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salantis [7]
2 years ago
10

What is the molar mass of C3N4? (Do NOT round this number.)

Chemistry
1 answer:
vichka [17]2 years ago
4 0

Answer:

92.06 g/mol

Explanation:

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A. element with the highest electronegativity.
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Answer:

Fluorine

Explanation:

Fluorine is assigned the oxidation number of -1 because it attracts the electrons in the bond more strongly than carbon does. Fluorine appears to have an extra electron, -1 oxidation number.

Fluorine is the most electronegative element on the periodic table.

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Initially, scientists described atoms as the smallest particles of matter. However, smaller particles within atoms were discover
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By designing new technology that would prove new information

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What is the formula of the hydride formed by germanium?
Nadusha1986 [10]

The atoms combine to form compounds to attain stability in nature. The combination of atoms takes place by sharing of electrons between the atoms or complete transfer of electrons from one atom to another. Generally, atoms combine to complete their octet, that is to possess eight electrons in their outer most shell (noble gas configurations) except hydrogen which can attain stability by two electrons in its outer most shell.

Since germanium has 4 electrons in its outer most shell so it needs 4 more electrons to complete its octet and attains the stability. Hydrogen has 1 electron in its outer most shell and it needs only 1 electron to attain stability so, each germanium will combine with 4 hydrogen atoms and thus forming GeH_4 molecule which is stable in nature.

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6 0
3 years ago
Suppose 0.981 g of iron (II) iodide is dissolved in 150. mL of a 35.0 m M aqueous solution of silver nitrate. Calculate the fina
yaroslaw [1]

Answer:

Final molarity of iodide ion C(I-) = 0.0143M

Explanation:

n = (m(FeI(2)))/(M(FeI(2))

Molar mass of FeI(3) = 55.85+(127 x 2) = 309.85g/mol

So n = 0.981/309.85 = 0.0031 mol

V(solution) = 150mL = 0.15L

C(AgNO3) = 35mM = 0.035M = 0.035m/L

n(AgNO3) = C(AgNO3) x V(solution)

= 0.035 x 0.15 = 0.00525 mol

(AgNO3) + FeI(3) = AgI(3) + FeNO3

So, n(FeI(3)) excess = 0.00525 - 0.0031 = 0.00215mol

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8 0
3 years ago
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