the cell structure represented by x is the nucleus
Answer:
i will think c-n=3 to NM=2
Element atomic number position
Ba 56 group 2, period 6
Ca 12 group 2, period 3
S 16 group 16, period 3
Si `14 group 14, period 3
Now, you need to know the properties of the different type of elements and the tendencies on the periodic table.
The metallic elements are, those placed on the left side of the periodic table, are the ones that release an electron more easily, so they will requiere less energy to give it up when forming chemical bonds.
The higher the metallic character the less the energy need to give up an electron.
The metallic character grows as the group number decreases (goes to the left) period increases (goes downward), so among the elements considered, Barium will require the least amount of energy to give un an electron when forming chemical bonds.
Answer:
The correct answer is 0, 235 mol
Explanation:
We use the formula PV =nRT. The normal conditions of temperature and pressure are 273K and 1 atm, we use the gas constant = 0, 082 l atm / K mol:
1 atm x 5, 25l = n x 0, 082 l atm / K mol x 273 K
n= 1 atm x 5, 25l /0, 082 l atm / K mol x 273 K
n= 0, 235 mol
Answer:
To increase the yield of H₂ we would use a low temperature.
For an exothermic reaction such as this, decreasing temperature increases the value of K and the amount of products at equilibrium. Low temperature increases the value of K and the amount of products at equilibrium.
Explanation:
Let´s consider the following reaction:
CO(g) + H₂O(g) ⇌ CO₂(g) + H₂(g)
When a system at equilibrium is disturbed, the response of the system is explained by Le Chatelier's Principle: <em>If a system at equilibrium suffers a perturbation (in temperature, pressure, concentration), the system will shift its equilibrium position to counteract such perturbation</em>.
In this case, we have an exothermic reaction (ΔH° < 0). We can imagine heat as one of the products. If we decrease the temperature, the system will try to raise it favoring the forward reaction to release heat and, at the same time, increasing the yield of H₂. By having more products, the value of the equilibrium constant K increases.
