Answer:
2.25 M is the final concentration of hydroxide ions ions in the solution after the reaction has gone to completion.
Explanation:
Moles of NaOH = 
Molarity of the nitric acid solution = 0.250 M
Volume of the nitric solution = 0.150 L
Moles of nitric acid = n
According to reaction, 1 mole of nitric acid recats with 1 mole of NaOH, then 0.0375 moles of nitric acid will react with :
of NaOH
Moles of NaOH left unreacted in the solution =
= 0.375 mol - 0.0375 mol = 0.3375 mol
1 mole of sodium hydroxide gives 1 mol of sodium ions and 1 mole of hydroxide ions.
Then 0.3375 moles of NaOH will give :
of hydroxide ion
The molarity of hydroxide ion in solution ;
2.25 M is the final concentration of hydroxide ions ions in the solution after the reaction has gone to completion.
Answer:
(D) (CH3CH2)2NH
Explanation:
In order to decide which base is strongest we need to calculate its PKb
PKb = -log [Kb]
A large Kb value and small PKb value gives the strongest base
Compound Kb PKb
(A) C6H5NH2 - 4 x 10^-10 9.349
(B) NH3 1.76x 10^-5 4.754
(C) CH3NH2 4.4x 10^-4 3.357
(D) (CH3CH2)2NH 8.6x 10^-4 3.066
(E) C5H5N 1.7x10^-9 8.77
Clearly (CH3CH2)2NH is the strongest base.
Temperature, salinity, and density are the group of factors are most important in determining the composition of ocean water.
a.)temperature, salinity, and density
<u>Explanation:</u>
The three fundamental factors that help in determining the composition of ocean water are temperature, salinity, and density. Temperature, saltiness, salinity, and density influence the thickness of seawater.
Enormous water masses of various densities are significant in the layering of the sea water (increasingly thick water sinks). As temperature builds water turns out to be less thick. As saltiness builds water gets denser. The temperature helps in deciding the pace of vanishing of the ocean.
Thermal energy travels<span> by conduction, convection, and radiation. It occurs when a cooler and warmer object touches each other. </span>
Answer:
20L is the new volume
Explanation:
In this case, moles and T° from the gas remain constant. This is the formula we must apply, to solve this:
P₁ . V₁ = P₂ . V₂
5 atm . 10 L = P₂ . 2.5L
P₂ = (5 atm . 10 L) / 2.5L →20L
