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Marina CMI [18]
3 years ago
8

Xenon forms several compounds with oxygen and fluorine. It is the most reactive non-radioactive noble gas because a. its large r

adius allows oxygen and fluorine to bond without being crowded. b. it has the highest electronegativity of these noble gases. c. it has the highest electron affinity of these noble gases. d. its effective nuclear charge is lower than the other noble gases. e. it has the lowest ionization energy of these noble gases.
Chemistry
1 answer:
muminat3 years ago
7 0

Answer:

d. its effective nuclear charge is lower than the other noble gases.

Explanation:

Xenon belongs to group O on the periodic table. Most of the elements here are unreactive.

Due to the large size of Xenon, the outermost electrons have very low effective nuclear charge. Effective nuclear charge is the effect of the positive charges of the nucleus on the electrons in orbits. This effect decreases outward as atomic shell increases.

Xenon has a very large atomic radius and there is weak a nuclear charge on the outermost electrons. The more electronegative elements would be able to attract some of its outermost electrons easily and form chemical bonds with xenon much more readily.

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mafiozo [28]
Number five is decrease
7 0
4 years ago
Some organic solvents do not work well in liquid-liquid aqueous extractions. Ethanol (HOCH2CH3) is a common inexpensive solvent,
erica [24]

Answer:

See explanation

Explanation:

Extraction has to do with the separation of the components of a mixture by dissolving the mixture in a set up involving two phases. One phase is the aqueous phase (beneath) while the other is the organic phase (on top). The solvents used for the two phases must not be miscible. Water commonly is used for the aqueous phase.

Ethanol is an important solvent in chemistry but the solvent is miscible with water in all proportions. As a result of this, ethanol is a poor solvent for carrying out extraction.

4 0
3 years ago
Consider the equilibrium reaction. 2 A + B − ⇀ ↽ − 4 C After multiplying the reaction by a factor of 2, what is the new equilibr
vredina [299]

Answer : The correct expression for equilibrium constant will be:

K_c=\frac{[C]^8}{[A]^4[B]^2}

Explanation :

Equilibrium constant : It is defined as the equilibrium constant. It is defined as the ratio of concentration of products to the concentration of reactants.

The equilibrium expression for the reaction is determined by multiplying the concentrations of products and divided by the concentrations of the reactants and each concentration is raised to the power that is equal to the coefficient in the balanced reaction.

As we know that the concentrations of pure solids and liquids are constant that is they do not change. Thus, they are not included in the equilibrium expression.

The given equilibrium reaction is,

4A+2B\rightleftharpoons 8C

The expression of K_c will be,

K_c=\frac{[C]^8}{[A]^4[B]^2}

Therefore, the correct expression for equilibrium constant will be, K_c=\frac{[C]^8}{[A]^4[B]^2}

4 0
3 years ago
A laboratory analysis of a sample finds it is composed of 38.8% carbon, 16.2% hydrogen, and 45.1% nitrogen. What is its empirica
Sladkaya [172]

Answer: The empirical formula for the given compound is CH_5N

Explanation : Given,

Percentage of C = 38.8 %

Percentage of H = 16.2 %

Percentage of N = 45.1 %

Let the mass of compound be 100 g. So, percentages given are taken as mass.

Mass of C = 38.8 g

Mass of H = 16.2 g

Mass of N = 45.4 g

To formulate the empirical formula, we need to follow some steps:

Step 1: Converting the given masses into moles.

Moles of Carbon =\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{38.8g}{12g/mole}=3.23moles

Moles of Hydrogen = \frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{16.2g}{1g/mole}=16.2moles

Moles of Nitrogen = \frac{\text{Given mass of nitrogen}}{\text{Molar mass of nitrogen}}=\frac{45.4g}{14g/mole}=3.24moles

Step 2: Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 3.23 moles.

For Carbon = \frac{3.23}{3.23}=1

For Hydrogen  = \frac{16.2}{3.23}=5.01\approx 5

For Oxygen  = \frac{3.24}{3.23}=1.00\approx 1

Step 3: Taking the mole ratio as their subscripts.

The ratio of C : H : N = 1 : 5 : 1

Hence, the empirical formula for the given compound is C_1H_5N_1=CH_5N

3 0
4 years ago
Using the following equation for the combustion of octane calculate the heat associated with the formation of 100.0 g of carbon
natali 33 [55]

Answer:

The right solution is "-602.69 KJ heat".

Explanation:

According to the question,

The 100.0 g of carbon dioxide:

= \frac{100.0 \ g}{114.33\  g/mole}

= 0.8747 \ moles

We know that 16 moles of CO_2 formation associates with -11018 kJ of heat, then

0.8747 moles CO_2 formation associates with,

= -\frac{0.8747}{16}\times 11018 \ KJ \ of \ heat

= -0.0547\times 11018

= -602.69 \ KJ \ heat

8 0
3 years ago
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