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horrorfan [7]
2 years ago
14

To be effective, collisions between atoms must occur with the proper orientation and also the proper amount of __.

Chemistry
1 answer:
Slav-nsk [51]2 years ago
5 0

To be effective, collisions between atoms must occur with the proper orientation and also the proper amount of energy.

<h3>What is the loss of kinetic energy due to the collision?</h3>

An inelastic collision is a collision in which there is a loss of kinetic energy. While the momentum of the system is conserved in an inelastic collision, the kinetic energy is not. This is because some kinetic energy has been transferred to something else.

In this case, is necessary have energy to be effective, so collisions between atoms must occur with the proper orientation.

See more abour energy at brainly.com/question/1932868

#SPJ1

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An element's atomic number is 112. How many electrons would an atom of this element have?
PIT_PIT [208]

Without any ionization, the element (Cn) would have 112 electrons because the atomic number of an element is the number of protons the element has and a neutral element has the same number of electrons as it does protons.

6 0
3 years ago
Determine the pH of a 2.8 ×10−4 M solution<br> of Ca(OH)2.
shepuryov [24]

Answer:

pH = 10.75

Explanation:

To solve this problem, we must find the molarity of [OH⁻]. With the molarity we can find the pOH = -log[OH⁻]

Using the equation:

pH = 14 - pOH

We can find the pH of the solution.

The molarity of Ca(OH)₂ is 2.8x10⁻⁴M, as there are 2 moles of OH⁻ in 1 mole of Ca(OH)₂, the molarity of [OH⁻] is 2*2.8x10⁻⁴M = 5.6x10⁻⁴M

pOH is

pOH = -log 5.6x10⁻⁴M

pOH = 3.25

pH = 14-pOH

<h3>pH = 10.75</h3>
3 0
3 years ago
I will give 25 points and make them brainlest if they solve this.
sdas [7]

Answer:

B

Explanation:

7 0
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What mass of oxygen contains the same number of molecules as 42g of nitrogen?
lakkis [162]

Answer:

Given: 42 g of N2

Solve for O2 mass that contains the same number of molecules to 42 g of N2.

Solve for the number of moles in 42 g of N2

1 mole of N2 = (14 * 2) g = 28 g so the number of moles in 42 g of N2 is equal to 42 g / 28 g per mole = 1.5 moles

Solve for mass of 1 mole of oxygen

1 mole of O2 = 16 g * 2 = 32 g per mole

Solve for the mass of 1.5 moles of oxygen

mass of 1.5 moles of O2 = 32 g per mole * 1.5 moles

mass of 1.5 moles of O2 = 48 g

So 48 g of O2 contains the same number of molecules as 42 g of N2

3 0
3 years ago
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