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Alja [10]
3 years ago
7

What is the difference between the atomic number and the mass number of an element?

Chemistry
1 answer:
sukhopar [10]3 years ago
6 0

Answer:

Atomic mass is associated with the number of neutrons and protons that are present in a particular nucleus of an element. Atomic number is usually the number of protons present in an element's nucleus. ... It is the total number of nucleons in the atom's nucleus

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1.Complete and balance each of the acid base neutralizations below. Identify the spectator ions.
Natalka [10]

Answer:

can you give me some hint

4 0
2 years ago
Which correctly summarizes the trend in electron affinity?
ra1l [238]

The answer is it tends to be more negative down a group. This is because as you go down the periodic table, the elements have more electron shells in their atoms. This makes the outermost shells less attracted to the nucleus due to their greater distances from the nucleus. Therefore, these shells are less likely to attract electrons (hence lower electron affinity) and are even more likely to lose electrons from their outer electron orbits.

4 0
4 years ago
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Part a cyclohexane has a freezing point of 6.50 ∘c and a kf of 20.0 ∘c/m. What is the freezing point of a solution made by disso
Andrei [34K]

The mathematical expression for the depression in freezing point is given as:

\Delta T_{f}= k_{f}m

where, \Delta T_{f} = depression in freezing point

k_{f} = molal depression constant    (20.0 ^{o}Ckg/mole)

m = molality

Molality is defined as the ratio of number of moles of solute to the kg of the solvent

Number of moles  =\frac{given mass in g}{molar mass}

given mass  of biphenyl solute = 0.925 g

Molar mass of biphenyl  = 154.21 g/mol

Put the values,

number of moles of biphenyl =\frac{0.925 g}{154.21 g/mol}

= 0.0059 mole

Molality =\frac{number of moles of biphenyl solute}{mass of the solvent in kg}

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Convert g into kg: 1 kg = 1000 g

Thus, mass of cyclohexane  = \frac{25.0}{1000}

= 0.025 kg

Now, put the values in the formula of molality:

Molality =\frac{0.0059 mole}{0.025 kg}

= 0.236 mole/kg or 0.236 m.

Calculate the depression in freezing point:

\Delta T_{f}= 20.0^{o}C kg/mole\times 0.236 mole/kg

= 4.72 ^{o}C

Now, \Delta T_{f} = T_{solvent} - T_{solution}

4.72 ^{o}C = 6.50^{o}C - T_{solution}

T_{solution} = 6.50^{o}C- 4.72 ^{o}C

= 1.78^{o}C

Hence, freezing point of the solution is 1.78^{o}C.



4 0
3 years ago
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