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.
Answer: 0.245 moles of oxygen must be placed in the container to exert the given pressure at the given temperature. The Ideal Gas Law equation gives the relationship among the pressure, volume, temperature, and moles of gas.
Further Explanation:
The Ideal Gas Equation is:
where:
P - pressure (in atm)
V - volume (in L)
n - amount of gas (in moles)
R - universal gas constant 
T - temperature (in K)
In the problem, we are given the values:
P = 2.00 atm (3 significant figures)
V = 3.00 L (3 significant figures)
n = ?
T = 25.0 degrees Celsius (3 significant figures)
We need to convert the temperature to Kelvin before we can use the Ideal Gas Equation. The formula to convert from degree Celsius to Kelvin is:
Therefore, for this problem,
Solving for n using the Ideal Gas Equation:
The least number of significant figures is 3, therefore, the final answer must have only 3 significant figures.
Learn More
1. Learn more about Boyle's Law brainly.com/question/1437490
2. Learn more about Charles' Law brainly.com/question/1421697
3. Learn more about Gay-Lussac's Law brainly.com/question/6534668
Keywords: Ideal Gas Law, Volume, Pressure
Answer:
Reagent A: PBr₃
Reagent B: Mg in Et₂O.
Explanation:
Hello,
In this case, your facing a problem in which a carboxylic acid is produced starting by an alcohol. More specifically, cyclopentanol must react with phosphorous tribromide in order to yield bromocyclopentane which is more likely to produce a carboxylic acid, therefore, reagent A is PBr₃.
On the other hand, by means of the production of the specified product, bromocyclopentane must react with carbon dioxide and magnesium in diethyl ether in acidic media to promote the production of the cyclopentanoic acid via the grignard reaction (substitution of the bromine by the carboxyle group), therefore, reagent B is Mg in Et₂O.
Best regards.
The reaction equation is:
<span>2CuO(s) + C(s) </span>→ <span>2Cu(s) + CO</span>₂<span>(g)
First, we determine the number of grams present in one ton of copper oxide. This is:
1 ton = 9.09 x 10</span>⁵ g
We convert this into moles by dividing by the molecular mass of copper oxide, which is:
9.09 x 10⁵ / 79.5 = 11,434 moles
Each mole of carbon reduces two moles of copper oxide, so the moles of carbon required are:
11,434 / 2 = 5,717 moles of Carbon required
The mass of carbon is then:
5,717 x 12 = 68,604 grams
The mass of coke is:
68,604 / 0.95 = 72,214 g
The mass of coke required is 7.22 x 10⁴ grams
An occluded front forms when a warm air mass is caught between two cooler air masses. The warm air mass is cut ofl or occluded' from the ground. The occluded warm front may cause clouds and precipitation. A swirling center of low air pressure is called a cyclone.
