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3 years ago

In which of these cases would it be most useful to use a mole to calculate the number of particles?

1 answer:

Ivan3 years ago

6 0

I believe it would be the last one because you can use the molar mass of HCl to find the number of moles, then use Avogadro’s number to find the number of atoms
Hope this helps!

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What electrically neutral atom will have an electron configuration of 1s2 2s2 2p4?

abruzzese [7]

Answer:- oxygen.

Explanations:- The electronic configuration is given and we are asked to figure out the electrically neutral atom that will have the electron configuration, $1s^22s^22p^4$ .

The sum of electrons for this electron configuration is 8. If we look at the periodic table then 8 is the atomic number of oxygen.

So, the electrically neutral atom for the given electron configuration is oxygen.

6 0

3 years ago

In the graph above for the dissociation of a strong acid, why do the bars for H3O+ and A- have the same height as the bar for HA

dimaraw [331]

Answer:

The concentration of HA is the same as concentration of H3O+ and A- produced.

Explanation:

The dissociation equation is given below:

HA(aq) + H2O (l) —> H3O+(aq) + A-(aq)

From the reaction above, we can see that the acid is monoprotic acid i.e it has only 1 ionisable hydrogen atom.

Now, from the balanced equation, we can see that the acid produced equal concentration of H3O+ and A-.

This account for the reason why the bars for H3O+ and A- have the same height as the bar for HA.

6 0

3 years ago

The CRC Handbook, a large reference book of chemical and physical data, lists two isotopes of indium (). The atomic mass of 4.28

ivanzaharov [21]

Answer: The mass of second isotope of indium is 114.904 amu

Explanation:

Average atomic mass of an element is defined as the sum of masses of each isotope each multiplied by their natural fractional abundance.

Formula used to calculate average atomic mass follows:

.....(1)

Let the mass of isotope 2 of indium be 'x'

For isotope 1:

Mass of isotope 1 = 112.904 amu

Percentage abundance of isotope 1 = 4.28 %

Fractional abundance of isotope 1 = 0.0428

For isotope 2:

Mass of isotope 2 = x amu

Percentage abundance of isotope 2 = [100 - 4.28] = 95.72 %

Fractional abundance of isotope 2 = 0.9572

Average atomic mass of indium = 114.818 amu

Putting values in equation 1, we get:

$114.818=[(112.904\times 0.0428)+(x\times 0.9572)]\\\\x=114.904amu$

Hence, the mass of second isotope of indium is 114.904 amu

4 0

3 years ago

As you add protons to the nucleus and you add more what happen

Lubov Fominskaja [6]

The more protons you add, the more positively charged the atom becomes
the charge of an atom determines what kind of an atom it is

7 0

3 years ago

If 1.50 L of 0.780 mol/L sodium sulfide is mixed with 1.00 L of a 3.31 mol/L lead(II) nitrate solution, what mass of precipitate

NARA [144]

Answer:

336.1 g of PbS precipitate

Explanation:

The equation of the reaction is given as;

Na2S(aq) + Pb(NO3)2(aq) ----> 2NaNO3(aq) + PbS(s)

Ionically;

Pb^2+(aq) + S^2-(aq) -----> PbS(s)

Number of moles of sodium sulphide= concentration of sodium sulphide × volume of sodium sulphide

Number of moles of sodium sulphide= 0.780 × 1.5 = 1.17 moles

Number of moles of lead II nitrate= concentration of lead II nitrate × volume of lead II nitrate

Number of moles of lead II nitrate= 3.31× 1.00= 3.31 moles

Then we determine the limiting reactant. The limiting reactant yields the least amount of product.

Since 1 moles of sodium sulphide yields 1 mole of lead II sulphide

1.17 moles of sodium sulphide also yields 1.17 moles of lead II sulphide

Hence sodium sulphide is the limiting reactant.

Thus mass of precipitate formed= amount of lead II sulphide × molar mass of sodium sulphide

Molar mass of lead II sulphide= 287.26 g/mol

Mass of lead II sulphide = 1.17 moles × 287.26 g/mol

Mass of lead II sulphide= 336.1 g of PbS precipitate

4 0

3 years ago