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Savatey [412]
3 years ago
7

Are the following bonds (A) ionic bonds or (B) covalent bonds?

Chemistry
1 answer:
qaws [65]3 years ago
4 0
Ionic  bond  are  formed  between  a   metal  and  a  non  metal  while  covalent  bonds  are  formed  between  two  non  metals.
        ionic  compounds  include
    Na-cl,  Mg-Br,  Ca-s,  Ba-F  (are  ionic  since their  bond  is  between a  metal  and  non metal)

covalent  bonds  include
 O-H,  N-O,  C-H,  F-F,  N-H ,  C-O,  F-F ( are  covalent  since  their  bond  is  between  two  non  metal
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katovenus [111]
<h3>1.</h3>

C) The volume of the gas is proportional to the number of moles of gas particles.

The Avogadro's law applies to ideal gases with constant pressure and temperature. By that law, the volume of an ideal gas is proportional to the number of moles of particles in that gas.

<h3>2.</h3>

B) The gas now occupies less volume, and the piston will move downward.

Boyle's Law applies to ideal gases with a constant temperature. The volume of an ideal gas is inversely related to its pressure. A high pressure drives gas particles together, such that they occupy less volume. The gas trapped inside the piston has a smaller volume. As a result, the the piston will move downward.

Alternatively, consider the forces acting on the piston. Both the atmosphere and gravity are dragging the piston down. In order for it to stay in place, the gas below it must exert a pressure to balance the two forces. Now the pressure from outside has increased. The gas inside needs to increase its pressure. It needs a smaller volume to create that extra pressure. As a result, its volume will decrease, and the piston will move downwards.

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Calculate the maximum solubility of silver carbonate, Ag2CO3 in g/L when in the presence of 0.057 M AgNO3. The solubility produc
Andreyy89

Answer:

Approximately 4.2 \times 10^{-7}\; \rm g \cdot L^{-1}.

Explanation:

Start by finding the concentration of \rm Ag_2CO_3 at equilibrium. The solubility equilibrium for

\rm Ag_2CO_3 \; (s) \rightleftharpoons 2\, Ag^{+}\; (aq) + {CO_3}^{2-}\; (aq).

The ratio between the coefficient of \rm Ag_2CO_3 and that of \rm Ag^{+} is 1:2. For

Let the increase in \rm {CO_3}^{2-} concentration be +x\; \rm mol \cdot L^{-1}. The increase in \rm Ag^{+} concentration would be +2\,x\; \rm mol \cdot L^{-1}. Note, that because of the 0.057\; \rm mol \cdot L^{-1}of \rm AgNO_3, the concentration of

  • The concentration of \rm Ag^{+} would be (0.057 + 2\, x) \; \rm mol\cdot L^{-1}.
  • The concentration of \rm {CO_3}^{2-} would be x\; \rm mol \cdot L^{-1}.

Apply the solubility product expression (again, note that in the equilibrium, the coefficient of \rm Ag^{+} is two) to obtain:

\begin{aligned}&\rm \left[Ag^{+}\right]^2 \cdot \left[{CO_3}^{2-}\right] = K_{\text{sp}} \\ & \implies (0.057 + x)^2\cdot x = 8.1 \times 10^{-12} \end{aligned}.

Note, that the solubility product of \rm Ag_2CO_3, K_{\text{sp}} = 8.1 \times 10^{-12} is considerably small. Therefore, at equilibrium, the concentration of

Apply this approximation to simplify (0.057 + x)^2\cdot x = 8.1 \times 10^{-12}:

0.057^2\, x \approx (0.057 + x)^2 \cdot x = 8.1 \times 10^{-12}.

\begin{aligned} x &\approx \frac{8.1 \times 10^{-12}}{0.057^2}\end{aligned}.

Calculate solubility (in grams per liter solution) from the concentration. The concentration of \rm Ag_2CO_3 is approximately \displaystyle \frac{8.1 \times 10^{-12}}{0.057^2}\; \rm mol\cdot L^{-1}, meaning that there are approximately \displaystyle n = \frac{8.1 \times 10^{-12}}{0.057^2}\; \rm mol of

\begin{aligned}m &= n \cdot M \\ &\approx \displaystyle \frac{8.1 \times 10^{-12}}{0.057^2} \; \rm mol\times 167.91\; g \cdot mol^{-1} \\ &\approx 4.2 \times 10^{-7}\; \rm g \end{aligned}.

As a result, the maximum solubility of \rm Ag_2CO_3 in this solution would be approximately 4.2 \times 10^{-7}\; \rm g \cdot L^{-1}.

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3 years ago
Describe the relationship between the mole and molar mass. Be prepared to describe the process for converting between moles and
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Answer:

The molar mass of any substance is the mass in grams of one mole of representative particles of that substance. ... In such a conversion, we use the molar mass of a substance as a conversion factor to convert mole units into mass units (or, conversely, mass units into mole units).

Explanation:

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it will sink

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