The concentration of the HI solution is 0.75M.
<h3>How do we calculate the required concentration?</h3>
Required concentration of any solution used in titration will be calculated by using the below equation as:
M₁V₁ = M₂V₂, where
M₁ & V₁ are the molarity and volume of CsOH.
M₂ & V₂ are the molarity and volume of HI.
On putting all values from the question, we get
M₂ = (1.9)(9.9) / (25) = 0.75M
Hence required concentration of HI solution is 0.75M.
To know more about concentration & volume, visit the below link:
brainly.com/question/24697661
#SPJ1
Given data:
Volume of HCl = 14.22 ml
Molarity of HCl = 2.97 M
mmoles of HCl = 14.22 * 2.97 = 42.2 mmoles
Volume of NaOH = 5.00 ml
Molarity of NaOH = 0.1055 M
mmoles of NaOH = 5.00 *.1055 = 0.5275 mmoles
Since HCl and NaOH combine in a 1:1 ratio
# moles of NaOH = # moles of excess HCl that is neutralized = 0.5275 moles
Now, the total moles of HCl taken = # mmoles HCl neutralized by antacid + # mmoles of excess HCl
42.2 = mmoles HCl neutralized by antacid + 0.5275
Therefore,
mmoles of HCl neutralized by antacid = 42.2 - 0.5275 = 41.6725 mmoles = 41.7 mmoles
Answer:
Explanation:
we know that specific heat is the amount of heat required to raise the temperature of substance by one degree mathmeticaly
Q=mcΔT
ΔT=T2-T1
ΔT=26.8-10.2=16.6
C for water is 4.184
therefore
Q=1.00*4.184*16.6
Q=69.4 j
now we have to covert joule into calorie
1 calorie =4.2 j
x calorie=69.4 j/2
so 69.4 j =34.7 calorie thats why 34.7 calorie heat is required to raise the temperature of water from 10.2 to 26.8 degree celsius