Answer:
A. elements
<em>I</em><em> hope</em><em> it's</em><em> helps</em><em> you</em>
Many compunds have a terminal carbonyl
Aldehyde, Ketone, Carboxylic acid, Amide, Imide, Acid anhydride are the first that come to my mind.
Answer:
0.13 g
Explanation:
mass of aluminum required = ( Dislocation length) / ( Dislocation density) × (density of metal)
3000 miles to cm ( 1 mile = 160934 cm) = 3000 miles × 160934 cm / 1 mile = 482802000 cm
density of Aluminium = 2.7 g /cm³
dislocation density of aluminum = 10¹⁰ cm³
mass of aluminum required = (482802000 cm × 2.7 g/cm³) / 10¹⁰ cm³ = 0.13 g
1. The reactivity among the alkali metals increases as you go down the group due to the decrease in the effective nuclear charge from the increased shielding by the greater number of electrons. The greater the atomic number, the weaker the hold on the valence electron the nucleus has, and the more easily the element can lose the electron. Conversely, the lower the atomic number, the greater pull the nucleus has on the valence electron, and the less readily would the element be able to lose the electron (relatively speaking). Thus, in the first set comprising group I elements, sodium (Na) would be the least likely to lose its valence electron (and, for that matter, its core electrons).
2. The elements in this set are the group II alkaline earth metals, and they follow the same trend as the alkali metals. Of the elements here, beryllium (Be) would have the highest effective nuclear charge, and so it would be the least likely to lose its valence electrons. In fact, beryllium has a tendency not to lose (or gain) electrons, i.e., ionize, at all; it is unique among its congeners in that it tends to form covalent bonds.
3. While the alkali and alkaline earth metals would lose electrons to attain a noble gas configuration, the group VIIA halogens, as we have here, would need to gain a valence electron for an full octet. The trends in the group I and II elements are turned on their head for the halogens: The smaller the atomic number, the less shielding, and so the greater the pull by the nucleus to gain a valence electron. And as the atomic number increases (such as when you go down the group), the more shielding there is, the weaker the effective nuclear charge, and the lesser the tendency to gain a valence electron. Bromine (Br) has the largest atomic number among the halogens in this set, so an electron would feel the smallest pull from a bromine atom; bromine would thus be the least likely here to gain a valence electron.
4. The pattern for the elements in this set (the group VI chalcogens) generally follows that of the halogens. The greater the atomic number, the weaker the pull of the nucleus, and so the lesser the tendency to gain electrons. Tellurium (Te) has the highest atomic number among the elements in the set, and so it would be the least likely to gain electrons.
What is the boiling point of a solution made by dissolving 1.0000 mole of sucrose in 1.0000 kg of water?
The change in Boiling Point of water can be calculated using this formula:
ΔTb = i * Kb * m
Where i is the van't hoff factor (the number of particles or ions), the kb is a constant (boiling point elevation constant) and m is the molality of the solution.
The kb for water is always 0.515 °C/m. Kb = 0.515 °C/m
The value for i in this case is 1. Since sucrose is a covalent compound and it doesn't dissociate into ions. i = 1
The molal concentration of the solution can be found using this formula:
molality = moles of sucrose/kg of water
molality = 1.000 mol / 1.000 kg of water
molality = 1 m
Now that we know all the values, we can use the formula to find the change in the boiling point of water:
ΔTb = i * Kb * m
ΔTb = 1 * 0.515 °C/m * 1 m
ΔTb = 0.515 °C
Finally, we are asked for the boiling point of the solution, not the change. The boiling point of water at atmospheric pressure is 100.00 °C. If the boiling point rises 0.515 °C when we prepare the solution. The boiling point of the solution is:
Boiling point solution = Boiling point of water + ΔTb
Boiling point solution = 100.000 °C + 0.515 °C
Boiling point solution = 100.515 °C
Answer: The boiling point of the solution is 100.515 °C.
