A larger glass has 100 mL of a liquid and a smaller glass has 50 mL of a liquid. Which glass of liquid would have more energy?

Identify the intermolecular attractions for dimethyl ether and for ethyl alcohol. Which molecule is expected to be more soluble

zheka24 [161]

Answer:

See explanation

Explanation:

All molecules possess the London dispersion forces. However London dispersion forces is the only kind of intermolecular interaction that exists in nonpolar substances.

So, the only kind of intermolecular interaction that exists in dimethyl ether is London dispersion forces.

As for ethyl alcohol, the molecule is polar due to the presence of polar O-H bond. In addition to London dispersion forces, dipole-dipole interactions and specifically hydrogen bonding also occurs between the molecules.

Because ethyl alcohol is polar, it is more soluble in water than dimethyl ether.

3 0

3 years ago

You mix 200. mL of 0.400M HCl with 200. mL of 0.400M NaOH in a coffee cup calorimeter. The temperature of the solution goes from

GuDViN [60]

Answer : The enthalpy of neutralization is, 56.012 kJ/mole

Explanation :

First we have to calculate the moles of HCl and NaOH.

$\text{Moles of HCl}=\text{Concentration of HCl}\times \text{Volume of solution}=0.400mole/L\times 0.200L=0.08mol$

$\text{Moles of NaOH}=\text{Concentration of NaOH}\times \text{Volume of solution}=0.400mole/L\times 0.200L=0.08mol$

The balanced chemical reaction will be,

$HCl+NaOH\rightarrow NaCl+H_2O$

From the balanced reaction we conclude that,

As, 1 mole of HCl neutralizes by 1 mole of NaOH

So, 0.08 mole of HCl neutralizes by 0.08 mole of NaOH

Thus, the number of neutralized moles = 0.08 mole

Now we have to calculate the mass of water.

As we know that the density of water is 1 g/ml. So, the mass of water will be:

The volume of water = $200mL+200L=400mL$

$\text{Mass of water}=\text{Density of water}\times \text{Volume of water}=1g/ml\times 400mL=400g$

Now we have to calculate the heat absorbed during the reaction.

$q=m\times c\times (T_{final}-T_{initial})$

where,

q = heat absorbed = ?

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

m = mass of water = 400 g

$T_{final}$ = final temperature of water = $27.78^oC=273+25.10=300.78K$

$T_{initial}$ = initial temperature of metal = $25.10^oC=273+27.78=298.1K$

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

$q=400g\times 4.18J/g^oC\times (300.78-298.1)K$

$q=4480.96J$

Thus, the heat released during the neutralization = -4480.96 J

Now we have to calculate the enthalpy of neutralization.

$\Delta H=\frac{q}{n}$

where,

$\Delta H$ = enthalpy of neutralization = ?

q = heat released = -4480.96 J

n = number of moles used in neutralization = 0.08 mole

$\Delta H=\frac{-4480.96J}{0.08mole}=-56012J/mole=-56.012kJ/mol$

The negative sign indicate the heat released during the reaction.

Therefore, the enthalpy of neutralization is, 56.012 kJ/mole

8 0

2 years ago

2 uniones iónicas y 2 covalentes, realizando las estructuras de Lewis correspondiente y averiguar la función del compuesto armad

posledela

Answer:

Cloruro de sodio y fluoruro de sodio.

Dióxido de carbono y monóxido de hidrógeno.

Explicación:

El cloruro de sodio y el fluoruro de sodio son los compuestos que tienen enlaces iónicos. Estos compuestos iónicos se utilizan para diferentes actividades de nuestra vida diaria. El cloruro de sodio se usa para cocinar y el fluoruro de sodio se usa en la pasta de dientes para limpiar nuestros dientes. El dióxido de carbono y el monóxido de hidrógeno son compuestos que tienen enlaces covalentes. El dióxido de carbono se usa en refrescos / refrescos y algunos otros líquidos que se pueden usar en la vida diaria. El monóxido de hidrógeno es el agua pura que bebemos todos los días en nuestra vida diaria y es muy importante para nuestra supervivencia.

7 0

3 years ago

Combustion vapor-air mixtures are flammable over a limited range of concentrations. The minimum volume % of vapor that gives a c

DochEvi [55]

Combustion equation of n-hexane:

2C₆H₁₄ + 19O₂ → 12CO₂ + 14H₂O

a)
Assuming we have 100 moles of air,
Oxygen = 20.9 moles
n-hexane required = 20.9/19 x 2
= 2.2 moles

LFL = Half of stoichometric amount = 2.2 / 2 = 1.1
LFL n-hexane = 1.1%

b)
1.1 volume percent required for LFL

1.1% x 1
= 0.0011 m³ of n-hexane required

6 0

3 years ago

Read 2 more answers

35.0 grams of nitrogen gas reacts with 60.0 grams of hydrogen gas: N2 + 3H2--> 2NH3

Misha Larkins [42]

Explanation:

Moles of N2 = 35.0g / (28g/mol) = 1.25mol

Moles of H2 = 60.0g / (2g/mol) = 30.0mol

Since 1.25mol * 3 < 30.0mol, nitrogen is limiting.

Moles of NH3 = 1.25mol * 2 = 2.50mol.

Mass of NH3 = 2.50mol * (17g/mol) = 42.5g.

30.0mol - 1.25mol * 3 = 26.25mol.

Excess mass of H2

= 26.25mol * (2g/mol) = 52.5g.

6 0

3 years ago