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Masteriza [31]
2 years ago
7

You have a balloon filled with hydrogen gas which keeps it at a constant pressure, regardless of its volume. The initial volume

of the gas is 736 mL at 15.0 C. The gas is heated until it’s volume is 2.28 L, what is it’s final temperature in degrees Celsius
Chemistry
1 answer:
Lelu [443]2 years ago
6 0

Answer:

619.2 °C

Explanation:

Given data:

Initial volume of gas = 736 mL

Initial temperature = 15°C

Final volume = 2.28 L

Final temperature = ?

Solution:

First of all we will convert the temperature into kelvin and mL into L.

Initial temperature = 15°C (15+273 = 288 k)

Initial volume of gas = 736 mL × 1 L/1000 mL = 0.736 L

The given problem will be solve through the Charles Law.

According to this law, The volume of given amount of a gas is directly proportional to its temperature at constant number of moles and pressure.

Mathematical expression:

V₁/T₁ = V₂/T₂

V₁ = Initial volume

T₁ = Initial temperature

V₂ = Final volume  

T₂ = Final temperature

Now we will put the values in formula.

V₁/T₁ = V₂/T₂

0.736 L / 288 k = 2.28 L / T₂

T₂  = 2.28 L× 288 k/0.736 L

T₂  =  656.64 K /0.736  

T₂  = 892.2 K

Kelvin to °C:

892.2 K - 273.15 = 619.2 °C

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Answer:

Explanation:

Given parameters:

Initial temperature T₁  = 25.2°C  = 25.2 + 273  = 298.2K

Initial pressure  = P₁  = 0.6atm

Final temperature = 72.4°C   = 72.4 + 273  = 345.4K

Unknown:

Final pressure = ?

Solution:

To solve this problem, we use an adaption of the combined gas law where the volume gas is fixed. This simplification results into:

                  \frac{P_{1} }{T_{1} }   = \frac{P_{2} }{T_{2} }

where P and T are temperatures, 1 and 2 are initial and final temperatures.

 Input the parameters and solve;

          \frac{0.6}{298.2}   = \frac{P_{2} }{345.4}  

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3 0
2 years ago
Lithium has an atomic mass of 6.941 amu. Lithium has two common isotopes. The one isotope has a mass of 6.015 amu and a relative
koban [17]

Answer:

The atomic mass of second isotope is 7.016

Explanation:

Given data:

Average Atomic mass of lithium = 6.941 amu

Atomic mass of first isotope = 6.015 amu

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Abundance of second isotope = ?

Atomic mass of other isotope = ?

Solution:

Total abundance = 100%

100 - 7.49 = 92.51%

percentage abundance of second isotope = 92.51%

Now we will calculate the mass if second isotope.

Average atomic mass of lithium = (abundance of 1st isotope × its atomic mass) +(abundance of 2nd isotope × its atomic mass)  / 100

6.941 = (6.015×7.49)+(x×92.51) /100

6.941 =  45.05235 + (x92.51) / 100

6.941×100 = 45.05235 + (x92.51)

694.1 - 45.05235   = (x92.51)

649.04765 = x 92.51

x = 485.583 /92.51

x = 7.016

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3 0
2 years ago
An atom whose valence electrons conform to the octet rule is:
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Answer: D. less likely to form any bond

Explanation:

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2 years ago
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6 0
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kykrilka [37]

the equation is p1 x v1 divided by T1 = p1 x v2 = T2 but since the pressure is kept constant you do not even need it so the equation would now be v1 divided by t1 = v2 divided by t2

2135 cm3 divided by 127 degrees celcius = x divided by 206

answer: 3460 cm3

7 0
2 years ago
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