Covalent compounds: N2, CCl4, SiO2 and AlCl3.
Ionic compounds: CaCl2 and LiBr.
Hope this helps!
They make the symbols individual
Answer:
Explanation:
The amine functional group is obtained by subsititution of one or more hydrogen atoms in the ammonia compound.
Ammonia is NH₃.
Then,
- by substituting one hydrogen you obtain R - NH₂.
- by substituting two hydrogens you obtain R' - NH - R''
- by subsituting the three hydrogens you obtain:
R'''
|
R' - N - R''
In this case, the three subsitutuents are silyl groups. The silyl group is derived form silane and is SiH₃. So, the tcompound <em>trisilylamine</em> is:
SiH₃
|
SiH₃ - N - SiH₃
Thus, you can count 3 hydrogen atoms for every silylgroup for a total of <u><em>9 hydrogen atoms in each molecule of trisilylamine.</em></u>
The number of atoms of K that are in 235 g of the compound is
2.57 x10^24 atoms
calculation
Step 1: find the moles of K2S
= moles = mass/molar mass
= 235 g/110 g/mol= 2.136 moles
Step 2: multiply 2.136 moles by no. of K atoms in K2S
= 2.136 x2 = 4.272 moles
Step 3: use the Avogadro's law to determine number of K atoms
that is according to Avogadro's law 1 mole = 6.02 x 10^23 atoms
4.272 moles= ? atoms
by cross multiplication
= (4.272 moles x 6.02 x10^23 atoms) / 1 mole = 2.57 x10^24 atoms
Question is incomplete, complete question is;
A 34.8 mL solution of 
(aq) of an unknown concentration was titrated with 0.15 M of NaOH(aq).
If it takes 20.4 mL of NaOH(aq) to reach the equivalence point of the titration, what is the molarity of 
? For your answer, only type in the numerical value with two significant figures. Do NOT include the unit.
Answer:
0.044 M is the molarity of (aq).
Explanation:
The reaction taking place here is in between acid and base which means that it is a neutralization reaction .
To calculate the concentration of acid, we use the equation given by neutralization reaction:
where,
are the n-factor, molarity and volume of acid which is
are the n-factor, molarity and volume of base which is NaOH.
We are given:
Putting values in above equation, we get:
0.044 M is the molarity of (aq).
