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3 years ago

Molecular compounds result from covalent bonding which are called _____

Chemistry

1 answer:

11Alexandr11 [23.1K]3 years ago

3 0

There are three which are:
dispersion/london/van der vaals forces
dipole/dipole
hydrogen bonding

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Formulae for calculating number of moles

Tresset [83]

You would use this number, 6.02×1023 (Avogadro's number) to convert from particles, atoms, or molecules to moles. Whenever you go to the mole, divide by Avogadro's number. When you go to the unit from moles, multiply by Avogadro's number.

4 0

3 years ago

Which substance is the oxidizing agent in the following reaction? Fe2S3 + 12HNO3 → 2Fe(NO3)3 + 3S + 6NO2 + 6H2O Which substance

aleksandr82 [10.1K]

Answer:

HNO3 is the oxidizing agent

Explanation:

Step 1:

The oxidizing agent is responsible to oxidize another. but itself undergoes a reduction .

Fe2S3:

Fe has an oxidation number of +3

S has an oxidation number of -2

HNO3:

O has an oxidation number of -2 ( we have 3 times O so this makes -6)

H has an oxidation number of +1

N has an oxidation number of +5

Fe(NO3)3

O has an oxidation number of -2 ( we have 3 times O so this makes -6)

Fe has an oxidation number of +3

Since NO3 has an oxidation number of -1; N has an oxidation number of +5

S alone has an oxidation number of 0

NO2:

O has an oxidation number of -2 (we have 2 times O, this makes -4)

N has an oxidation number of +4

Fe doesn't change from oxidation number. It stays +3

N goes from +5 to +4 → this is a reduction

S goes from -2 to 0 → this is an oxidation

The reducing agent is the compound that contributes the oxidized species (S).

The oxidizing agent contributes the reducing species (N)

The answer is HNO3

5 0

3 years ago

T = 409.5 K, P = 1.50 atm: V = ?L

Molodets [167]

Explanation:

T = 409.5 K, P = 1.50 atm: V = 22.4 L The ideal gas law is: PV = nRT where. P = pressure. V = volume n = number of moles.

7 0

2 years ago

Diatomic iodine [I2] decomposes at high temperature to form I atoms according to the reaction I2(g)⇌2I(g), Kc = 0.011 at 1200∘C

umka2103 [35]

Answer:

The concentration of I at equilibrium = 3.3166×10⁻² M

Explanation:

For the equilibrium reaction,

I₂ (g) ⇄ 2I (g)

The expression for Kc for the reaction is:

$K_c=\frac {\left[I_{Equilibrium} \right]^2}{\left[I_2_{Equilibrium} \right]}$

Given:

$\left[I_2_{Equilibrium} \right]$ = 0.10 M

Kc = 0.011

Applying in the above formula to find the equilibrium concentration of I as:

$0.011=\frac {\left[I_{Equilibrium} \right]^2}{0.10}$

So,

$\left[I_{Equilibrium} \right]^2=0.011\times 0.10$

$\left[I_{Equilibrium} \right]^2=0.0011$

$\left[I_{Equilibrium} \right]=3.3166\times 10^{-2}\ M$

<u>Thus, The concentration of I at equilibrium = 3.3166×10⁻² M</u>

3 0

3 years ago

Nitroglycerin (C3H5N3O9) is a powerful explosive. Its decomposition may be represented by 4C3H5N3O9 → 6N2 + 12CO2 + 10H2O + O2

almond37 [142]

Answer:

the guy infront of me

Explanation:

8 0

2 years ago