Answer:
A warm shallow sea.
Explanation:
A warm, shallow sea invaded the Big Bend during the Cretaceous Period, some 135 million years ago, providing the setting for deposition of lime mud and the remains of sea-dwelling organisms such as clams and snails. Limestone layers formed from those shallow muds are now visible throughout much of the Big Bend.
Salt water have less cohesion than distilled water. This is because most of the atoms are already bonded to each other so there are less atoms to be able to let cohesion occur. Hope this answers the question.
The answer is true, particles in the gaseous state are the furthest apart
Answer:
a) Acidic buffer
b) No buffer
c) Acidic buffer
d) Basic buffer
e) Basic buffer
Explanation:
a) 75.0 mL of 0.10 M HF ; 55.0 mL of 0.15 M NaF
-Acidic buffer
Mixing of 75.0 mL of 0.10 HF and 55.0 mL of 0.15 mL NaF results in acidic buffer. HF/NaF is a buffer of weak acid and its conjugate base. F- is the conjugate base of acid,HF.
b.) 150.0 mL of 0.10 M HF ; 135.0 mL of 0.175 M HCl-No buffer
Mixing HF and HCl will not results in a buffer. Both are acids, and no conjugate base is present.
c.) 165.0 mL of 0.10 M HF ; 135.0 mL of 0.050 M KOH-Acidic buffer
HF reacts with KOH to form KF. F- is a conjujate base of HF. As volume and concentration of HF is more than KOH, therefore, HF will remain after reaction with KOH. HF/KF will be a buffer of weak acid and its conjugate base.
d.) 125.0 mL of 0.15 M CH3NH2 ; 120.0 mL of 0.25 M CH3NH3Cl -Basic buffer
CH3NH2/CH3NH3+ is a buffer of weak base and its conjugate acid.
e.) 105.0 mL of 0.15 M CH3NH2 ; 95.0 mL of 0.10 M HCl-Basic buffer
CH3NH2 is a weak base and HCl is a strong acid. CH3NH2 reacts with HCl to form its conjugate acid CH3NH3+. Volume and concentration of CH3NH2 is more as compared to HCl and hence, will remain in the soution after reactionf with HCl.
CH3NH3+/CH3NH2 is a buffer of weak base and its conjugate acid.
