The answer is (2) A bond is formed and energy is released. The left side of equation is I atom and the right side of equation is I2 molecule. So the bond is formed between I atom to form I2 molecule. And forming bond will release energy while breaking bond will absorb energy.
Answer:
1) After adding 15.0 mL of the HCl solution, the mixture is before the equivalence point on the titration curve.
2) The pH of the solution after adding HCl is 12.6
Explanation:
10.0 mL of 0.25 M NaOH(aq) react with 15.0 mL of 0.10 M HCl(aq). Let's calculate the moles of each reactant.
There is an excess of NaOH so the mixture is before the equivalence point. When HCl completely reacts, we can calculate the moles in excess of NaOH.
NaOH + HCl ⇒ NaCl + H₂O
Initial 2.5 × 10⁻³ 1.5 × 10⁻³ 0 0
Reaction -1.5 × 10⁻³ -1.5 × 10⁻³ 1.5 × 10⁻³ 1.5 × 10⁻³
Final 1.0 × 10⁻³ 0 1.5 × 10⁻³ 1.5 × 10⁻³
The concentration of NaOH is:
![[NaOH]=\frac{1.0 \times 10^{-3} mol }{25.0 \times 10^{-3} L} =0.040M](https://tex.z-dn.net/?f=%5BNaOH%5D%3D%5Cfrac%7B1.0%20%5Ctimes%2010%5E%7B-3%7D%20mol%20%7D%7B25.0%20%5Ctimes%2010%5E%7B-3%7D%20L%7D%20%3D0.040M)
NaOH is a strong base so [OH⁻] = [NaOH].
Finally, we can calculate pOH and pH.
pOH = -log [OH⁻] = -log 0.040 = 1.4
pH = 14 - pOH = 14 - 1.4 = 12.6
Answer:
c
Explanation:
I think it is right not sure but hope it helped
The weight in grams = 7.93 g
Given volume = 2.00 
Given density = 0.242 g/
We need to find the Mass(weight) in grams.
To find the weight in grams we need to keep in mind that the volume and density must use the same volume unit for cancellation. So that the volume units will cancel out, leaving only the mass units.
The unit of given volume is 
and unit of volume in density is , so first we need to change the unit of volume from 
to so that the volume units will cancel out, leaving only the mass units.
1 = 16.39 (given conversion)
units get cancel out leaving the unit.
Mass = Density X Volume.
Density = 0.242 g/ and Volume = 32.78
Mass = 7.93 grams (g)
Answer:
237.8L of water would need to be added.
Explanation:
The first thing to do is to identify that the equation to be used is M1V1=M2V2. (This equation works because it turns everything into moles which can then be compared).
Then figure out what information you have and what is being found. In this case:
M1 = 54.7 M
V1 = 1092 mL = 1.092 L
M2 = 0.25 M
V2 = unknown
Then solve the equation for whatever you are trying to find.
M1V1=M2V2
V2=M1V1/M2
Now you need to plug everything in.
V2=(54.7M*1.091L)/0.25M
V2=238.93L
That means that the solution needs a volume of 238.7L to gain a molarity of 0.25M but the starting solution already had a volume of 1.092 L meaning that to find the amount of solvent that needs to be added you just subtract the starting volume by the volume that the solution needs to be.
238.93L - 1.091L = 237.8L
Therefore the answer is that 237.8L needs to be added to a 1.092L 54.7M NaCl solution to make the concentration 0.25M.
I hope this helps. Let me know if anything is unclear.
