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3 years ago

The density (d) of a substance is an intensive property that is defined as the ratio of its mass (m) to its volume (v).

Chemistry

2 answers:

vovangra [49]3 years ago

8 0

Answer:

Density is an intensive property

Explanation:

The intensive properties don´t depend on mass or size, if you have a liter of water, this litter have a density equal 1000 kg/m^3. If zo have 250 liters of water, the density is 1000 kg/m^3 too.

The density does not change with the addition or substraction of matter.

I hope I have been clear.

Bas_tet [7]3 years ago

5 0

Answer with explanation:

Extensive properties depend on the total matter while intensive properties do not, mass and volume are extensive properties because a bigger mass and a bigger volume are indicative of a more massive object, on the other hand, a more colorful or denser object is not an indication of a more massive object. Density, in particular, is a ratio so it will always be the same regardless of the size of the object, therefore, it is not directly related with the total amount of mass; to better understand this concept imagine you have a bag of sand and you add more sand to the bag, you are increasing the mass and the volume but because both increases the ration m/V do not.

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What is the molarity of a solution that contains 125 g nacl in 4.00 l solution?

bogdanovich [222]

Answer:

[NaCl[ = 0.535M

Explanation:

We determine the moles of solute:

125 g . 1 mol/ 58.45 g = 2.14 moles

Molarity (mol/L) → 2.14 mol / 4L → 0.535M

Molarity is a sort of concentration that indicates the moles of solute in 1L of solution

4 0

4 years ago

PLEASE HELP ME I NEED HELP

Crazy boy [7]

Answer:

5 moles S x (36.02 g S/mole S) = 180.1 grams of S

Explanation:

The periodic table has mass units for every element that can be correlated with the number of atoms of that element. The relationship is known as Avogadro's Number. This number, 6.02x $10^{23}$ , is nicknamed the mole, which scientists found to be a lot more catchy, and easier to write than 6.02x $10^{23}$ . The mole is correlated to the atomic mass of that element. The atomic mass of sulfur, S, is 36.02 AMU, atomic mass units. But it can also be read as 36.02 grams/mole.

This means that 36.02 grams of S contains 1 mole (6.02x $10^{23}$ ) of S atoms.

This relationship holds for all the elements. Zinc, Zn, has an atomic mass of 65.38 AMU, so it has a "molar mass" of 65.38 grams/mole. ^5.38 grams of Zn contains 1 mole of Zn atoms.

And so on.

5.0 moles of Sulfur would therefore contain:

(5.0 moles S)*(36.02 grams/mole S) = 180.1 grams of S

Note how the units cancel to leaves just grams. The units are extremely helpful in mole calculations to insure the correct mathematical operation is done. To find the number of moles in 70 g of S, for example, we would write:

(70g S)/(36.02 grams S/mole S) = 1.94 moles of S. [Note how the units cancel to leave just moles]

4 0

2 years ago

If Jonathan went skateboarding from 4:00 PM to 4:30 PM and traveled 450 meters, what was his average speed?

mrs_skeptik [129]

20 m Increase by five

8 0

3 years ago

How atomic radius changes across a period and down a group in the periodic table?

allochka39001 [22]

Answer:

An atom gets larger as the number of electronic shells increase; therefore the radius of atoms increases as you go down a certain group in the periodic table of elements. In general, the size of an atom will decrease as you move from left to the right of a certain period.

Explanation:

3 0

3 years ago

The energy E of the electron in a hydrogen atom can be calculated from the Bohr formula: E = R_y/n^2 In this equation R_y stands

Katarina [22]

Answer:

The wavelength of the line in the emission line spectrum of hydrogen caused by the transition of the electron for the given energy levels is $5.23\times 10^{-5} m$

Explanation:

Given :

The energy E of the electron in a hydrogen atom can be calculated from the Bohr formula:

$E=\frac{R_y}{n^2}$

$R_y=2.18\times 10^{-18} J$ = Rydberg energy

n = principal quantum number of the orbital

Energy of 11th orbit = $E_{11}$

$E_{11}=\frac{2.18\times 10^{-18} J}{11^2}=1.80\times 10^{-20} J$

Energy of 10th orbit = $E_{10}$

$E_{10}=\frac{2.18\times 10^{-18} J}{10^2}=2.18\times 10^{-20} J$

Energy difference between both the levels will corresponds to the energy of the wavelength of the line which can be calculated by using Planck's equation.

$E'=E_{10}-E_{11}=2.18\times 10^{-20} J-1.80\times 10^{-20} J$

$=E'=0.38\times 10^{-20} J$

$\lambda =\frac{hc}{E'}$ (Planck's' equation)

$\lambda = \frac{6.626\times 10^{-34} Js\times 3\times 10^8 m/s}{0.38\times 10^{-20} J}$

$\lambda = 5.2310\times 10^{-5} m\approx 5.23\times 10^{-5} m$

The wavelength of the line in the emission line spectrum of hydrogen caused by the transition of the electron for the given energy levels is $5.23\times 10^{-5} m$

3 0

3 years ago