Across a period I.E increases progressively from left to right
Explanation:
The trend of the first ionization energy is such that across a period I.E increases from left to right due to the decreasing atomic radii caused by the increasing nuclear charge. This not compensated for by successive electronic shells.
- Ionization energy is a measure of the readiness of an atom to lose an electron.
- The lower the value, the easier it is for an atom to lose an electron.
- Elements in group I tend to lose their electrons more readily whereas the halogens hold most tightly to them.
- The first ionization energy is the energy needed to remove the most loosely bonded electron of an atom in the gaseous phase.
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Answer:
Four substitution products are obtained. The carbocation that forms can react with either nucleophile (H2O or CH3OH) from either the top or bottom side of the molecule
Explanation:
An SN1 reaction usually involves the formation of a carbocation in the slow rate determining step. This carbocation is now attacked by a nucleophile in a subsequent fast step to give the desired product.
However, the product is obtained as a racemic mixture because the nucleophile may attack from the top or bottom of the carbocation hence both attacks are equally probable.
The attacking nucleophile in this case may be water or CH3OH
Answer:
130 Liters
Explanation:
if 1 mol is 22.4 L, then 5.8 mol is 130 L (129.92 but use sig figs)
Answer:
the Germany scientist Alfred Wagner describes fold mountains that are found in The world are the same also geological similarity that occur between this continent are the same for example the rock that are found in brazil are the same of that of south africa.
The reaction between oxygen, O2, and hydrogen, H2, to produce water can be expressed as,
2H2 + O2 --> 2H2O
The masses of each of the reactants are calculated below.
2H2 = 4(1.01 g) = 4.04 g
O2 = 2(16 g) = 32 g
Given 1.22 grams of oxygen, we determine the mass of hydrogen needed.
(1.22 g O2)(4.04 g H2 / 32 g O2) = 0.154 g of O2
Since there are 1.05 grams of O2 then, the limiting reactant is 1.22 grams of oxygen.
<em>Answer: 1.22 g of oxygen</em>
