As current flows through a system, what happens to the voltage available at each load?

What is the molecular geometry if you have 3 single bonds and 1 lone pair around the central atom?

Elodia [21]

Answer:

There are 2 double bond units and 1 lone pair, which will try to get as far apart as possible - taking up a trigonal planar arrangement. Because the lone pair isn't counted when you describe the shape, SO2 is described as bent or V-shaped.

Explanation:

There are 2 double bond units and 1 lone pair, which will try to get as far apart as possible - taking up a trigonal planar arrangement. Because the lone pair isn't counted when you describe the shape, SO2 is described as bent or V-shaped.

7 0

3 years ago

If an acid has a ka=1.6x10^-10, what is the acidity of the solution?

LUCKY_DIMON [66]

The information given in the question is not enough to determine the acidity of the solution. This is because, acidity can only be found with the equation: pH = -log [H+].
In order to determine the acidity of the solution, the half titration point value is needed, this will make it possible to determine the value of H30+. If the half point titration value is known, then Ka will be equivalent to pH and the value will be evaluated using the equation: - log (1.6 * 10^-10).

5 0

3 years ago

1. In the investigation of an unknown alcohol, there was a positive Jones test and a negative Lucas test. What deductions may be

pogonyaev

Answer:

1. When observing a positive test for the jones reagent and negative for the Lucas test, it indicates that it is in the presence of a primary alcohol.

Jones reagent behaves like a strong oxidant, where it transforms the primary alcohols into carboxylic acids and the secondary alcohols into ketones. Tertiary alcohols do not react.

With the Lucas test, tertiary alcohols react immediately producing turbidity, while secondary alcohols do so in five minutes. Primary alcohols do not react significantly with Lucas reagent at room temperature.

2. No reaction (See the attached drawing)

3. (see the attached drawing)

7 0

3 years ago

A container of gas is initially at 0.25 atm and 0 ˚C. What will the pressure be at 125 ˚C?

Pani-rosa [81]

Answer:

0.37atm

Explanation:

Given parameters:

Initial pressure = 0.25atm

Initial temperature = 0°C = 273K

Final temperature = 125°C = 125 + 273 = 398K

Unknown:

Final pressure = ?

Solution:

To solve this problem, we use a derivative of the combined gas law;

$\frac{P1}{T1}$ = $\frac{P2}{T2}$

P and T are pressure and temperature

1 and 2 are initial and final values

$\frac{0.25}{273}$ = $\frac{P2}{398}$

P2 = 0.37atm

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

3 years ago

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