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Bingel [31]
3 years ago
14

A light bulb contains 0.0560 g of argon gas with a pressure of 1.34 atm at 26.0 c. The light bulb is then turned on and, after a

period of time, the temperature rises to 179c. Calculate the final pressure the light bulb
Chemistry
1 answer:
jenyasd209 [6]3 years ago
7 0

Answer:

Final pressure of light bulb is 2.03 atm.

Explanation:

Given data:

Initial pressure = 1.34 atm

initial temperature = 26 °C

Finial temperature = 179 °C

Final pressure = ?

Formula:

P1/T1 = P2/T2

Solution:

First of all we will convert the °C into kelvin.

179 +273 = 452 K

26 +273 = 299 K

P2 = P1×T2/T1

P2 = 1.34 atm . 452 K / 299 K

P2 = 605.68 atm/ 299

P2 = 2.03 atm

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dolphi86 [110]

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4 0
3 years ago
Lithium has two stable isotopes with masses of 6.01512 amu and 7.01600 amu. The average molar mass of Li is 6.941 amu. What is t
Dvinal [7]

Answer :  The percent abundance of Li isotope-1 and Li isotope-2 is, 6.94 % and 93.1 % respectively.

Explanation :

Average atomic mass of an element is defined as the sum of masses of each isotope each multiplied by their natural fractional abundance.

Formula used to calculate average atomic mass follows:

\text{Average atomic mass }=\sum_{i=1}^n\text{(Atomic mass of an isotopes)}_i\times \text{(Fractional abundance})_i   .....(1)

Let the fractional abundance of Li isotope-1 be 'x' and the fractional abundance of Li isotope-2 will be '100-x'

For Li isotope-1 :

Mass of Li isotope-1 = 6.01512 amu

Fractional abundance of Li isotope-1 = x

For Li isotope-2 :

Mass of Li isotope-2 = 7.01600 amu

Fractional abundance of Li isotope-2 = 100-x

Average atomic mass of Li = 6.941 amu

Putting values in equation 1, we get:

6.941=[(6.01512\times x)+(7.01600\times (100-x))]

By solving the term 'x', we get:

x=694.048

Percent abundance of Li isotope-1 = \frac{694.048}{100}=6.94\%

Percent abundance of Li isotope-2 = 100 - x = 100-6.94 = 93.1 %

6 0
3 years ago
Consider the equation below. Upper C a upper C upper O subscript 3 (s) double-headed arrow upper C a upper O (s) plus upper C up
stiks02 [169]

Answer:K subscript e q equals StartFraction StartBracket upper C upper O subscript 2 EndBracket StartBracket upper C a upper O EndBracket over StartBracket upper C a upper C upper O subscript 3 EndBracket EndFraction

Explanation: the answer has it's root in Law of mass action which states that; the rate of a chemical reaction is directly proportional to the product of the concentrations of the reactants raised to their respective stoichiometric coefficients.

8 0
3 years ago
Read 2 more answers
Which 2 resonance forms destablize the carbocation intermediate if bezonitrile undergoes chlronation at the ortho or para positi
Anna11 [10]

The question is incomplete, the complete question is shown in the image attached

Answer:

A and B

Explanation:

The electrophilic substitution of arenes yields a cation intermediate. The positive charge of the cation is delocalized over the entire ring.

The -CN group directs incoming electrophiles to the ortho/para position. The resonance structures for the chlorination of benzonitrile are shown in the question.

Recall that -CN is an electron withdrawing group. The resonance forms that destablize the carbocation intermediate are those in which the -CN group is directly attached to the carbon atom bearing the positive charge as in structures A and B.

8 0
3 years ago
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