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BaLLatris [955]

3 years ago

13

When water is pushed out of the way it is said to be____________________.

1 answer:

Travka [436]3 years ago

4 0

I’m not sure but I thin is transpiration

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Cell Structure and Functions Activity Worksheet

ICE Princess25 [194]

Answer:

u still need this?

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3 years ago

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Convert 22 grams of copper Il chloride (Cucl2) to moles. <br>with work please

Fittoniya [83]

Hope this helps!

https://youtu.be/MT5tKijKZwQ

5 0

3 years ago

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A mixture of compounds containing diethylamine, phenol, ammonia, and acetic acid is separated using liquid-liquid extraction as

umka2103 [35]

Answer:

I would expect to extract the acetic acid.

Explanation:

In the first step, since we are adding a concentrated acid,<u> it will react with the bases present in the mixture (diethylamine and ammonia) </u><u>forming salts</u><u>, </u><u>which are soluble in water</u>. Therefore, after draining the aqueous layer, we will have phenol and acetic acid left in the organic layer.

In the second step, we are adding a diluted base, so it will react with a strong acid. This compound is acetic acid, and its salt will be present in the aqueous layer. Phenol will be left on the organic layer.

7 0

3 years ago

hammer [34]

Answer:

in the excited state

Explanation:

Because in excited state an atom has more energy

5 0

3 years ago

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The balanced combustion reaction for C6H6 is 2C6H6(l)+15O2(g)⟶12CO2(g)+6H2O(l)+6542 kJ If 8.600 g C6H6 is burned and the heat pr

jenyasd209 [6]

Answer : The final temperature of the water is, $22.5166^oC$

Explanation : Given,

Mass of benzene = 8.600 g

Molar mass of benzene = 78 g/mole

First we have to calculate the moles of benzene.

$\text{Moles of benzene}=\frac{\text{Mass of benzene}}{\text{Molar mass of benzene}}=\frac{8.600g}{78g/mol}=0.1103mole$

Now we have to calculate the energy of combustion.

The given balanced chemical reaction is:

$2C_6H_6(l)+15O_2(g)\rightarrow 12CO_2(g)+6H_2O(l)+6542 kJ$

According to reaction,

As, 2 moles of benzene gives 6542 kJ of energy on combustion.

So, 0.1103 mole of benzene gives $\frac{6542 kJ}{2}\times 0.1103=360.7913kJ$ of energy on combustion.

Now we have to calculate the final temperature of the water.

Formula used : $q_w=m_w\times c_w\times \Delta T=m_w\times c_w\times (T_{final}-T_{initial})$

where,

$q_w$ = heat released = 360.7913 kJ = 36079.13 J

$m_w$ = mass of water = 5691 g

$c_w$ = specific heat of water= $4.18J/g^oC$

$T_{final}$ = final temperature = ?

$T_{initial}$ = initial temperature = $21^oC$

Now put all the given values in the above formula, we get:

$36079.13J=5691g\times 4.18J/g^oC\times (T_{final}-21^oC)$

$T_{final}=22.5166^oC$

Therefore, the final temperature of the water is, $22.5166^oC$

8 0

4 years ago