Answer:
0.0375 moles HBr
Explanation:
we are given;
- Molarity of HBr solution as 0.15 M
- Volume of the solution as 250 mL
We are required to determine the number of moles;
We need to know that;
Molarity = Moles ÷ Volume
Therefore;
Moles of HBr = Molarity of HBr solution × Volume of solution
Thus;
Moles of HBr = 0.15 M × 0.25 L
= 0.0375 Moles
Thus, the number of moles of HBr solute is 0.0375 moles
Answer:
Chemical bonds are how atoms, and even molecules join together.
Explanation:
There are two main types of primary chemical bonds. While secondary links relate to molecules, primary ties are atom to atom. This answer explains basic primary bonds only.
One must comprehend what a valence shell is before I proceed. The outer electron orbital of an atom is known as the valence shell. Most of the time (except from hydrogen), atoms desire to have 8 electrons in their valence shell, thus they form bonds with other atoms to accomplish this.
<em>All bonds result in a new chemically different molecule. Now, the two types are:</em>
- Covalent: When two atoms combine their electrons to fill their valence shells. The atoms are joined together by this "sharing."
- Ionic: When one atom <em>transfers</em> an electron to another atom in order to fulfill the valence electron requirement. Because electrons have a negative charge, the atom that <em>produced </em>them gains a positive charge as a result of losing its negative charge. The atom that received the electron therefore acquires a negative charge. Because opposing charges attract, it seems sense that the charged atoms bind as a result.
Answer:For atoms and molecules, the width of spectral lines is governed mainly by the broadening of the energy levels of the atoms or molecules during interactions with surrounding particles and by the broadening of the spectral lines as a result of the Doppler effect.
Explanation:
Answer:
The total pressure is 27.8 atm
Explanation:
From the ideal gas equation,
PV = nRT
P (total pressure) = nRT/V
n (total moles of gases) = (6/1 moles of hydrogen) + (15.2/14 moles of nitrogen) + (16.8/4 moles of helium) = 6+1.1+4.2 = 11.3 moles
R = 0.082057L.atm/gmol.K, T = 27°C = 27+273K = 300K, V = 10L
P = 11.3×0.082057×300/10 = 27.8 atm
