Answer:
- They are highly reactive metals
- They have low electro negativity
- They have low ionization energy
- They don't exist alone in nature
- They have low densities
Explanation:
Alkali metals are the elements in group 1 of the periodic table. They include Sodium, Lithium, Potassium e.t.c.
Due to the fact they have one atom in their outermost shell, they are very unstable because they easily react with other elements and are therefore don't exist alone in nature but combined with other elements for this same reason.
Since alkali metals don't easily attract other elements due to it's lone pair in the outer most shell, it can be said to have low electro negativity.
Also, they don't need energy to discharge their electrons since they are highly reactive due to their lone pair in the outermost shell and so we say they have low ionization energy.
Due to this reason, they also have low densities.
Answer:
The standard enthalpy of formation of NOCl(g) at 25 ºC is 105 kJ/mol
Explanation:
The ∆H (heat of reaction) of the combustion reaction is the heat that accompanies the entire reaction. For its calculation you must make the total sum of all the heats of the products and of the reagents affected by their stoichiometric coefficient (number of molecules of each compound that participates in the reaction) and finally subtract them:
Enthalpy of the reaction= ΔH = ∑Hproducts - ∑Hreactants
In this case, you have: 2 NOCl(g) → 2 NO(g) + Cl₂(g)
So, ΔH=
Knowing:
- ΔH= 75.5 kJ/mol
= 90.25 kJ/mol
= 0 (For the formation of one mole of a pure element the heat of formation is 0, in this caseyou have as a pure compound the chlorine Cl₂)
=?
Replacing:
75.5 kJ/mol=2* 90.25 kJ/mol + 0 -
Solving
-=75.5 kJ/mol - 2*90.25 kJ/mol
-=-105 kJ/mol
=105 kJ/mol
<u><em>The standard enthalpy of formation of NOCl(g) at 25 ºC is 105 kJ/mol</em></u>
Answer:
44.8 L of O2 will react (option D)
Explanation:
Step 1: Data given
Number of moles of SO2 = 4.00 moles
STP = Pressure = 1 atm and temperature = 273 K
Step 2: The balanced equation
2 SO2(g) + O2(g) → 2 SO3(g)
Step 3: Calculate moles of O2
For 2 moles SO2, we need 1 mol O2 to produce 2 moles SO3
For 4.00 moles SO2 we need 4.00 / 2 = 2.00 moles O2
Step 4: Calculate volume of O2
For 1 mol we have a volume of 22.4 L
V = (n*R*T)/ p
V = (2.00 * 0.08206 * 273)/p
V = 44.8 L
For 2.00 moles we have a volume of 2*22.4 = 44.8 L
44.8 L of O2 will react (option D)
Answer:
Down Syndrome- is when you have an extra 21st chromosome.
Turner Syndrome- is when one of the x chromosome is missing or partly there.
Explanation:
There are many more and you can find a complete list online.
Answer:
II
Explanation:
We must have a good idea of the fact that there are two mechanisms that come into play when we are discussing about the addition of hydrogen halides to alkenes. The first is the ionic mechanism and the second is the radical mechanism.
The ionic mechanism is accounted for by the Markovnikov rule while the radical mechanism occurs in the presence of peroxides and is generally referred to as anti Markovnikov addition.
The intermediate in anti Markovnikov addition involves the most stable radical, in this case, it is a tertiary radical as shown in the images attached. The most stable radical is II hence it leads to the major product shown in the other image.
