There are naturally occuring elements.

If Marie and Calvin dissolve 50 grams of KBr in 100 grams of water at 90oC, the solution is

hram777 [196]

Answer:

The solution is 50 %wt

Explanation:

50% wt is a sort of concentration and means, that 50 g of solute (in this case, the potassium bromide) dissolved in 100 g of water.

It is the same to say, that there are 50g of KBr for every 100g of H₂O

8 0

3 years ago

a) Whatis the composition in mole fractions of a solution of benzene and toluene that has a vapor pressure of 35 torr at 20 °C?

iogann1982 [59]

Answer:

molar composition for liquid

xb= 0.24

xt=0.76

molar composition for vapor

yb=0.51

yt=0.49

Explanation:

For an ideal solution we can use the Raoult law.

Raoult law: in an ideal liquid solution, the vapor pressure for every component in the solution (partial pressure) is equal to the vapor pressure of every pure component multiple by its molar fraction.

For toluene and benzene would be:

$P_{B}=x_{B}*P_{B}^{o}$

$P_{T}=x_{T}*P_{T}^{o}$

Where:

$P_{B}$ is partial pressure for benzene in the liquid

$x_{B}$ is benzene molar fraction in the liquid

$P_{B}^{o}$ vapor pressure for pure benzene.

The total pressure in the solution is:

$P= P_{T}+ P_{B}$

And

$1=x_{B}+x_{T}$

Working on the equation for total pressure we have:

$P=x_{B}*P_{B}^{o} + x_{T}*P_{T}^{o}$

Since $x_{T}=1-x_{B}$

$P=x_{B}*P_{B}^{o} + (1-x_{B})*P_{T}^{o}$

We know P and both vapor pressures so we can clear $x_{B}$ from the equation.

$x_{B}=\frac{P- P_{T}^{o}}{ P_{B}^{o} - P_{T}^{o}}$

$x_{B}=\frac{35- 22}{75-22} = 0.24$

So

$x_{T}=1-0.24 = 0.76$

To get the mole fraction for the vapor we know that in the equilibrium:

$P_{B}=y_{B}*P$

$y_{T}=1-y_{B}$

So

$y_{B} =\frac{P_{B}}{P}=\frac{ x_{B}*P_{B}^{o}}{P}$

$y_{B}=\frac{0.24*75}{35}=0.51$

$y_{T}=1-0.51=0.49$

Something that we can see in these compositions is that the liquid is richer in the less volatile compound (toluene) and the vapor in the more volatile compound (benzene). If we take away this vapor from the solution, the solution is going to reach a new state of equilibrium, where more vapor will be produced. This vapor will have a higher molar fraction of the more volatile compound. If we do this a lot of times, we can get a vapor that is almost pure in the more volatile compound. This is principle used in the fractional distillation.

7 0

3 years ago

the quantity of antimony in an ore can be determined by an oxidation-reduction titration with an oxidizing agent. The ore is dis

Basile [38]

Answer:

BrO₃⁻(aq) + 3Sb³⁺(aq) + 6H⁺(aq) → Br⁻(aq) + 3Sb⁵⁺(aq) + 3H₂O(l)

Explanation:

At a redox equation, one substance is being oxidized (losing electrons), and the other is being reduced (gaining electrons). In the given reaction:

BrO₃⁻(aq) + Sb³⁺(aq) → Br⁻(aq) + Sb⁵⁺(aq)

When it's at an acidic solution, it must be ions H⁺ on the reactant, which will form water with the oxygen, so the complete reaction is:

BrO₃⁻(aq) + Sb³⁺(aq) + H⁺(aq) → Br⁻(aq) + Sb⁵⁺(aq) + H₂O(l)

As we can see, the antimony is being oxidized (go from +3 to +5), and the Bromo is being reduced. The oxidation number of brome in the reactant, knowing that the oxidation number of O is -2, is:

x + 3*(-2) = -1

x = +5

So, it's going from +5 to -1, and the half-reactions are:

BrO₃⁻(aq) + 6e⁻ → Br⁻(aq)

Sb³⁺(aq) → Sb⁵⁺(aq) + 2e⁻

The number of electrons must be the same, so the second equation must be multiplied by 3:

3Sb³⁺(aq) → 3Sb⁵⁺(aq) + 6e⁻

Thus, the equation will be:

BrO₃⁻(aq) + 3Sb³⁺(aq) + H⁺(aq) → Br⁻(aq) + 3Sb⁵⁺(aq) + H₂O(l)

Now, we verify the amount of the elements, which must be equal on both sides. So, we multiply H₂O by 3, and H⁺ by 6, and the balanced reaction will be:

BrO₃⁻(aq) + 3Sb³⁺(aq) + 6H⁺(aq) → Br⁻(aq) + 3Sb⁵⁺(aq) + 3H₂O(l)

6 0

3 years ago

What is the vapor pressure of a liquid at 305.03 K if its ∆Hvap = 28.9 kJ/mol and its normal boiling point is 341.88 K?

ASHA 777 [7]

Answer: The vapor pressure of the liquid is 0.293 atm

Explanation:

To calculate the vapor pressure of the liquid, we use the Clausius-Clayperon equation, which is:

$\ln(\frac{P_2}{P_1})=\frac{\Delta H_{vap}}{R}[\frac{1}{T_1}-\frac{1}{T_2}]$