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Chemistry

1 answer:

-Dominant- [34]3 years ago

4 0

Answer:

The answer is SiO2

Explanation:

Silocon dioxide is written without a 1 after the silocon and with a 2 after the oxygen.

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Two gas-filled tanks have the same volume, temperature, and pressure. They are identical in every way except that one is filled

Tems11 [23]

Answer:

The tank with O₂ weighs more.

Explanation:

We can find the mass of gas using the ideal gas equation.

$P.V=n.R.T=\frac{m}{M} .R.T\\m=\frac{P.V.M}{R.T}$

Considering the pressure (P), volume (V), temperature (T) and ideal gas constant (R) are the same, we can establish that:

m ∝ M

The mass is directly proportional to the molar mass. The molar mass of O₂ (32 g/mol) is higher than the molar mass of N₂ (28 g/mol). Therefore, the tank with O₂ weighs more.

8 0

3 years ago

Hydrogen bonds are in the family of Van der Waals forces. They are weaker than ionic and covalent bonds but they cause interesti

-Dominant- [34]

Answer:

The answer is B. Van der Waals forces are weaker than ionic and covalent bonds.

Explanation:

In general, if we arrange these molecular forces from the strongest to weakest, it would be like this:

Covalent bonds > Ionic bonds > Hydrogen bonds > Dipole-Dipole Interactions > Van der Waals forces

Covalent bonds are known to have the strongest and most stable bonds since they go deep and into the inter-molecular state. A diamond is an example of a compound with this characteristic bond.

Ionic bonds are the next strongest molecular bond following covalent bonds. This is due to the protons and electrons causing an electro-static force which results to the strong bonds. An example would be Sodium Chloride (NaCl), which when separated is Na⁺ and Cl⁻.

Van der Waals forces, also known as Dispersion forces, are the weakest type of molecular bonds. They are only formed through residual molecular attractions when molecules pass by each other. It doesn't even last long due to the uneven electron dispersion. It can be made stronger by adding more electrons in the molecule. This kind of molecular bonds appear in non-polar molecules such as carbon dioxide.

HOPE THIS HELPS!!!!!!!!!!!!!!

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7 0

3 years ago

Draw the major organic substitution product(s) for (2R,3S)-2-bromo-3-methylpentane reacting with the given nucleophile. Indicate

Andrew [12]

Answer:

(2R,3S)-2-ethoxy-3-methylpentane

and

(2S,3S)-2-ethoxy-3-methylpentane

Explanation:

For this case, we will have $CH_3CH_2O^-$ as nucleophile. Also, this compound is also in excess. So, we will have as solvent $CH_3CH_2OH$ a protic solvent. Therefore the Sn1 reaction would be favored.

The first step would be the carbocation formation followed by the attack of the nucleophile. In this case both isomers would be produced: R and S (see figure).