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Lina20 [59]
4 years ago
14

An aqueous antifreeze solution is 60.0% ethylene glycol (HOCH2CH2OH) by mass and has a density of 1.06 g/mL. Calculate the molal

ity of ethylene glycol. Enter your answer to 1 decimal place.
Chemistry
1 answer:
galina1969 [7]4 years ago
4 0

Answer:

[HOCH₂CH₂OH] = 24.1 m

Explanation:

Ethylene glycol → HOCH₂CH₂OH

60% by mass means that 60 g of ethylene glycol are contained in 100 g of solution.

Solution mass = Solute mass + Solvent mass

100 g = 60 g + Solvent mass

Solvent mass = 40 g

Molality are the moles of solute contained in 1kg of solvent.

We determine the moles of solute → 60 g . 1mol/62 g = 0.967 moles

We convert the mass of solvent from g to kg → 40 g . 1kg/1000 g = 0.04 kg

Molality → 0.967 mol / 0.04 kg = 24.1 m

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KengaRu [80]

Answer:I’m not for sure

Explanation:

8 0
3 years ago
Which characteristic is found in liquids and gases? A. Has a fixed volume B. Particles stay in a fixed position C. No particle m
STALIN [3.7K]

Answer:

The answer to your question is letter D.

Explanation:

A. Has a fixed volume This is not the right answer, liquids and gases take the shape of the container in which they are.

B. Particles stay in a fixed position This answer is wrong, this characteristic is  of solids but not of liquids and gases.

C. No particle movement This characteristic is also of solids, in liquids and gases the particles can move.

D. Takes the shape of the container. This is the right answer, liquids and gases take the shape of the container.

6 0
3 years ago
Read 2 more answers
2, 3-diethyl–4-octene  what is structure of this pls help​
pantera1 [17]

Answer:

(CH2CH3).

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CH-CH-CH-CH=CH-CH2-CH2-CH3

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(CH2CH3)

2, 3-diethyl–4-octene

i am not sure.

4 0
3 years ago
Air is compressed from an inlet condition of 100 kPa, 300 K to an exit pressure of 1000 kPa by an internally reversible compress
ElenaW [278]

Answer:

(a) W_{isoentropic}=8.125\frac{kJ}{mol}

(b) W_{polytropic}=7.579\frac{kJ}{mol}

(c) W_{isothermal}=5.743\frac{kJ}{mol}

Explanation:

Hello,

(a) In this case, since entropy remains unchanged, the constant k should be computed for air as an ideal gas by:

\frac{R}{Cp_{air}}=1-\frac{1}{k}  \\\\\frac{8.314}{29.11} =1-\frac{1}{k}\\

0.2856=1-\frac{1}{k}\\\\k=1.4

Next, we compute the final temperature:

T_2=T_1(\frac{p_2}{p_1} )^{1-1/k}=300K(\frac{1000kPa}{100kPa} )^{1-1/1.4}=579.21K

Thus, the work is computed by:

W_{isoentropic}=\frac{kR(T_2-T_1)}{k-1} =\frac{1.4*8.314\frac{J}{mol*K}(579.21K-300K)}{1.4-1}\\\\W_{isoentropic}=8.125\frac{kJ}{mol}

(b) In this case, since n is given, we compute the final temperature as well:

T_2=T_1(\frac{p_2}{p_1} )^{1-1/n}=300K(\frac{1000kPa}{100kPa} )^{1-1/1.3}=510.38K

And the isentropic work:

W_{polytropic}=\frac{nR(T_2-T_1)}{n-1} =\frac{1.3*8.314\frac{J}{mol*K}(510.38-300K)}{1.3-1}\\\\W_{polytropic}=7.579\frac{kJ}{mol}

(c) Finally, for isothermal, final temperature is not required as it could be computed as:

W_{isothermal}=RTln(\frac{p_2}{p_1} )=8.314\frac{J}{mol*K}*300K*ln(\frac{1000kPa}{100kPa} ) \\\\W_{isothermal}=5.743\frac{kJ}{mol}

Regards.

8 0
3 years ago
Convert 533 cm/s to units of meters per minute. Show the unit analysis by dragging the conversion factors into the unit‑factor s
lukranit [14]

Answer:

319.8 m/min

Explanation:

533 cm/s

We can convert 533 cm/s to m/min by doing the following:

First, we shall convert 533 cm/s to m/s. This can be obtained as illustrated below:

Recall:

100 cm/s = 1 m/s

Therefore,

533 cm/s = 533 cm/s /100 cm/s × 1 m/s

533 cm/s = 5.33 m/s

Finally, we shall convert 5.33 m/s to m/min. This can be obtained as follow:

1 m/s = 60 m/min

Therefore,

5.33 m/s = 5.33 m/s / 1 m/s × 60 m/min

5.33 m/s = 319.8 m/min

Therefore, 533 cm/s is equivalent to 319.8 m/min

8 0
3 years ago
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