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Mrac [35]
2 years ago
11

A sample of krypton has a volume of 6.00 L, and the pressure is 0.960 atm. If the final temperature is 55.0°C, the final volume

is 7.70 L, and the final pressure is 1.25 atm, what was the initial temperature of the krypton?
Chemistry
1 answer:
MissTica2 years ago
5 0

193.38 K was the initial temperature of the krypton.

Explanation:

Data given:

Initial volume of the krypton gas = 6 litres

initial pressure of the krypton gas = 0.960 atm

initial temperature of the krypton gas = ?

final volume of the krypton gas  = 7.70 litres

final pressure of the Krypton gas  = 1.25 atm

final temperature of the krypton gas = 55 degrees or 273.25+55 = 323.15 K

Applying the  Combined Gas Laws:

\frac{P1V1}{T1} = \frac{P2V2}{T2}

Rearranging the equation:

T1  = \frac{P1V1T2}{P2V2}

Putting the value in the equation:

T1 = \frac{0.960 X 6 X 323.15}{1.25 X 7.70}

T1 = 193.38 K

Initial temperature of the krypton gas is 193.78 K

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Answer:

Wavelength of this beam of light: \rm 4.39\times 10^{-7}\; m.

Explanation:

The speed of light in vacuum is approximately \rm 2.998\times 10^{8}\;m \cdot s^{-1}.

Light behaves like a wave. The wavelength of a wave is equal to the distance that it travels (in the given medium) in each period of oscillation.

On the other hand, the frequency of a wave is the number of periods in unit time. 1\rm \; Hz means one oscillation per second. The frequency of this particular wave is \rm 6.83\times 10^{14}\; Hz. In other words, there are 6.83\times 10^{14} oscillations in each second.

The period of oscillation will be equal to

\displaystyle t = \frac{1}{f} = \frac{1}{\rm 6.83\times 10^{14}\; s^{-1}}.

In that period of time, a beam of light in vacuum would have traveled  

\displaystyle \rm 2.998\times 10^{8}\; m\cdot s^{-1} \times \frac{1}{\rm 6.83\times 10^{14}\; s^{-1}} = 4.39\times 10^{-7}\; m.

In other words, if this beam of light of frequency \rm 6.83\times 10^{14}\; Hz is in vacuum, its wavelength will be equal to \rm 4.39\times 10^{-7}\; m.

8 0
2 years ago
What is the limiting reactant for the reaction below given that you start with 10.0 grams of Al (molar mass 26.98 g mol-1) and 1
mezya [45]

Answer:

Al

Explanation:

4 Al  +  3 O₂  →  2 Al₂O₃

You need to figure out which one has the smaller mole ratio.  Convert both substances from grams to moles.

(10.0 g Al)/(26.98 g/mol) = 0.3706 mol Al

(19.0 g O₂)/(32.00 g/mol) = 0.5938 mol O₂

Now, use the mole ratios of reactant to product to see which substance produces the least amount of product.

(0.3706 mol Al) × (2 mol Al₂O₃/4 mol Al) = 0.1853 mol Al₂O₃

(0.5938 mol O₂) × (2 mol Al₂O₃/3 mol O₂) = 0.3958 mol Al₂O₃

Since aluminum produces the least amount of product, this is the limiting reagent.

4 0
2 years ago
How many lead (Pb) atoms will be generated when 5.38 moles of ammonia react according to the following equation: 3PbO+2NH3→3Pb+N
jasenka [17]

Answer:

4.86×10^23 molecule of Pb

Explanation:

Based on that equation, for every 2 moles of ammonia, you get 3 moles of lead.

So:

2 mol NH3/ 3 mol Pb

Using this ratio we can find the amounts of either molecule. Given 5.38 mol NH3:

(5.38 NH3)(3 Pb/ 2 NH3) = (5.38)(3/2) mol Pb = 8.07 mol Pb

Then, we just need to use Avagadro's number to get the number of molecules.

(8.07)(6.02×10^23) = 4.86×10^23 molecule of Pb

4 0
2 years ago
Read 2 more answers
Describe two shortcomings of the pennium model for isotopes
umka21 [38]

let me know when u find out plz because i would like to know as well its one of my chemistry qustions in an assiment. :)

3 0
2 years ago
What is the symbol for the isotope of 58 co that possesses 33 neutrons?
Leona [35]
Element with an atomic number of 58 is actually Cerium, so the symbol should be Ce, not Co because that is Cobalt which has an atomic number of 27. With that being said, the notation for isotopes is the symbol of the element with a superscript and a subscript that are aligned. The superscript represents the mass number.

Mass number = protons + neutrons = 58 + 33 = 91

The subscript is the atomic number which is 58. The notation is written in the picture attached.

6 0
2 years ago
Read 2 more answers
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