The following are neutral salts: NANO3, BaBr2 AND Ba[OH]2.
Neutral salts are those salts that are formed when a strong base react with a strong acid. In those neutralization reactions, the base completely neutralize the acid to form a neutral salts.
Answer:
the oldest rocks are in the center of the dome structure.
Explanation:
There are 10 hydrogen atoms that bind and there are 2 pairs of free electrons in the non-binding O atom
<h3>Further explanation</h3>
Aldehydes are alkane-derived compounds containing carbonyl groups (-CO-) where one bond binds to an alkyl group while another binds to a hydrogen atom.
The general structure is R-CHO with the molecular formula :

Naming is generally the same as the alkane by replacing the suffix with -al
Butanal or butyraldehyde is an aldehyde which has 4 C atoms
Inside the structure there are 3 atoms involved in bonding:
- 1. Atom C with 4 valence electrons, requires 4 electrons to reach the octet
- 2. Atom O with 6 valence electrons, requires 2 electrons to reach the octet
- 3. Atom H with 1 valence electron, requires 1 electron to reach a duplet
In describing Lewis's structure the steps that can be taken are:
- 1. Count the number of valence electrons from atoms in a molecule
- 2. Give each bond a pair of electrons
- 3. The remaining electrons are given to the atomic terminal so that an octet is reached
- 4. The remaining electrons that still exist in the central atom
- 5. If the central atom is not yet octet, free electrons are drawn to the central atom to form double bonds
In the Butanal structure (C₄H₈O) there is 1 double bond of the functional group (-CHO) between the C atom and the O atom
<h3>Learn more:
</h3>
Adding electron dots
brainly.com/question/6085185
Ionic bonding
brainly.com/question/1603987
Formal charge
brainly.com/question/7190235
Keywords: butanal, aldehyde, Lewis structure, a valence electron
Answer:
pH = 11.216.
Explanation:
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In this case, according to the ionization of ammonia in aqueous solution:

We can set up its equilibrium expression in terms of x as the reaction extent equal to the concentration of each product at equilibrium:
![Kb=\frac{[NH_4^+][OH^-]}{[NH_3]} \\\\1.80x10^{-5}=\frac{x*x}{0.150-x}](https://tex.z-dn.net/?f=Kb%3D%5Cfrac%7B%5BNH_4%5E%2B%5D%5BOH%5E-%5D%7D%7B%5BNH_3%5D%7D%20%5C%5C%5C%5C1.80x10%5E%7B-5%7D%3D%5Cfrac%7Bx%2Ax%7D%7B0.150-x%7D)
However, since Kb<<<1 we can neglect the x on bottom and easily compute it via:

Which is also:
![[OH^-]=1.643x10^{-3}M](https://tex.z-dn.net/?f=%5BOH%5E-%5D%3D1.643x10%5E%7B-3%7DM)
Thereafter we can compute the pOH first:

Finally, the pH turns out:

Regards!
Lar mass of Ca<span> = 40.08 </span>grams/mole 77.4 g Ca<span> * ( 1 </span>mole Ca<span>/ 40.08 ... n = m / M 1mol </span>Ca<span>weights 40 gmol-1 n = 77,4 / 40 = 1.93 </span>mol<span>.</span>