The density of a sample of metal is calculated using these three different sets of data: 1.2 g/mL, 1.4 g/mL, and 1.1 g/mL. If th
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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
Read more about the neutralization:
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Answer:
The molarity is 0.56
Explanation:
In a mixture, the chemical present in the greatest amount is called a solvent, while the other components are called solutes. Then, the molarity or molar concentration is the number of moles of solute per liter of solution.
In other words, molarity is the number of moles of solute that are dissolved in a given volume.
The Molarity of a solution is determined by:
Molarity is expressed in units ().
Then you must know the number of moles of Cu(NO₂)₂. For that it is necessary to know the molar mass. Being:
- Cu: 63.54 g/mol
- N: 14 g/mol
- O: 16 g/mol
the molar mass of Cu(NO₂)₂ is:
Cu(NO₂)₂= 63.54 g/mol + 2*(14 g/mol + 2* 16 g/mol)= 155.54 g/mol
Now the following rule of three applies: if 155.54 g are in 1 mole of the compound, 225 g in how many moles are they?
moles= 1.45
So you know:
- number of moles of solute= 1.45 moles
- volume=2.59 L
Replacing in the definition of molarity:
Molarity= 0.56
<u><em>The molarity is 0.56</em></u><u><em></em></u>
Answer:- Formula of the hydrate is and it's name is Iron(III)sulfate pentahydrate.
Solution:- As per the given information, there is 18.4% water in the hydrate. If we assume the mass of the hydrate as 100 grams then there would be 18.4 grams of water and 81.6 grams of Iron(III)sulfate present in the hydrate.
Molar mass for Iron(III)sulfate is 399.88 gram per mol and the molar mass for water is 18.02 gram per mol.
We will calculate the moles of Iron(III)sulfate and water present in the compound on dividing their grams by their molar masses as:
=
=
Now, the next step is to calculate the mol ratio and for this we divide the moles of each by the least one of them means whose moles are less. Here, the moles of Iron(III)sulfate are less than moles of water. So, we divide the moles of each by 0.204.
= 1
= 5
There is 1:5 mol ratio between Iron(III)sulfate and water. So, the formula of the hydrate is and the name of the hydrate is Iron(III)sulfate pentahydrate.