Answer:
two atoms
Explanation:
In its stable molecular form, oxygen exists as two atoms and is written O2. to distinguish it from an atom of oxygen O, or ozone, a molecule of three oxygen atoms, O3. Even though each of these is all oxygen, combining atoms of the same element may give very different properties to the molecule.
Answer:
Random particle motion in liquids and gases is a difficult concept for in temperature, the particles move faster as they gain kinetic energy.
Explanation:
Answer:
The answer to your question is 25.9 g of KCl
Explanation:
Data
Grams of KCl = ?
Volume = 0.75 l
Molarity = 1 M
Formula

Solve for number of moles

Substitution
Number of moles = 1 x 0.75
Simplification
Number of moles = 0.75 moles
Molecular mass KCl = 39 + 35.5 = 34.5
Use proportions to find the grams of KCl
34.5 g of KCl ---------------- 1 mol
x ---------------- 0.75 moles
x = (0.75 x 34.5) / 1
x = 25.9 g of KCl
Answer:
The new force will be \frac{1}{100} of the original force.
Explanation:
In the context of this problem, we're dealing with the law of gravitational attraction. The law states that the gravitational force between two object is directly proportional to the product of their masses and inversely proportional to the square of a distance between them.
That said, let's say that our equation for the initial force is:
![F = G\frac{m_1m_2}{R^2}The problem states that the distance decrease to 1/10 of the original distance, this means:[tex]R_2 = \frac{1}{10}R](https://tex.z-dn.net/?f=F%20%3D%20G%5Cfrac%7Bm_1m_2%7D%7BR%5E2%7D%3C%2Fp%3E%3Cp%3EThe%20problem%20states%20%20that%20%20the%20distance%20decrease%20to%201%2F10%20of%20the%20original%20distance%2C%20this%20means%3A%3C%2Fp%3E%3Cp%3E%5Btex%5DR_2%20%3D%20%5Cfrac%7B1%7D%7B10%7DR)
And the force at this distance would be written in terms of the same equation:

Find the ratio between the final and the initial force:

Substitute the value for the final distance in terms of the initial distance:

Simplify:

This means the new force will be \frac{1}{100} of the original force.