How many moles of oxygen are required to react completely with 5.00 grams of hydrogen?

What is the boiling point elevation constant, Kb, of diethyl ether if 38.2 g of the nonelectrolyte benzophenone, C6H5COC6H5, dis

kakasveta [241]

Answer: the boiling point elevation constant is $1.73^0C/m$

Explanation:

Elevation in boiling point is given by:

$\Delta T_b=i\times K_b\times m$

$\Delta T_b=T_b-T_b^0=$ = Elevation in boling point

i= vant hoff factor = 1 (for non electrolyte)

$K_b$ =boiling point constant = ?

m= molality

$\Delta T_b=i\times K_b\times \frac{\text{mass of solute}}{\text{molar mass of solute}\times \text{weight of solvent in kg}}$

Weight of solvent (diethylether)= 330 g = 0.33 kg

Molar mass of solute (benzophenone)= 182 g/mol

Mass of solute (benzophenone) = 38.2 g

$(35.7-34.6)^0C=1\times K_b\times \frac{38.2g}{182g/mol\times 0.33kg}$

$K_b=1.73^0C/m$

Thus the boiling point elevation constant is $1.73^0C/m$

3 0

3 years ago

Which of the following represents the least number of molecues?

Mars2501 [29]

Answer:

Answer: a) 20g of H2O (18.02 g/mol) molecules=6.68x10^23

Explanation:

In order to find the amount of molecules of each of the options, we need to follow the following equation.

$molecules=\frac{mass(g)x6.022x10^{23}(molecules/mol) }{atomic weight(g/mol)}$

So, let´s get the number of molecules for each of the options.

$a) molecules=\frac{20(g)x6.022x10^{23}(molecules/mol) }{18.02(g/mol)}=6.68x10^{23}molecules$

$b) molecules=\frac{77(g)x6.022x10^{23}(molecules/mol) }{16.06(g/mol)}=2.89x10^{24}molecules$

$c) molecules=\frac{68(g)x6.022x10^{23}(molecules/mol) }{42.09(g/mol)}=9.73x10^{23}molecules$

$d) molecules=\frac{100(g)x6.022x10^{23}(molecules/mol) }{44.02(g/mol)}=1.37x10^{24}molecules$

$d) molecules=\frac{84(g)x6.022x10^{23}(molecules/mol) }{20.01(g/mol)}=2.53x10^{24}molecules$

the smalest number is in option a)

Best of luck.

7 0

3 years ago

The heat of vaporization of a liquid is 84.0 J/g. How many joules of heat would it take to completely vaporize 172 g of this liq

Aliun [14]

Answer:

14,448 J of heat would it take to completely vaporize 172 g of this liquid at its boiling point.

Explanation:

The heat Q that is necessary to provide for a mass m of a certain substance to change phase is equal to Q = m*L, where L is called the latent heat of the substance and depends on the type of phase change.

During the evaporation process, a substance goes from a liquid to a gaseous state and needs to absorb a certain amount of heat from its immediate surroundings, which results in its cooling. The heat absorbed is called the heat of vaporization.

So, it is called "heat of vaporization", the energy required to change 1 gram of substance from a liquid state to a gaseous state at the boiling point.

In this case, being: