Elements and compounds are usually more dense in their solid than in their

A chemical reaction was used to produce 2.95 moles of copper(II) bicarbonate, Cu(HCO3)2.

BARSIC [14]

Answer:

About 547 grams.

Explanation:

We want to determine the mass of copper (II) bicarbonate produced when a reaction produces 2.95 moles of copper (II) bicarbonate.

To do so, we can use the initial value and convert it to grams using the molar mass.

Find the molar mass of copper (II) bicarbonate by summing the molar mass of each individual atom:

$\displaystyle \begin{aligned} \text{MM}_\text{Cu(HCO$_3$)$_2$} &= (63.55 + 2(1.01)+2(12.01)+6(16.00))\text{ g/mol} \\ \\ &=185.59\text{ g/mol} \end{aligned}$

Dimensional Analysis:

$\displaystyle 2.95\text{ mol Cu(HCO$_3$)$_2$}\cdot \frac{185.59 \text{ g Cu(HCO$_3$)$_2$}}{1 \text{ mol Cu(HCO$_3$)$_2$}} \Rightarrow 547 \text{ g Cu(HCO$_3$)$_2$ }$

In conclusion, about 547 grams of copper (II) bicarbonate is produced.

8 0

3 years ago

Describe what happens to an ionic compound when it has been fully dissolved in water

Marat540 [252]

When you immerse an ionic compound in water, the ions are attracted to the water molecules, each of which carries a polar charge. If the attraction between the ions and the water molecules is great enough to break the bonds holding the ions together, the compound dissolves.

6 0

3 years ago

PLEASE HELP IM SO CONFUSED How does potassium need to be modified either on site or in a factory to make it useful. This should

3241004551 [841]

Potassium is not found free in nature but is found in the form of potash. Potash is the ore of potassium and this ore is mined from deep down the earth or can sometimes be found on the surface. Potash was mostly formed as sea water receded and left deposits.

Potash is usually in the form of potassium salts such potassium chloride and potassium sulphate. The potash is mined then taken to the factory where it is crushed and purified by removing such impurities as clay.

The now purified potassium salts are subjected to a process called electrolysis where potassium metal is obtained from its salt.

5 0

3 years ago

Calculate the mass of beryllium (Be) needed to completely react with 18.9 g nitrogen gas (N2) to produce Bez N2, which is the on

choli [55]

Answer:

C = 18.29 g

Explanation:

Given data:

Mass of beryllium needed = ?

Mass of nitrogen = 18.9 g

Solution:

Chemical equation:

3Be + N₂ → Be₃N₂

now we will calculate the number of moles of nitrogen:

Number of moles = mass/molar mass

Number of moles = 18.9 g/ 28 g/mol

Number of moles = 0.675 mol

Now we will compare the moles of nitrogen and Be from balance chemical equation.

N₂ : Be

1 : 3

0.675 : 3/1×0.675 = 2.03 mol

Number of moles of Be needed are 2.03 mol.

Mass of Beryllium:

Mass = number of moles × molar mass

Mass = 2.03 mol × 9.01 g/mol

Mass = 18.29 g

4 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