3 H2O

5 0

3 years ago