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3 years ago

Why do elements in group 8A (the noble gases) tend not to gain or lose electrons?

Chemistry

1 answer:

Neko [114]3 years ago

5 0

Answer:

The noble gases have 8 valence electrons in their outermost electron shell. In other words, they have full out shells. These elements are highly stable.

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Under which conditions does a sample of the same mass of carbon dioxide have the lowest

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The answer is B, as a solid and a low temperature would result in the lowest entropy value

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3 years ago

3. Crystalline structural unit of barium metal is a body-centered cubic cell. The edge length of the unit cell is 5.02x10-8 cm.

tiny-mole [99]

Answer:

The Avogadro's number is $N_A = 6.02289 *10^{23}$

Explanation:

From the question we are told that

The edge length is $L = 5.02 * 10^{-8} \ cm= \frac{5.02 * 10^{-8} }{100} = 5.02 * 10^{-10}$

The density of the metal is $\rho = 5.30\ g/cm^3 = 5.30 * \frac{g}{cm^3} * \frac{1*10^6}{1*10^3} = 5.30 *10^3kg/m^3$

The molar mass of Ba is $Z = 137.3 \ g/mol = \frac{137.3}{1000} = 0.1373 \ kg / mol$

Generally the volume of a unit cell is

$V = L^3$

substituting value

$V = [5.02 *10^{-10}]^3$

$V = 1.265*10^{-28}\ m^3$

From the question we are told that 68% of the unit cell is occupied by Ba atoms and that the structure is a metal which implies that the crystalline structure will be (BCC),

The volume of barium atom is

$V_a = \frac{V}{2} * 0.68$

substituting value

$V_a = \frac{ 1.265*10^{-28}}{2} * 0.68$

$V_a = 4.301 *10^{-29} \ m^3$

The Molar mass of barium is mathematically represented as

$Z = N_A V_a * \rho$

Where $N_A$ is the Avogadro's number

$N_A = \frac{ Z}{ V_a * \rho}$

substituting value

$N_A = \frac{ 0.1373}{ 4.301*10^{-29} * 5.3*10^{3}}$

$N_A = 6.02289 *10^{23}$

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3 years ago

Ñspent fuelî that can no longer be used to create energy.<br> nuclear waste ?

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Spent fuel that can no longer be used to create energy is waste.

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3 years ago

Which type of energy is stored in the nucleus of an atom?

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Nuclear energy is energy in the nucleus (core) of an atom. Atoms are tiny particles that make up every object in the universe. There is enormous energy in the bonds that hold atoms together. Nuclear energy can be used to make electricity.

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3 years ago

Read 2 more answers

What volume, in mL, of carbon dioxide gas is produced at STP by the decomposition of 0.242 g calcium carbonate (the products are

damaskus [11]

Answer:

54.21 mL.

Explanation:

We'll begin by calculating the number of mole in 0.242 g calcium carbonate, CaCO3.

This is illustrated below:

Mass of CaCO3 = 0.242 g

Molar mass of CaCO3 = 40 + 12 +(16x3) = 40+ 12 + 48 = 100 g/mol

Mole of CaCO3 =?

Mole = mass /Molar mass

Mole of CaCO3 = 0.242/100

Mole of CaCO3 = 2.42×10¯³ mole.

Next, we shall write the balanced equation for the reaction. This is given below:

CaCO3 —> CaO + CO2

From the balanced equation above,

1 mole of CaCO3 decomposed to produce 1 mole CaO and 1 mole of CO2.

Next, we shall determine the number of mole of CO2 produced from the reaction.

This can be obtained as follow:

From the balanced equation above,

1 mole of CaCO3 decomposed to produce 1 mole of CO2.

Therefore,

2.42×10¯³ mole of CaCO3 will also decompose to produce 2.42×10¯³ mole of CO2.

Therefore, 2.42×10¯³ mole of CO2 were obtained from the reaction.

Finally, we shall determine volume occupied by 2.42×10¯³ mole of CO2.

This can be obtained as follow:

1 mole of CO2 occupies 22400 mL at STP.

Therefore, 2.42×10¯³ mole of CO2 will occupy = 2.42×10¯³ x 22400 = 54.21 mL

Therefore, 54.21 mL of CO2 were obtained from the reaction.

7 0

3 years ago