Answer:
The O atom will tend to attract the electrons.
Explanation:
The electronegativity of O (3.5) is much higher than H (2.1), which means it is more likely to attract electrons. The higher the electronegativity, the more attractive.
Answer:
Increase the pressure of the gas
Explanation:
According to the Pressure law, for a fixed mass of gas, at a constant volume (V), the pressure (P) is directly proportional to the absolute temperature (T).
From the kinetic molecular theory, gases are composed of particles which are in constant motion, colliding with themselves as well as with the walls of their container.
When the temperature of these gas molecules is increased, the molecules acquire more kinetic energy and the rate of collisions increases. Since the container cannot expand, the increase in pressure is due to the increase in collisions between the molecules of the gas as well as with the walls of their container.
Burning a magnesium ribbon in the air is an addition reaction while heating potassium manganate 7 is a decomposition reaction.
<h3>Addition and decomposition reactions</h3>
Magnesium burns in air to produce magnesium oxide as follows:
Potassium manganate 7 burns to produce multiple products as follows:
Thus, the MgO will be heavier than Mg. On the other hand, 
will be less heavy than 
.
More on reactions can be found here: brainly.com/question/17434463
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Hello!
To find the number of atoms in 2.822 moles of nickel, we need to multiply it by Avogadro's number. Avogadro's number is 6.02 x 10^23 atoms.
2.822 moles x (6.02 x 10^23) ≈ 1.698844 x 10^24
Therefore, there are about 1.70 x 10^24 atoms (according to the number of significant figures) in 2.822 moles of nickel.
Answer:
Electrons
Explanation:
In an atom there would be three subatomic particles: Neutrons, electrons, protons. The smallest and lightest in terms of mass is electrons. This is because the nucleus is comprised of the protons and the neutrons, these have a greater mass than electrons as electrons has very little mass that can considered to be 0.
