How does viscosity affect boiling point?

Choose the formula for calculating percent composition.

kondor19780726 [428]

Answer:

The equation for percent composition is (mass of element/molecular mass) x 100.

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6 0

2 years ago

Which choice best describes what would happen if a cube with a density of 0.2 g/mL were placed in a container of water?

Andrew [12]

Answer:

A. It would float with about 80% of the cube below the surface of the water and 20% above the surface.

Explanation:

The choice that best describes what happens to cube of the given density value is that it would float with about 80% of the cube would be below the surface of the water and 20% above the surface.

Density is the mass per unit volume of a substance. The more mass a body has relative to volume, the great it's density. In short, density is directly proportional to mass and inversely related to volume.

The density of water is 1g/mL

If the density of the cube were to be the same with that of water, the substance will just mix up with water .

Here the density is less than that of water.

The density is 0.2g/mL

Therefore, 20% will stay afloat and 80% will be below the surface of the water.

5 0

2 years ago

Give regents and mechanism

Kryger [21]

Answer:

Reagents: 1) $BH_3$ 2) $H_2_O2$ , $OH^-$

Mechanism: Hydroboration

Explanation:

In this case, we have a hydration of alkenes reaction. But, in this example, we have an anti-Markovnikov reaction. In other words, the "OH" is added in the least substituted carbon. Therefore we have to choose an anti-Markovnikov reaction: "hydroboration".

The first step of this reaction is the addition of borane ( $BH_3$ ) to the double bond. Then in the second step, we have the deprotonation of the hydrogen peroxide, to obtain the peroxide anion. In the third step, the peroxide anion attacks the molecule produced in the first step to produce a complex compound in which we have a bond " $BH_2-O-OH$ ". In step number 4 we have the migration of the C-B bond to oxygen. Then in step number 5, we have the attack of $OH^-$ on the $O-BH_2$ to produce an alkoxide. Finally, the water molecule produce in step 2 will protonate the molecule to produce the alcohol.