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3 years ago

Calculate number of silver atoms in

Chemistry

1 answer:

Veseljchak [2.6K]3 years ago

6 0

Answer:

$1.84x10^{24}atoms$

Explanation:

Hello there!

In this case, according to the given information, it turns out possible for us to calculate the number of silver atoms in 330 grams of a silver bracelet by knowing 1 mole of silver has a mass of 108 grams and also contains 6.022x10²³ atoms; thus, the mathematical setup is shown below:

$330g*\frac{1mol}{108g}*\frac{6.022x10^{23}atoms}{1mol}$

Which is equal to:

$1.84x10^{24}atoms$

Regards!

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Which group 4A element has the highest ionization energy?

Gelneren [198K]

The ionizatin energy is the is the energy required to remove one electron from from a gaseous atom.

There is a clear trend for the first ionization energies (the ionization energy to remove the first electron) of the atoms.

Inside a group (a column of the periodic table) the ionization energy increases downwards .

Then, the element that has the greatest atomic number in the group has the highest ionization energy (it is more difficult to form its ion).

As per this trend, in the group 4A lead, Pb (atomic number 82) is the natural element with the highest ionization energy. But there is one artificial element in this group, whose atomic number is 114. It is called Flerovium and as per the rule it shall have a higher ionization energy.

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For H3PO4, Ka1 = 7.3 x 10^-3, Ka2 = 6.2 x 10^-6, and Ka3 = 4.8 x 10^-13. A 0.10 M aqueous solution of Na3PO4 therefore would be

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Answer:

The aqueous solution of $Na_{3} Po_{4}$ therefore would be "strongly basic."

Explanation:

Strongly basic because the $Na_{3} Po_{4}$ is completely an ionic compound. These compound can be 100% splitted into metal ions and hydroxide ion when used in the aqueous solution. Each moles dissolves which can give rise to mole of hydrogen ions when used in the solution. Whereas strong bases are not soluble in water as well. in the presence of water equilibrium can also be achieved. A strong base has the capability which can just remove the proton from the molecule.

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What hypothesis led to the discovery of the proton?

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When a neutral hydrogen atom loses an electron, a positively-charged particle should remain.

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3 years ago

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What is the molecular formula of a compound that is 51.02 % carbon, 13.80 % hydrogen, and 35.18 %

elena-14-01-66 [18.8K]

Answer:

Molecular formula = C₄H₁₂O₂

Explanation:

Given data:

Percentage of hydrogen = 13.80%

Percentage of carbon = 51.02%

Percentage of oxygen = 35.18%

Molecular formula = ?

Solution:

Number of gram atoms of H = 13.80 / 1.01 = 13.6

Number of gram atoms of O = 35.18 / 16 = 2.2

Number of gram atoms of C = 51.02 / 12 = 4.25

Atomic ratio:

C : H : O

4.25/2.2 : 13.6/2.2 : 2.2/2.2

2 : 6 : 1

C : H : O = 2 : 6 : 1

Empirical formula is C₂H₆O.

Molecular formula:

Molecular formula = n (empirical formula)

n = molar mass of compound / empirical formula mass

Empirical formula mass = C₂H₆O = 46 g/mol

n = 88 / 46

n = 2

Molecular formula = n (empirical formula)

Molecular formula = 2 (C₂H₆O)

Molecular formula = C₄H₁₂O₂

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3 years ago

The rock in a particular iron ore deposit contains 82 % Fe2O3 by mass. How many kilograms of the rock must be processed to obtai

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Answer:

Mass of rock = 4533.4 kg

Explanation:

Percent composition is percentage by the mass of element present in the compound.

Given that the $Fe_2O_3$ is 82 % by mass.

It means that 82 kg of $Fe_2O_3$ is present in 100 kg of rock

To find the mass of rock which contains 2600 kg of iron

2600 kg = 2600*1000 g

2 moles of iron are present in 1 mole of $Fe_2O_3$

Molar mass of iron = 55.845 g/mol

Mass of 2 moles = 55.845 * 2 = 111.69 g/mol

Molar mass of $Fe_2O_3$ = 159.69 g/mol

It means,

111.69 g of iron are present in 159.69 g of $Fe_2O_3$

1 g of iron is present in 159.69/111.69 g of $Fe_2O_3$

2600*1000 g is present in (159.69/111.69)*2600*1000 g of $Fe_2O_3$

Mass of $Fe_2O_3$ = 3717378.47 g =3717.38 kg

Thus,

82 kg of $Fe_2O_3$ is present in 100 kg of rock

1 kg of $Fe_2O_3$ is present in 100/82 kg of rock

3717.38 kg of $Fe_2O_3$ is present in (100/82)*3717.38 kg of rock

Mass of rock = 4533.4 kg

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