Name the reagent if any that you would add to a solution of cobalt iii chloride

Four acids are analyzed and there KA values are determined which of the following values represents the strongest acid

Advocard [28]

x= the coefficients in front of the substance in the balanced chemical equation

[H+]= the concentration of hydrogen ions

[A-]= the concentration of the other ion that broke off from the H+

[HA]= the un-disassociated acid concentration

The higher the Ka value, the greater amount of disassociation of the reactants into products. As for acids, they will break down to form H+ ions. The more the H+ ions, the stronger acidity of the solution. Thus since A has the highest Ka value, that represents the strongest acid.

You can determine the Ka value from a number of ways. If equilibrium concentrations are given of a certain acid solution, you can find the proportion of the concentration of ions to the concentration of the remaining HA molecules, using the equation above. Also, pH and KpH can be used in a number of ways. This gets more complicated and depends on the situation, and requires more advanced equations.

Hope this helped a little, its obviously not my best work

6 0

3 years ago

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The nuclear fission process releases neutrons and

yuradex [85]

the answer is c. energy

3 0

3 years ago

HURRY PLEASE

katrin2010 [14]

Answers:

Question 1:

The diagram for gallium will have flat, horizontal lines at 30 and 2204 °C.

Questoin 2:

The diagram for methane will have a diagonal line representing the liquid phase between -183°C and -162°C.

Question 3:

For gold, the boiling point corresponds to the y-value at 2856 °C of the top horizontal line

Question 4.

For nitrogen, the line at -210°C will be flat, which represents the change from a solid to a liquid

Explanations:

Question 1:

The diagram for gallium will have flat, horizontal lines at 30 and 2204 °C.

The table shows that the melting point of gallium is 30°C and its boiling point is 2204°C.

Melting point is the temperature at which the substace changes its state from solid to liquid. During that change, the temperature of the substance does not change, because the heat supplied is used to accomplish the phase change. So, the temperature is constant and that means that portion of the diagram is flat.

The same is valid during boiling: the temperature remains constant while the substance is passing from liquid to gas at the boiling point.

Questoin 2:

The diagram for methane will have a diagonal line representing the liquid phase between -183°C and -162°C.

Between the melting (-183°C) and boiling (-162°C) points of methane, its temperature will increase more or less linearly, which is represented with a diagonal (slant) line between those points. During this interval the heat is used to increase the temperature and no phase of change happens.

Question 3:

For gold, the boiling point corresponds to the y-value at 2856 °C of the top horizontal line

The table shows that the boiling point of gold is 2,856°C.

In a temperature-vs.-time diagram the temperature is represented on the vertical axis (y-value) and the time is represented on the horizontal axis.

Since, the temperature of the substance does not change during boiling, the line during the time that this change of phase is happening is flat. And since this temperatue is higher than the melting temperature, this is the top horizontal line in the diagram.

Question 4.

For nitrogen, the line at -210°C will be flat, which represents the change from a solid to a liquid

The table shows that the melting point of nitrogen is -210°C, that means that the temperature will remain constant at -210°C while the substance is absorbing heat to pass from solid to liquid.

In conclusion, you must remember that all the phase changes, melting (from solid to liquid), freezing (from liquid to solid), boilng (from liquid to gas), and condensing (from gas to liquid) happens at constant temperature, and so the temperature - vs. - time diagrams show flat lines (constant y-values) during those intervals of time.

4 0

2 years ago

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How many moles of oxygen are present in 33.6l of the gas at 1atm and 0c

andreev551 [17]

The answer is: 1.5 moles of oxygen are present.

V(O₂) = 33.6 L; volume of oxygen.

p(O₂) = 1.0 atm; pressure of oxygen.

T = 0°C; temperature.

Vm = 22.4 L/mol; molar volume at STP (Standard Temperature and Pressure).

At STP one mole of gas occupies 22.4 liters of volume.

n(O₂) = V(O₂) ÷ Vm.

n(O₂) = 33.6 L ÷ 22.4 L/mol.

n(O₂) = 1.50 mol; amount of oxygen.

5 0

3 years ago

The equations given in the problem introduction can be added together to give the following reaction: overall: N2O4→2NO2 However

Murljashka [212]

Answer:

Reverse the $2 NO_2 \longrightarrow 2 NO + O_2$ reaction

Explanation:

Reactions:

$2 NO_2 \longrightarrow 2 NO + O_2$

$N_2O_4 \longrightarrow 2 NO + O_2$

Overall:

$N_2O_4 \longrightarrow 2 NO_2$

As can be seen, in the overall reaction we have $N_2O_4$ in the reactants like in the second reaction and $NO_2$ in the products. The $NO_2$ is in the first reaction but as a reactant so we need to reverse that reaction:

$2 NO + O_2 \longrightarrow 2 NO_2$

$N_2O_4 \longrightarrow 2 NO + O_2$

Combining:

$N_2O_4 + 2 NO + O_2\longrightarrow 2 NO + O_2 + 2 NO_2$

$N_2O_4 \longrightarrow 2 NO_2$

4 0

3 years ago