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lina2011 [118]
3 years ago
10

When comparing the two elements P and I , the element with the higher first ionization energy is PP When comparing the two eleme

nts P and I , the element with the higher first ionization energy is blank based on periodic trends alone. based on periodic trends alone.
Chemistry
1 answer:
Anni [7]3 years ago
6 0

Answer:

Explanation:

The first ionization energy varies in a predictable way across the periodic table. The ionization energy decreases from top to bottom in groups, and increases from left to right across a period. Thus, helium has the largest first ionization energy, while francium has one of the lowest.

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The jet stream and ocean currents most commonly influence which of the following?
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Arrange the procedural steps, from start to finish, that are required to prepare indigo from nitrobenzaldehyde and acetone in ba
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Acetone has α-hydrogens (on both sides) and thus can be deprotonated to give a nucleophilic enolate anion. The aldehyde carbonyl is much more electrophilic than that of a ketone, and therefore reacts rapidly with the enolate.

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5 0
1 year ago
One litre of hydrogen at STP weight 0.09gm of 2 litre of gas at STP weight 2.880gm. Calculate the vapour density and molecular w
expeople1 [14]

Answer:

we know, at STP ( standard temperature and pressure).

we know, volume of 1 mole of gas = 22.4L

weight of 1 Litre of hydrogen gas = 0.09g

so, weight of 22.4 litres of hydrogen gas = 22.4 × 0.09 = 2.016g ≈ 2g = molecular weight of hydrogen gas.

similarly,

weight of 2L of a gas = 2.88gm

so, weight of 22.4 L of the gas = 2.88 × 22.4/2 = 2.88 × 11.2 = 32.256g

hence, molecular weight of the gas = 32.256g

vapor density = molecular weight/2

= 32.256/2 = 16.128g

hence, vapor density of the gas is 16.128g.

Explanation:

4 0
2 years ago
What is the mass of oxygen in 3.34 g of potassium permanganate?
3241004551 [841]

Answer:

1.35 g

Explanation:

Data Given:

mass of Potassium Permagnate (KMnO₄) = 3.34 g

Mass of Oxygen: ?

Solution:

First find the percentage composition of Oxygen in Potassium Permagnate (KMnO₄)

So,

Molar Mass of KMnO₄ = 39 + 55 + 4(16)

Molar Mass of KMnO₄ = 158 g/mol

Calculate the mole percent composition of  Oxygen in Potassium Permagnate (KMnO₄).

Mass contributed by Oxygen (O) = 4 (16) = 64 g

Since the percentage of compound is 100

So,

                        Percent of Oxygen (O) = 64 / 158 x 100

                        Percent of Oxygen (O) = 40.5 %

It means that for ever gram of Potassium Permagnate (KMnO₄) there is 0.405 g of Oxygen (O) is present.

So,

for the 3.34 grams of Potassium Permagnate (KMnO₄) the mass of Oxygen will be

                  mass of Oxygen (O) = 0.405 x 3.34 g

                  mass of Oxygen (O) = 1.35 g

5 0
3 years ago
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