Answer:
The molarity of the solution is 0,12 M.
Explanation:
We calculate the molarity, which is a concentration measure that indicates the moles of solute (in this case KOH) in 1000ml of solution (1 liter):
0,25 L solution----- 0,030 moles of KOH
1 L solution----x= (1 L solution x 0,030 moles of KOH)/0,25 L solution
x= 0,12 moles of KOH ---> <em>The solution is 0,12 M</em>
Hydrogen - 7.44%
Carbon - (100-7.44)% = 92.56%
Lets take 100 g of benzene, then we have
Hydrogen - 7.44 g
Carbon - 92.56 g
n - number of moles
n(H) = 7.44g *1 mol/1.0g = 7.44 mol
n(C) = 92.56 g* 1mol/12.0 g ≈ 7.713 mol
n(C) : n(H) = 7.713 mol : 7.44 mol = 1:1
Empirical formula is CH.
M(CH) = (12.0+ 1.00) g/mol = 13.0 g/mol
M (benzene) = 78.1 g/mol
M (benzene)/M(CH)= 78.1 g/mol/13.0 g/mol = 6
So, molecular formula of benzene is C6H6.
Answer:
When the electron changes levels, it decreases energy and the atom emits photons. The photon is emitted with the electron moving from a higher energy level to a lower energy level. The energy of the photon is the exact energy that is lost by the electron moving to its lower energy level.
Explanation:
The correct answer is e. 3.57×10³
Al³+(aq) + 3e→AL(s)
4.00g of AL=4g/26.98 g/mol= 0.1483 mol
t=znF/1 where t is time in seceonds.
Z= valency number of ions of the substance or electrons which are transferred per ion
F= Faraday's constant
I = electric current in'A'CA C/s
t=(3×0.1483 mol ×96485 C/mol) /12(C15)
t=3577 second = 3.5 ×10³s
Answer:
Option B
Explanation:
The name of the molecule is;
1,1,1,3,3,3-Hexachloropropanone
Now, we can see that it contains propanone which is also called acetone. So it will have a double bond with Carbon(C) and Oxygen (O).
Option B has that double bond between C and O that indicates it propanone or acetone.
Thus, it is the correct option.
