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2 years ago

What is the key factor in defining a reaction as an oxidation-reduction reaction?

Chemistry

1 answer:

kari74 [83]2 years ago

6 0

Answer:

Transfer of electrons between atoms

Explanation:

Just because it has the prefix "oxi" it doesn't necessarily mean that oxygen exists as a reactant.

Making and breaking ionic compounds happen right before or after redox reactions after the atoms or ions are modified into one another.

precipitation occurs during the redox rxns involved in electrolysis & daniel cells. but it is not a key factor.

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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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25,000cg is equal to _g

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Answer:

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Explanation:

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