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4 years ago

Perform each of the following unit conversions using the conversion factors given below: 1 atm = 760 mmHg = 101.325 kPa

2 answers:

5 0

unit coversation
1.429 atm - 1086mmhg

9361 pa-9.36 KPa - 70.21 mmhg

725 mmhg -0.95 atm- 96.26 kpa

calculation

(a) 1 atm = 760 mmhg
1.429 atm = ?
1.429 x760/1 = 1086.34 mm hg

(B) 1 mmhg = 101.325 kpa
? =9361 KPa
9361 x1 /101.25 =70.21 mmhg

760 mm hg= 101.325 KPa
70.21 mm hg=?

70.21 x101.325/760 = 9.36 Kpa

(C ) 1 atm = 760 mmhg
? = 725
= 725 x1/ 760=0.95 atm

1 atm = 101.325 kpa
0.95 =?

0.95 x101.325/1 = 96.26 KPa

Leto [7]4 years ago

4 0

1atm= 760mmhg= 101.325kPa

1.429atm= 1,086 mmhg

9,361 Pa= 9.361 kPa= 70.21 mmhg

725 mmhg= 954 atm= 96.7 kPa

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Consider a transition of the electron in the hydrogen atom from n=3 to n=7.

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Answer:

For a: The wavelength of light is $1.005\times 10^{-6}m$

For b: The light is getting absorbed

Explanation:

For a:

To calculate the wavelength of light, we use Rydberg's Equation:

$\frac{1}{\lambda}=R_H\left(\frac{1}{n_i^2}-\frac{1}{n_f^2} \right )$

Where,

$\lambda$ = Wavelength of radiation

$R_H$ = Rydberg's Constant = $1.097\times 10^7m^{-1}$

$n_f$ = Higher energy level = 7

$n_i$ = Lower energy level = 3

Putting the values in above equation, we get:

$\frac{1}{\lambda }=1.097\times 10^7m^{-1}\left(\frac{1}{3^2}-\frac{1}{7^2} \right )\\\\\lambda =1.005\times 10^{-6}m$

Hence, the wavelength of light is $1.005\times 10^{-6}m$

For b:

There are two ways in which electrons can transition between energy levels:

Absorption spectra: This type of spectra is seen when an electron jumps from lower energy level to higher energy level. In this process, energy is absorbed.
Emission spectra: This type of spectra is seen when an electron jumps from higher energy level to lower energy level. In this process, energy is released in the form of photons.

As, the electron jumps from lower energy level to higher energy level. The wavelength is getting absorbed.

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3 years ago

Draw a condensed structure of each of the following

MrRa [10]

The condensed structure of given compounds are shown in the below image.

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In 1,2-propanediol compound hydroxyl functional group is present on the 1st and 2nd position.
In Ethyl methyl ether, ether functional group is present.
In Dichloromethane, two atoms of chlorine groups are present.

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2 years ago

What is the word equation of photodecomposition?

Oksi-84 [34.3K]

Answer:

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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.

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3 years ago

Read 2 more answers

aseous methane will react with gaseous oxygen to produce gaseous carbon dioxide and gaseous water . Suppose 5.5 g of methane is

maksim [4K]

Answer:

There is 9.6 grams of CO2 produced

Explanation:

Step 1: Data given

Mass of methane = 5.50 grams

Molar mass of methane = 16.04 g/mol

Mass of oxygen = 13.9 grams

Molar mass of oxygen = 32.0 g/mol

Step 2: The reaction

CH4(g) + 2O2(g) → CO2(g) + 2H2O(g)

Step 3: Calculate number of moles

Moles = mass / molar mass

Moles methane = 5.50 grams / 16.04 g/mol

Moles methane = 0.343 moles

Moles oxygen = 13.9 grams / 32.0 g/mol

Moles oxygen = 0.434 moles

For 1 mol CH4 we need 2 moles O2 to produce 1 mol CO2 and 2 moles H2O

O2 is the limiting reactant. It will completely react (0.434 moles).

There will react 0.434/2 = 0.217 moles CH4

There will remain 0.343-0.217 = 0.126 moles CH4

There will be produced 0.434 moles of H2O and

0.434/2 =0.217 moles of CO2

Step 4: Calculate mass of products

Mass = moles * molar mass

Mass CO2 = 0.217 moles ¨44.01 g/mol

Mass CO2 = 9.6 grams

Mass H2O = 0.434 moles * 18.02

Mass H2O = 7.8 grams

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