V1 = 2.00 L
<span>T1 = 25 + 273 = 298 K </span>
<span>V2 = 6.00 L </span>
<span>T2 = ? </span>
<span>Assuming the pressure is to remain constant, then </span>
<span>V1/T1 = V2/T2 </span>
<span>T2 = T1V2/V1 = (298)(6)/(2) = 894 deg K</span>
Answer:
See the explanation
Explanation:
In this case, in order to get an <u>elimination reaction</u> we need to have a <u>strong base</u>. In this case, the base is the phenoxide ion produced the phenol (see figure 1).
Due to the resonance, we will have a more stable anion therefore we will have a less strong base because the negative charge is moving around the molecule (see figure 2).
Finally, the phenoxide will attack the <u>primary carbon</u> attached to the Cl. The C-Cl bond would be broken and the C-O would be produced <u>at the same time</u> to get a substitution (see figure 1).
Answer:
Binary compound
Explanation:
Binary compounds:
The compounds which are made up of the atoms of only two elements are called binary compounds.
For example:
The following compounds are binary:
HCl
H₂O
NH₃
HCl is binary because it is composed of only hydrogen and chlorine. Ammonia is also binary compound because it is made up of only two elements nitrogen and hydrogen.
water is also binary because it is also made up of only two elements hydrogen and oxygen.
SF₆ is binary compound because it consist of atoms of only two elements i.e, sulfur and fluorine.
Answer:
Use pOH = -log₁₀ [OH-]
Explanation:
pOH can be calculated from the concentration of hydroxide ions using the formular below:
pOH = -log₁₀ [OH-]
The pOH is the negative logarithm of the hydroxide ion concentration.
Answer:- 544.5 mL of water need to be added.
Solution:- It is a dilution problem. The equation used for solving this type of problems is:

where,
is initial molarity and
is the molarity after dilution. Similarly,
is the volume before dilution and
is the volume after dilution.
Let's plug in the values in the equation:



Volume of water added = 907.5mL - 363mL = 544.5 mL
So, 544.5 mL of water are need to be added to the original solution for dilution.