A compound is formed when two substances are chemically combined. H2O, or water, is a compound as it is composed of 2 hydrogen atoms and 1 oxygen atoms.
Explanation:
Let us assume that the given data is as follows.
mass of barium acetate = 2.19 g
volume = 150 ml = 0.150 L (as 1 L = 1000 ml)
concentration of the aqueous solution = 0.10 M
Therefore, the reaction equation will be as follows.

Hence, moles of
=
.......... (1)
As, No. of moles =
Hence, moles of
will be calculated as follows.
No. of moles =
=
(molar mass of
is 255.415 g/mol)
= 
Moles of
= 
= 0.01715 mol
Hence, final molarity will be as follows.
Molarity = 
= 
= 0.114 M
Thus, we can conclude that final molarity of barium cation in the solution is 0.114 M.
Answer:
D. ionic bond
Explanation:
Due to electron deficiency in a metal cation, they cannot form a covalent bond beacuse it means to share electrons. By contrast metal cation seek for electrons. In an ionic bond, one atom give electrons, while another atom recevie electron. Because of that, this is the better option to metal cations.
Answer:

Explanation:
Hello there!
In this case, sine the solution of this problem require the application of the Raoult's law, assuming heptane is a nonvolatile solute, so we can write:

Thus, we first calculate the mole fraction of chloroform, by using the given masses and molar masses as shown below:

Therefore, the partial pressure of chloroform turns out to be:

Regards!
pH=4.625
The classification of this sample of saliva : acid
<h3>Further explanation</h3>
The water equilibrium constant (Kw) is the product of concentration
the ions:
Kw = [H₃O⁺] [OH⁻]
Kw value at 25° C = 10⁻¹⁴
It is known [OH-] = 4.22 x 10⁻¹⁰ M
then the concentration of H₃O⁺:
![\tt 10^{-14}=4.22\times 10^{-10}\times [H_3O^+]\\\\(H_3O^+]=\dfrac{10^{-14}}{4.22\times 10^{-10}}=2.37\times 10^{-5}](https://tex.z-dn.net/?f=%5Ctt%2010%5E%7B-14%7D%3D4.22%5Ctimes%2010%5E%7B-10%7D%5Ctimes%20%5BH_3O%5E%2B%5D%5C%5C%5C%5C%28H_3O%5E%2B%5D%3D%5Cdfrac%7B10%5E%7B-14%7D%7D%7B4.22%5Ctimes%2010%5E%7B-10%7D%7D%3D2.37%5Ctimes%2010%5E%7B-5%7D)
pH=-log[H₃O⁺]
Saliva⇒acid(pH<7)