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mamaluj [8]
3 years ago
11

A student is working with a 4.95 L sample of carbon dioxide at 22.8°C and 0.956 atm. The student compresses the gas to 1.75 L an

d increases the temperature of the gas to 65.3°C. What will the resulting pressure of the CO 2 gas be after the student made these changes?
Chemistry
1 answer:
myrzilka [38]3 years ago
5 0

 The resulting  pressure   is 3.093 atm


   <u><em>calculation</em></u>

The  resulting pressure is calculated  using the combined gas equation

That is   P₁V₁/T₁ =P₂V₂/T₂   where,

P₁=0.956 atm

V₁=4.95 L

T₁=22.8 c  in to kelvin= 22.8 +273 =295.8 k

P₂=? atm

V₂=1.75 L

T₂=65.3 c  into kelvin = 65.3 +273  =338.3 K

make P₂ the subject  of the formula by multiplying  both side of the formula  by T₂/V₂

P₂ =T₂P₁V₁ / V₂T₁

P₂  = [(338.3 k x 0.956 atm  x 4.95 L) / ( 1.75 L x 295.8 K)] =3.093 atm


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Heterogeneous mixtures are made of different substances that remain physically seperate. An example would be mixing sand and sugar together.
the answer would be soil
The answer is B soil
6 0
3 years ago
The length breadth and thickness of a brick is 18cm 8 cm and 5cm respictively find the area of the widest part of rhe brick​
Sunny_sXe [5.5K]

Answer:

144cm²

Explanation:

Given dimensions:

     Length x breadth x thickness

        18cm  x   8cm    x    5cm

The widest part of this figure will be the face containing the length and the breadth.

The breadth is the width of the figure;

Area of the widest part  = length x breadth = 18cm x 8cm  = 144cm

The area of the widest part of the figure is 144cm²

4 0
3 years ago
During photosynthesis, plant cells use sunlight as an energy source; animal cells cannot do this. What substance do Animal cells
8090 [49]

Answer:

Chlorophyll

Explanation:

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6 0
3 years ago
The pH of a 0.65M solution of hydrofluoric acid HF is measured to be 1.68. Calculate the acid dissociation constant Ka of hydrof
sergey [27]

Answer:

Kₐ = 6.7 x 10⁻⁴

Explanation:

First lets write the equilibrium expression, Ka , for the dissociation of hydrofluoric acid:

HF + H₂O     ⇄   H₃O⁺ +   F⁻

Kₐ = [ H₃O⁺ ] [ F⁻ ] /[ [ HF ]

Since we are given the pH we can calculate the  [ H₃O⁺ ]  ( pH = - log [ H₃O⁺ ] , and because the acid dissociates into a 1: 1  relation , we will also have [F⁻ ]. The  [ HF ] is given in the question so we have all the information that is needed to  compute Kₐ.

pH = -log [ H₃O⁺ ]

1.68 = - log [ H₃O⁺ ]

Taking antilog to both sides of this equation:

10^-1.68 = [ H₃O⁺ ] ⇒ 2.1 X 10⁻²  M= [ H₃O⁺ ]

[ F⁻ ] = 2.1 X 10⁻² M

Solving for Kₐ :

Kₐ = ( 2.1 X 10⁻² ) x  ( 2.1 X 10⁻² ) / 0.65 = 6.7 x 10⁻⁴  

(Rounded to two significant figures, the powers of 10 have infinite precision )

4 0
3 years ago
A 2.75 kg sample of a substance occupies a volume of 250.0 cm3. Find its density in g/cm3.
satela [25.4K]
Density is 0.011 g/cm3.
5 0
3 years ago
Read 2 more answers
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