Nitric acid + mg(no3)35.0 ml of 0.255 m nitric acid is added to 45.0 ml of 0.328 m mg(no3)2. what is the concentration of nitrat
2 answers:
Answer: The accurate option is a.
The nitrate ion concentration in the final mixture is approximately 0.48056 M.
Explanation:
Calculating nitrate ions from 35.0 mL of 0.255 M nitric acid:
Calculated moles of nitric acid = 0.008925 mol
1 mole of nitric acid yields 1 mole of nitrate ions,
Therefore, 0.008925 moles of nitric acid will produce 0.008925 moles of nitrate ions.
Resulting in moles of nitrate in 35.00 mL of the solution = 0.008925 mole..(1)
Calculating nitrate ions from 45.0 mL of 0.328 M magnesium nitrate:
Calculated moles of magnesium nitrate = 0.01476 mol
1 mole of magnesium nitrate generates 2 moles of nitrate ions,
Hence, 0.01476 moles of magnesium nitrate results in:
of nitrate ions
Thus, moles of nitrate ions in 45.00 mL of solution = 0.02952 mol...(2)
Total moles of nitrate ions =
0.008925 mol+0.02952 mol = 0.038445 mol
Total volume after mixing = 35.00 mL+ 45.00 mL =
80.00 mL = 0.080 L
Final solution's nitrate ion concentration:
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Answer:
1. Ionic compound- 
2. Polar molecular compound- 
Explanation:
Magnesium (Mg), with atomic number 12, has an electron configuration of . The outermost shell possesses 2 electrons, thus it loses these 2 electrons to become
ions. Bromine (Br), a nonmetal with atomic number 35, has an electron configuration of
. Its outermost shell holds 7 electrons, allowing it to accept one electron and thus forms
. Hence, the magnesium ion and bromide ion bond together to form an ionic compound
.
Phosphorus (P), also a nonmetal, bonds with bromine covalently. Due to differing electronegativities, they produce polar covalent compounds like .
1. The luminol stock solution has a molarity of 1.431 M.
2. In 2.00 L of the diluted spray, there are 0.12 moles of luminol.
3. The volume of the stock solution from Part A that contains the same number of moles present in the diluted solution from Part B is 83.86 ml.
Additional Information
Stoichiometry in Chemistry focuses on the quantitative aspects of chemical reactions, which includes calculations related to volume, mass, and the count of ions, molecules, and elements.
Key concepts in stoichiometry include:
- 1. Relative atomic mass
- 2. Relative molecular mass
This refers to the relative atomic mass of a molecule.
- 3. Mole
A mole represents the number of particles in a substance equivalent to the number of atoms in 12 grams of carbon-12.
1 mole = 6.02 × 10²³ particles.
The quantity of moles can also be derived by dividing mass (in grams) by either the relative mass of an element or the relative mass of a molecule.
Luminol (C₈H₇N₃O₂) is utilized for detecting blood traces at crime scenes, due to its reaction with iron found in blood.
To prepare a luminol stock solution, 19.0 g of luminol is mixed into a total volume of 75.0 mL of water.
Thus, the molarity is calculated as:
- 1. Moles of Luminol
- the relative molecular mass of Luminol:
= 8.C + 7.H + 3.N + 2.16
= 8.12 + 7.1 + 3.14 + 2.16
= 177 grams/mol.
Thus, we have:
moles = grams / relative molecular mass.
moles = 0.1073.
2. Molarity (M)
M = moles / volume
M = 1.431.
- b. The concentration of luminol in the spray bottle is 6.00 × 10⁻² M. Therefore, in a 2 L solution, the number of moles is:
moles = M × volume
moles = 6 × 10⁻² × 2
moles = 0.12.
- c. The molarity of the stock solution (Part A) is 1.431 M.
The diluted solution (Part B) contains 0.12 moles of luminol.
To find the volume of the stock solution (Part A) that has the same moles as the diluted solution (Part B):
volume = moles / M
volume = 0.08386 L = 83.86 mL.
Further Learning
moles of water you can generate
the amount of each atom in the chemical's formula
the proportion of hydrogen to oxygen atoms in 2 L of water
Keywords: mole, volume, molarity, Luminol, relative molecular mass
