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2 years ago

What is occurring in the image below?

Chemistry

1 answer:

DiKsa [7]2 years ago

3 0

The addictive quality is a reduction in vapour pressure.

The vapour pressure at a liquid's normal boiling point is the same as the ordinary atmospheric pressure, which is 1 atmosphere, 760 Torr, 101.325 kPa, or 14.69595 psi.

The pressure that results from liquids evaporating is known as vapour pressure. Surface area, intermolecular forces, and temperature are three often occurring variables that affect vapour press.

lower vapour pressure

raising the boiling point

Low-temperature depression

Osmotic force

They are all dependent on the solute; when you increase the solute, the colligative property and the ratio you added may change.

The Van't Hoff Factor is another option to examine (i). the number of dissolved ions. The colligative property will be further altered if the solute is ionic.

Learn more about Vapor pressure here brainly.com/question/14949391

SPJ1

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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

1. What kinds of metals are REALLY reactive? why?

Olin [163]

Answer:

1. Alkali metals (group 1)

2. halogens (Group 17)

3. noble gasses (group 18)

Explanation:

1. alkali metals only have one valence electron meaning that they really want to lose that one valence electron to get a full octet.

2. halogens have 7 valence electrons meaning that they just need to gain 1 to get a full octet.

3. Nobel gasses already have a full octet meaning that they don't want to react. (atoms only react to get a full octet)

I hope this helps. Let me know if anything is unclear.

8 0

3 years ago

True or false: the part of photosynthesis that does not require light is the Calvin cycle

Mumz [18]

False False False False

5 0

4 years ago

Read 2 more answers

Based on current information, cosmologists expect the universe to _____.

Jlenok [28]

They expect the Universe to continue expanding. Right now, the evidence is that the expansion is speeding up. They also expect entropy to continue increasing, so in the distant future, there will not be any free energy left to support life, for example.

8 0

3 years ago

27.8 mL solution of 0.797 M HCHO2 with 0.928 M NaOH. What is the pH for the solution at the equivalence point in the titration?

kati45 [8]

Answer:

8.69 is the pH at the equivalence point

Explanation:

Formic acid, HCHO₂, reacts with NaOH as follows:

HCHO₂ + NaOH → NaCHO₂ + H₂O.

At the equivalence point you will have in the reaction just NaCHO₂ and H₂O. The concentration of NaCHO₂ will be:

<em />

<em>Moles: </em>0.0278L * 0.797mol/L = 0.02216moles

To reach the equivalence point it is necessary to add:

0.02216mol * (1L / 0.928mol) = 0.0239L

Total volume in the equivalence point:

0.0278L + 0.0239L = 0.0517L

Concentration: 0.02216moles / 0.0517L = 0.429M

The equilibrium of NaCHO₂, CHO₂⁻, in water is:

CHO₂⁻(aq) + H₂O(l) ⇄ OH⁻(aq) + HCHO₂(aq)

Where Kb, 5.56x10⁻¹¹ is defined as:

5.56x10⁻¹¹ = [OH⁻] [HCHO₂] / [CHO₂⁻]

In the equilibrium, it is produced X OH⁻ and HCHO₂, and as concentration of NaCHO₂ is 0.429M:

5.56x10⁻¹¹ = [X] [X] / [0.429M]

2.383x10⁻¹¹ = X²

4.88x10⁻⁶ = X = [OH⁻]

As pOH = -log [OH⁻]

pOH = 5.31

And pH = 14 - pH

pH = 8.69 is the pH at the equivalence point

3 0

4 years ago