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quester [9]
4 years ago
5

Which method would increase the solubility of a gas?

Chemistry
2 answers:
Irina18 [472]4 years ago
4 0

Answer:

increasing the pressure

Explanation:

UNO [17]4 years ago
3 0
Temperature and Pressure One way to increase the solubility of a gas is to decrease the temperature of the liquid.  The solubility of a gas in a liquid is usually temperature dependent, although it depends on the particular combination of which gas and which liquid.  Usually the solubility of a gas goes down with increasing temperature (think of warm carbonated beverages going flat). 

<span>The other way to increase the solubility is to increase the pressure of the gas.  The higher the pressure of the gas above the liquid, the more will dissolve.  Again, think of a carbonated beverage:  when it is sealed it doesn't go flat because it is under pressure, but when open to air, it will go flat. </span>
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If 2 objects have the same mass of 100 grams and object A has a volume of 50cm 3 and object B has a volume of 200 cm 3, which of
kenny6666 [7]

Answer:

The object A will be having the greater density compared to object B.

Explanation:

It is known that density of any object is defined as the mass of any object occupying a given volume. So the ratio of mass and volume will help to determine the density of any object. Density=Mass/Volume

From the above equation, it can be seen that the density of any object is directly proportional to the mass of the object and inversely proportional to the volume occupied by the object.

So in the present context, the mass of objects A and B are same and it is 100 g. Thus, the density of object A and object B will be influenced by their volume. As it is given that the volume of object A is 50 cm3 and object B is 100 cm3, then depending upon the relationship of volume and density, the density of both the objects can be determined. As the object with higher volume will be having lesser density as volume is inversely proportional to density. Thus, in the given case the volume of object B is greater than object A and so the object A will be having greater density compared to object B.

4 0
3 years ago
(a) What is the basis of the approximation that avoids using the quadratic formula to find an equilibrium concentration?
rusak2 [61]

The approximation is valid because  is very small.

Calculation of  concentration:

Since

0.85 M        0    0

(0.85-x)M    x      x

Now the value of x should be

x = 0.0000229

So based on this, the above concentration should be determined.

Now you will solve using the quadratic formula instead of iterations, to show that the same value of x is obtained either way. using the quadratic equation to calculate [h3o+] in 0.00250 m hno2, what are the values of a, b, c and x , where a, b, and c are the coefficients in the quadratic equation ax2+bx+c=0, and x is [h3o+]? recall that ka=4.5×10−4 .

a: 1

b: 4.5x10⁻⁴

c: 1.125x10⁻⁶

[H₃O⁺] = 0.000859M

As HNO₂ is a weak acid, its equilibrium in water is:

HNO₂(aq) + H₂O(l) ⇄ H₃O⁺(aq) + NO₂⁻(aq)

Equilibrium constant, ka, is defined as:

ka = 4.5x10⁻⁴ = [H₃O⁺] [NO₂⁻] / [HNO₂] (1)

Equilibrium concentration of each specie are:

[HNO₂] = 0.00250M - x

[H₃O⁺] = x

[NO₂⁻] = x

Replacing in (1):

4.5x10⁻⁴ = x × x / 0.00250M - x

1.125x10⁻⁶ - 4.5x10⁻⁴x = x²

0 = x² + 4.5x10⁻⁴x - 1.125x10⁻⁶

As the quadratic equation is ax² + bx + c = 0

Coefficients are:

a: 1

b: 4.5x10⁻⁴

c: 1.125x10⁻⁶

Now, solving quadratic equation:

x = -0.0013 → False answer, there is no negative concentrations.

x = 0.000859

As [H₃O⁺] = x; [H₃O⁺] = 0.000859M

To know more about Equilibrium constant

brainly.com/question/19340344

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7 0
2 years ago
What Kelvin temperature is the same as -13° Celsius?
zavuch27 [327]

Answer:

Degrees Kelvin = Degrees Celsius + 273.15

Degrees Kelvin = -13 + 273.15

= 260.15 °

Explanation:

8 0
3 years ago
If a cell wants to tell another cell what to do, it will send the message through
horsena [70]

Answer:

it will send the message through the axon

6 0
3 years ago
A solution has a hydrogen ion concentration of 0.001 M, what is the pH of the solution
LenKa [72]
PH= −log
10
​
[H
+
]
= −log
10
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(0.001)
= −log
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(10
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= −(−3)log
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3 years ago
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