Answer:
<h2>Density = 3.9 g/ mL</h2>
Explanation:
The density of a substance can be found by using the formula
From the question
mass = 43 g
volume = 11.1 mL
Substitute the values into the above formula and solve for the density
That's
We have the final answer as
Density = 3.9 g/ mL to one decimal place
Hope this helps you
The IUPAC name of the above mentioned compound is "2-Chloro-4,4-dimethylpentane"
<u>Exaplanation:</u>
- Since the above organic compound is an compound with only one saturated bond, It can be considered as a single bond compound, and hence we can conclude that as alkane.
- It also has 5 carbon atoms, so it is termed as pentane.
- From right to left we have to number the atoms, and 2 nd carbon atom contain Cl atom so it is termed as 2-Chloro and in the 4th position carbon atom contains 2 methyl groups, so it is termed as, 2-Chloro-4,4-dimethylpentane.
It will be nitrogen.
The energy associated with the formation of 1 mol of gaseous cations is The ionization energy. 1 mole of electrons will form 1 mole of gaseous atoms under standard conditions. The ionization energy decreases as you go down a group on the periodic table.
Explanation:
The given data is as follows.
Current (I) = 3.50 amp, Mass deposited = 100.0 g
Molar mass of Cr = 52 g
It is known that 1 faraday of electricity will deposit 1 mole of chromium. As 1 faraday means 96500 C and 1 mole of Cr means 52 g.
Therefore, 100 g of Cr will be deposited by "z" grams of electricity.
z = 
= 185576.9 C
As we know that, Q = I × t
Hence, putting the given values into the above equation as follows.
Q = I × t
185576.9 C = 
t = 53021.9 sec
Thus, we can conclude that 100 g of Cr will be deposited in 53021.9 sec.
Answer:
D-block
Explanation:
Transition metals are metals of the d-block elements located between group 2 and 3 in the periodic table. They are called d-block element because the filling of electrons in their electronic configuration ends in the d-orbital.
Now, they have variable oxidation state because of their partially filled 3d orbital i.e the 3d orbital is not completely filled.
