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3 years ago

Why do some objects float why others sink

2 answers:

5 0

It will sink. This means whether or not an object will float or sink depends on its own density and the density of the liquid it is placed in. In the case of water, an object with a density less than 1 g/cm3 will float. The closer its density is to 1 g/cm3, the more of it will sit below the water level.

Leona [35]3 years ago

4 0

Because so objects are denser than water and some are less dense than water

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Classify each of the coordination compounds according to the coordination number.

amid [387]

Coordination number has to do with the number of ligands that is attched to the central metal atom/ion.

<h3>What is coordination number?</h3>

The term coordination number has to do with the number of ligands that is attched to the central metal atom/ion. The coordination sphere contains this central metal along with associated ligands.

The question is incomplete. However, if I have something like, K3[Cr(H2O)6], the coordination number of the complex is six.

Learn mlore about coordination number: brainly.com/question/16236454

3 0

1 year ago

A gas has a volume of 1.75L at -23°C and 150.0 kPa. At what temperature would the gas occupy 1.30L at 210.0 kPa?

Nastasia [14]

Answer:

At -13 $^{0}\textrm{C}$ , the gas would occupy 1.30L at 210.0 kPa.

Explanation:

Let's assume the gas behaves ideally.

As amount of gas remains constant in both state therefore in accordance with combined gas law for an ideal gas-

$\frac{P_{1}V_{1}}{T_{1}}=\frac{P_{2}V_{2}}{T_{2}}$

where $P_{1}$ and $P_{2}$ are initial and final pressure respectively.

$V_{1}$ and $V_{2}$ are initial and final volume respectively.

$T_{1}$ and $T_{2}$ are initial and final temperature in kelvin scale respectively.

Here $P_{1}=150.0kPa$ , $V_{1}=1.75L$ , $T_{1}=(273-23)K=250K$ , $P_{2}=210.0kPa$ and $V_{2}=1.30L$

Hence $T_{2}=\frac{P_{2}V_{2}T_{1}}{P_{1}V_{1}}$

$\Rightarrow T_{2}=\frac{(210.0kPa)\times (1.30L)\times (250K)}{(150.0kPa)\times (1.75L)}$

$\Rightarrow T_{2}=260K$

$\Rightarrow T_{2}=(260-273)^{0}\textrm{C}=-13^{0}\textrm{C}$

So at -13 $^{0}\textrm{C}$ , the gas would occupy 1.30L at 210.0 kPa.

5 0

3 years ago

acid-catalyzed hydration of 1-methylcyclohexene gives two alcohols. The major product does not undergo oxidation, while the mino

noname [10]

Answer:

Major product does not undergo oxidation since it is a tertiary alcohol whereas minor product undergoes oxidation to ketone as it is secondary alcohol.

Explanation:

Hello,

In this case, given the attached picture, the hydration of the 1 methylcyclohexene yields to alcohols; 1-methylcyclohexan-1-ol and 1-methylcyclohexan-2-ol. Thus, since the OH in the 1-methylcyclohexan-1-ol (major product) is bonded to a tertiary carbon (bonded with other three carbon atoms) it is not able to increase the number of oxygen bonds (oxidation) as it already attained the octet whereas the 1-methylcyclohexan-2-ol (minor product) is able to undergo oxidation to ketone as the carbon bonded to it is secondary (bonded with other two carbon atoms), so one extra bond the oxygen is allowed to be formed to carbonyl.

Best regards.