A neutralization titration is a chemical response this is used to decide the composition of an answer and what kind of acid or base is in it. This is a way of volumetric analysis and the formula is ().
Utilize the titration method of in view that we're given the concentrations of every compound and the quantity of . Let: M1 = 0.138M, V1 = x, M2 = 0.205M, V2 = 26.0 ML.
- M1 = initial mass
- V1= initial volume
- M2 = final mass
- V2= final volume
- (0.138)(V1) = (0.205)x(26.0)
- V2=(0.205)x(26.0)\ 0.138
- V2 = 47.10 M/L
- The final value of Volume needed for neutralization of nitric acid solution is V2 = 47.10 M/L
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Answer:
H =2; I = 2; J = 2
Explanation:
Carbon is element 6 in the Periodic Table.
Start at element 1 (H) and count from left to right until you reach element 6 (C).
You get the electron configuration
C: 1s² 2s²2p².
Thus,
H =2; I = 2; J = 2
Answer:
Explanation:
The first one is CrO. The Chromium has the same charge as the oxygen so mol numbers are dropped.
The Second one is CrO2 The two oxygens have a charge of 2(-2) = -4. To balance this, the Chromium must have a charge of +4 Cr(Iv)O2
The third one is can be set up like this
Cr + 3(-2) = 0
Cr - 6 = 0
Cr = 6
Therefore the formula is Cr(vi)O3
The last one is a bit tricky. Follow this carefully. There are 2 Crs and 3Os.
The formula looks like this
2Cr + 3(-2) = 0
2Cr - 6 = 0
2Cr = 6
Cr = 3
The formula is Cr(iii)2 O3
Answer:
MgCl2 + 2AgNO3 → 2AgCl + Mg(NO3)2
Explanation:
I'm assuming you want to balance it so...
The first thing I see is that there are two chlorines on the reactant side and one on the product side
Adding a coefficient of 2 would get 2AgCl2
Now there are two silvers on the reactant side, so add a 2 to AgNO3 on the products side. Now they are all balanced.
If that is not what you are looking for let me know!
Answer:
2.893 x 10⁻³ mol NaOH
[HCOOH] = 0.5786 mol/L
Explanation:
The balanced reaction equation is:
HCOOH + NaOH ⇒ NaHCOO + H₂O
At the endpoint in the titration, the amount of base added is just enough to react with all the formic acid present. So first we will calculate the moles of base added and use the molar ratio from the reaction equation to find the moles of formic acid that must have been present. Then we can find the concentration of formic acid.
The moles of base added is calculated as follows:
n = CV = (0.1088 mol/L)(26.59 mL) = 2.892992 mmol NaOH
Extra significant figures are kept to avoid round-off errors.
Now we relate the amount of NaOH to the amount of HCOOH through the molar ratio of 1:1.
(2.892992 mmol NaOH)(1 HCOOH/1 NaOH) = 2.892992 mmol HCOOH
The concentration of HCOOH to the correct number of significant figures is then calculated as follows:
C = n/V = (2.892992 mmol) / (5.00 mL) = 0.5786 mol/L
The question also asks to calculate the moles of base, so we convert millimoles to moles:
(2.892992 mmol NaOH)(1 mol/1000 mmol) = 2.893 x 10⁻³ mol NaOH