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ivann1987 [24]
3 years ago
12

What is the volume of 80.0 g of ether if the density of ether is 0.70 g/mL

Chemistry
1 answer:
Serggg [28]3 years ago
6 0
Density = 0.70 g/mL

Mass = 80.0 g

Volume = ?

V = m / D

V = 80.0 / 0.70

V = 114.28 mL

hope this helps!

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Why is magma so hot?
Eduardwww [97]

Answer: Friction in the center of the earth makes melts rocks and makes magma

Explanation:

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3 years ago
The heat of vaporization of 1–pentanol is 55.5 kJ/mol, and its entropy of vaporization is 148 J/K•mol. What is the approximate b
tensa zangetsu [6.8K]

Answer:

375 K

Explanation:

Using the experssion shown below as:

\Delta G^0=\Delta H^0_{vap}-T\Delta S^0_{vap}

At vaporization point, the liquid and the gaseous phase is in the equilibrium.

Thus,

\Delta G^0=0

So,

Delta H^0_{vap}=T\Delta S^0_{vap}

Given that:

Delta H^0_{vap}=55.5\ kJ/mol

Also, 1 kJ = 10³ J

So,

Delta H^0_{vap}=55500\ J/mol

\Delta S^0_{vap}=148\ J/K.mol

So, temperature is :

T=\frac{Delta H^0_{vap}}{\Delta S^0_{vap}}

T=\frac{55500}{148}

<u>T= 375 K</u>

3 0
3 years ago
Write a mechanism for the esterification of propanoic acid with 18O-labeled ethanol. Show clearly the fate of the 18O label. (b)
tatuchka [14]

Answer:

See explanation and images attached

Explanation:

a) In the mechanism for the acid catalysed esterification of propanoic acid using ethanol, we can see that the first step is the protonation of the acid followed by nucleophillic attack of the alcohol. Loss of water and consequent deprotonation regenerates the acid catalyst. We can see the fate of the 18O labelled ethanol in the mechanism shown.

b)  In the second mechanism, an unnamed ester is hydrolysed using an acid catalyst. The attack of the acid and subsequent nucleophillic attack of water labelled with 18O leads to the incorporation of this 18O into the product acid as shown in the mechanism attached to this answer.

5 0
3 years ago
Which colligative property is employed when salt is put on an icy sidewalk to melt ice?
ankoles [38]
<span>The correct answer is 'freezing point depression'. Colligative properties depend on the concentration of molecules of a solute. Examples of other colligative properties are boiling point elevation or vapour pressure lowering. The salt causes ice on the side walk to melt because it lowers the freezing point. </span>
4 0
3 years ago
Read 2 more answers
Suppose the gas resulting from the sublimation of 1.00 g carbon dioxide is collected over water at 25.0◦c into a 1.00 l containe
jeyben [28]

Answer:

0.55 atm

Explanation:

First of all, we need to calculate the number of moles corresponding to 1.00 g of carbon dioxide. This is given by

n=\frac{m}{M_m}

where

m = 1.00 g is the mass of the gas

Mm = 44.0 g/mol is the molar mass of the gas

Substituting,

n=\frac{1.00 g}{44.0 g/mol}=0.0227 mol

Now we can find the pressure of the gas by using the ideal gas law:

pV=nRT

where

p is the gas pressure

V = 1.00 L is the volume

n = 0.0227 mol is the number of moles

R = 0.082 L/(atm K mol) is the gas constant

T = 25.0 C + 273 = 298 K is the temperature of the gas

Solving the formula for p, we find

p=\frac{nRT}{V}=\frac{(0.0227 mol)(0.082 L/(atm K mol))(298 K)}{1.00 L}=0.55 atm

8 0
4 years ago
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