Answer:
A. Atoms randomly crashing into each other
E. The force of one object pushing against the force of another object.
Explanation:
Forces can be classified into two categories based upon the the mode of transfer or application:
1. Contact forces
2. Non-contact forces
Contact forces are the ones which require the physical contact of the matter to get transferred and tend to create the affect. Whereas non-contact forces have the field property which transfers the affect of force from one point to another without any physical contact of the matter or the medium.
- Atoms crashing onto each other have some mass and velocity which upon collision impacts the other atoms exerts a contact force.
- The interaction between the charged particles due to their charges is always due to the electric field be it electron or proton, be it within an atom or out of an atom.
- The force between any two objects pushing or pulling each other is also possible only due to contact.
There are different formula you need to keep in mind when solving for [OH-]
Given that pH = 6.10
pH + pOH = 14
6.10 + pOH = 14
pOH = 7.9
[OH-] = 10^(-pOH)
[OH-] = 10^(-7.9)
[OH-] = 0.000000013
[OH-] = 1.3 x 10^-8
<h2>
<u>Answer: [OH-] = 1.3 x 10^-8</u></h2>
Answer:
The minimum volume of the container is 0.0649 cubic meters, which is the same as 64.9 liters.
Explanation:
Assume that ethane behaves as an ideal gas under these conditions.
By the ideal gas law,
,
.
where
is the pressure of the gas,
is the volume of the gas,
is the number of moles of particles in this gas,
is the ideal gas constant, and
is the absolute temperature of the gas (in degrees Kelvins.)
The numerical value of will be 
if , , and are in SI units. Convert these values to SI units:
;- shall be in cubic meters,
; 
.
Apply the ideal gas law:
.
Answer:
520ML and apparently I need to put more in this answer
Explanation:
brainly.com
Answer:
1.36 × 10³ mL of water.
Explanation:
We can utilize the dilution equation. Recall that:
Where <em>M</em> represents molarity and <em>V</em> represents volume.
Let the initial concentration and unknown volume be <em>M</em>₁ and <em>V</em>₁, respectively. Let the final concentration and required volume be <em>M</em>₂ and <em>V</em>₂, respectively. Solve for <em>V</em>₁:
Therefore, we can begin with 0.640 L of the 2.50 M solution and add enough distilled water to dilute the solution to 2.00 L. The required amount of water is thus:
Convert this value to mL:
Therefore, about 1.36 × 10³ mL of water need to be added to the 2.50 M solution.
