The density of a 41.2% m/m solution of ammonia in water is 0.708 g/mL. Calculate the concentration of the solution in mol/L

Explain the digestive process in an essay.

fiasKO [112]

Refer to attachment for your answer

8 0

2 years ago

Which example is the site of reduction when the dry cell is operating

Simora [160]

Answer:

A battery contains electrochemical cells that can store chemical energy to be converted to electrical energy. A dry-cell battery stores energy in an immobilized electrolyte paste, which minimizes the need for water. Common examples of dry-cell batteries include zinc-carbon batteries and alkaline batteries.

Explanation: i hope this helps sorry if it didnt

4 0

3 years ago

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Approximately 1 mL of two clear, colorless solutions-0.1 M Ba(NO3)2 and 0.1 M Na2SO4- were combined. Upon mixing, a thick, milky

murzikaleks [220]

Answer: balanced chemical equation: $Ba(NO_3)_2(aq)+Na_2SO_4(aq)\rightarrow BaSO_4(s)+2NaNO_3(aq)$

Net ionic equation : $Ba^{2+}(aq)+SO_4^{2-}(aq)\rightarrow BaSO_4(s)$

Explanation:

A double displacement reaction is one in which exchange of ions take place. The salts which are soluble in water are designated by symbol (aq) and those which are insoluble in water and remain in solid form are represented by (s) after their chemical formulas.

The balanced chemical equation is:

$Ba(NO_3)_2(aq)+Na_2SO_4(aq)\rightarrow BaSO_4(s)+2NaNO_3(aq)$

Spectator ions are defined as the ions which does not get involved in a chemical equation or they are ions which are found on both the sides of the chemical reaction present in ionic form.

$Ba^{2+}(aq)+2NO_3^-(aq)+2Na^+(aq)+SO_4^{2-}(aq)\rightarrow BaSO_4(s)+2Na^+(aq)+2NO_3^-(aq)$

The ions which are present on both the sides of the equation are sodium and nitrate ions and hence are not involved in net ionic equation.

Hence, the net ionic equation is $Ba^{2+}(aq)+SO_4^{2-}(aq)\rightarrow BaSO_4(s)$

4 0

3 years ago

What physical effect can change the boiling point of a substance?

Artemon [7]

Pressure can affect the boiling pressure of a substance

as when pressure increases the particles are closer together and so require more energy to boil therefore increasing the substances boiling point

hope that helps

5 0

2 years ago

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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