<h3>Answer:</h3>
The pressure will decrease by half.
<h3>Explanation:</h3>
Data Given:
P₁ = 101.3 kPa
V₁ = X
V₂ = 2X
P₂ = ??
Let us suppose the initial volume is 5 L and the final volume when doubled is 10 L. Then According to Boyle's Law, " The Volume of a given mass of gas at constant temperature is inversely proportional to the applied Pressure". Mathematically the initial and final states of gas are given as,
P₁ V₁ = P₂ V₂ ----------- (1)
Solving equation 1 for P₂,
P₂ = P₁ V₁ ÷ V₂
Putting values,
P₂ = (101.3 kPa × 5 L) ÷ 10 L
P₂ = 50.65 kPa
<h3>Result:
</h3>
As the volume is doubled, the pressure is decreased from 101.3 kPa to 50.65 kPa or it has decreased by half.
The increase in Atomic radius contributes to the decrease in ionization energy down a group. As you go down a group the Atomic radius increases meaning more energy levels are being added, and there are more electron shieldings. So, therefore, electron shielding blocks the pull of the nucleus from the valence electrons and it makes easier for the atom to lose its outermost electron. So the atom needs less ionization energy to remove its valence electron.
Answer:
True
Explanation:
Becuse the humans are dipositing billions of plastic and it has just grown over the years.
C) medium.
The medium of a wave is any substance that carries the wave, or through which the wave travels.Ocean waves are carried by water, sound waves are carried by air, and. the seismic waves of an earthquake are carried by rock and soil.
Answer:
The given statement is true.
Explanation:
Initially, the addition of borane to the alkene takes place in the form of a concerted reaction owing to the dissociation of the bond and subsequent formation, which occurs at a similar time. After that the Anti Markovnikov supplementation of boron takes place. The addition of this atom takes place with the less substituted carbon of the alkene that then substitutes the molecule of hydrogen on the more substituted carbon.
Then through the donation of a pair of electrons from the hydrogen peroxide ion, the process of oxidation takes place resulting in the formation of trialkylborane. After this realignment of an R group with its pair of bonding, electrons take place with adjacent oxygen resulting in the withdrawal of a hydroxide ion. Eventually, the trialkylborate reacts with the aqueous NaOH to generate alcohol and sodium borate as the side product.