Any buffer exists in this equilibrium
HA <=>
In a buffer, there is a large reservoir of both the undissociated acid (HA) and its conjugate base (
)
When a strong acid is added, it reacts with the large reservoir of the conjugate base (
) forming a salt and water. Since this large reservoir of the conjugate base is used, the ph does not alter drastically, but instead resist the pH change.
Answer:
Mass
Step-by-step explanation:
Usually, you plot the independent variable along the horizontal (x) axis and the dependent variable along the vertical (y) axis.
Marcia's teacher plotted the mass of the sample along the x-axis and volume along the y-axis.
The mass is the independent variable, because that is <em>what the teacher varied</em>.
The volume is the <em>dependent variable</em>, because it <em>depends</em> on the mass.
Sample number is <em>wrong</em>, because it is not a variable.
Substance is <em>wrong</em>, because all samples consist of the same substance.
Density is <em>wrong</em>, because it is constant. It is the slope of the graph.
The change in temperature had the greatest effect at changing the volume of the balloon.
<h3>What are the gas laws?</h3>
The gas laws are used to describe the parameters that has to do with gases.
Given that;
P1 = 98.5 kPa
T1 = 18oC or 291 K
V1 = 74.0 dm3
P2 = 7.0 kPa
V2 = ?
T2 = 18oC or 291 K
P1V1/T1 = P2V2/T2
P1V1T2 =P2V2T1
V2= P1V1T2/P2T1
V2 = 98.5 kPa * 74.0 dm3 * 291 K/ 7.0 kPa * 291 K
V2 = 1041.3 dm3
When;
V1 = 1041.3 dm3
T1 = 291 K
V2 = ?
T2 = 80oC or 353 K
V1/T1 = V2/T2
V1T2 = V2T1
V2 = V1T2/T1
V2 = 1041.3 dm3 * 353 K/291 K
V2 = 1263 dm3
The change in temperature had the greatest effect at changing the volume of the balloon.
Given that
V1 = 100 cm^3
T1 = 273 K
P1 = 1.01 * 10^5 Pa
V2 = ?
P2 = 3.00 x 10^-4 Pa
T2 = -180oC or 255 K
V2= P1V1T2/P2T1
V2 = 1.01 * 10^5 Pa * 100 cm^3 * 255 K / 3.00 x 10^-4 Pa * 273 K
V2 = 3.14 * 10^10 cm^3
Learn more about gas laws:brainly.com/question/12669509
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Answer:
If you see in the image above, there is an unbalance force applied while playing tug of war. Since it is 1 vs 2, there is a greater net force in the right side then the left side. If it was 2 vs 2 or 1 vs 1, then they are appling balance force. You can also see in the picture that the arrows are pointing outwards (--->) rather then inwards (<---) because you are pulling the rope not pushing the rope. If you add one person on the left side, then the newtons which is 20N will become to 35N and will be balanced, but since there in only 1 person, there is less force on the left side, the newtons gets subtracted having only 20N. Since you are pulling the rope, the friction is opposite (<---). Since you are pulling the rope, you are using Kinetic force and the rope stays in potential force since it stays constant.
Hope this helps, thank you :) and I am not sure about magnitude I think you can that since there is greater force on the right side, there is more magnitude there.
Density is defined as the ratio of mass to the volume.
Density = 
(1)
Mass of water = 10 grams
Mass of acetone = 10 grams
Density of water = 1 
Density of acetone = 0.7857
Put the value of density of water and its mass in equation (1)
1 = 
Volume of water = 10 
Put the value of density of acetone and its mass in equation (1)
0.7857 =
Volume of acetone = 12.72
Thus, volume of acetone is more than volume of water because the density of acetone is lower.
