What number equals atomic mass

Chemistry

2 answers:

ElenaW [278]3 years ago

8 0

Answer:

The total number of nucleons - protons and neutrons - in the nucleus of an atom is called the Mass Number.

Explanation:

.Atomic mass is the weighted average mass of an atom of an element based on the relative natural abundance of that element's isotopes.

mario62 [17]3 years ago

8 0

Answer:

The mass number of the atom (M) is equal to the sum of the number of protons and neutrons in the nucleus. The number of neutrons is equal to the difference between the mass number of the atom (M) and the atomic number (Z).

Hope this helps! Mark brainliest please :)

You might be interested in

Fix the formula for Mg3F Write out the full correct formula?

LenKa [72]

Answer:

$MgF_2$

Explanation:

Hello!

In this case, since, when we see an ionic binary salt like that formed by magnesium (metal) and fluorine (nonmetal), in order to set up the correct molecular formula we need to check out the periodic table in order to identify the suitable oxidation states, since metals remain positively charged as they lose electrons and nonmetals negatively charged as they gain electrons.

In such a way, since the oxidation state of magnesium is +2 and that of fluorine is -1 we write:

$Mg^{2+}F^-$

Next, we need to exchange the oxidation states as subscripts without the charge to obtain:

$MgF_2$

Which is the corrected molecular formula for magnesium fluoride.

Best regards!

3 0

3 years ago

The standard entropy of Pb(s) at 298.15 K is 64.80 J K–1 mol–1. Assume that the heat capacity of Pb(s) is given by: CP,m(Pb, s)

aalyn [17]

Answer:

$s_{Pb(l),500\°C}=100.83\frac{J}{mol*K}$

Explanation:

Hello,

In this case, for the calculation of the standard entropy of liquid lead at 500 °C (773.15 K), starting by solid lead 298.15 K we need to consider three processes:

1. Heating of solid lead at 298.15 K to 600.55 K (melting point).

2. Melting of solid lead to liquid lead.

3. Heating of liquid lead at 600.55 K (melting point) to 773.15 K.

Which can be written in terms of entropy by:

$s_{Pb(l),500\°C}=s_{Pb(s),298.15K}+s_1+s_2+s_3$

Whereas each entropy is computed as follows:

$s_1=\int\limits^{600.55K}_{298.15K} {\frac{22.13 + 0.01172 T + 1.00 x 10^{-5} T^2}{T} } \, dT =20.4\frac{J}{mol*K}\\\\\\s_2=\frac{4770\frac{J}{mol} }{600.55K}= 7.94\frac{J}{mol*K}\\\\\\s_3=\int\limits^{773.15K}_{600.55K} {\frac{32.51-0.00301T}{T} } \, dT=7.69\frac{J}{mol*K}$