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4 years ago

How many atoms are in a sample of 72.8 g calcium (Ca)?

2 answers:

8 0

Atomic mass Ca=40.078 a.m.u

40.078 g--------------------6.02x10²³ atoms
72.8 g ---------------------- ??

72.8 x (6.02x10²³ ) / 40.078 =

4.38x10²⁵ / 40.078 = 1.093x10²⁴ atoms

HOPE THIS HELPED :D

xxMikexx [17]4 years ago

5 0

Atomic mass Ca = 40.078 a.m.u

40.078 g -------------------- 6.02x10²³ atoms
72.8 g ---------------------- ??

72.8 x ( 6.02x10²³) / 40.078 =

4.38x10²⁵ / 40.078 = 1.093x10²⁴ atoms

hope this helps!

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calculate density of a rectangular solid, which has a mass of 25.71 g. It is 2.30 cm long, 4.01 cm wide and 1.82 cm high.

Levart [38]

Density is a property of the substances that is obtained by dividing its mass by the volume. For a rectangular solid, the volume may be solved by the following equation,

V = L x W x H

Substituting the given values for the dimension,

V = (2.30 cm) x (4.01 cm) x (1.82 cm) = 16.78786 cm³

Calculating for the density,

Density = mass / volume

Density = 25.71 cm / <span>16.78786 cm³ = 1.53 grams per cm</span>³

Thus, the density of the given solid is approximately 1.53 grams per cm³.

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3 years ago

3.58 x 1023 molecules of propane C8H8 would occupy how much space at STP

Lynna [10]

Answer: At STP, a mole of gas takes up 22.4 Liters. The 22.4 Liters/mole quantity can be derived from the Ideal Gas Law, PV = nRT, plugging in STP conditions for P and T, and solving for V/n, which gets 22.4 Liters/mole.

Explanation:

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3 years ago

2 AIPO4 + 3 BaCl2 --> 2 AICIz + 1 Baz(PO4)2

givi [52]

Answer:

12.3 moles of $AlCl_{3}$ are produced by reacting 18.4 moles of $BaCl_{2}$ .

Explanation:

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According to balanced equation:

3 molecules of $BaCl_{2}$ produce 2 molecules of $AlCl_{3}$

So $3N_{A}$ molecules of $BaCl_{2}$ produce $2N_{A}$ molecules of $AlCl_{3}$ ( $N_{A}$ is avogadro number).

We know 1 mol of molecule = $N_{A}$ number of molecules.

So 3 moles of $BaCl_{2}$ produce 2 moles of $AlCl_{3}$ .

Hence 18.4 moles of $BaCl_{2}$ produce $(\frac{2}{3}\times 18.4)$ moles of $AlCl_{3}$ or 12.3 moles of $AlCl_{3}$ .

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4 years ago

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GuDViN [60]

The most common method astronomers use to determine the composition of stars, planets, and other objects is spectroscopy. This process utilizes instruments with a grating that spreads out the light from an object by wavelength. This spread-out light is called a spectrum. Every element has a unique fingerprint that allows researchers to determine what it is made of.

The fingerprint often appears as the absorption of light. Every atom has electrons, and these electrons like to stay in their lowest-energy levels. But when photons carrying energy hit an electron, they can push it to higher energy levels. This is absorption, and each element’s electrons absorb light at specific wavelengths related to the difference between energy levels in that atom. But the electrons want to return to their original levels, so they don’t hold onto the energy for long. When they emit the energy, they release photons with exactly the same wavelengths of light that were absorbed in the first place. An electron can release this light in any direction, so most of the light is emitted in directions away from our line of sight. Therefore, a dark line appears in the spectrum at that particular wavelength.

Because the wavelengths at which absorption lines occur are unique for each element, astronomers can measure the position of the lines to determine which elements are present in a target. The amount of light that is absorbed can also provide information about how much of each element is present.

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3 years ago

A reddish-orange element used for electrical wiring

dsp73

Copper is the answer to ur question

4 0

4 years ago