The moments immediately after the universe began were in some ways similar the
1 answer:
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Answer: The bonds are intermediate between double and single bonds
Explanation:
A closer look at the diagram below shows that the bonds in sulphur IV oxide are intermediate between double and single bonds. Hence they do not have the exact bond angle of single bonds. This is why the bond angle is not exactly 120°. There are two resonance structures in the diagram that clearly show this point.
1. The molarity of the stock solution of luminol = 1,431 M
2. 0.12 moles of luminol is present in 2.00 L of the diluted spray
3. The volume of the stock solution (Part A) would contain the number of moles present in the diluted solution (Part B): 83.86 ml
<h3>Further explanation</h3>
Stoichiometry in Chemistry studies about chemical reactions mainly emphasizing quantitative, such as the calculation of volume, mass, amount, which is related to the number of ions, molecules, elements, etc.
In stoichiometry included :
- 1. Relative atomic mass
- 2. Relative molecular mass
is the relative atomic mass of the molecule
- 3. mole
1 mole is the number of particles contained in a substance with the same number of atoms in 12 gr C-12
1 mole = 6.02.10²³ particles
While the number of moles can also be obtained by dividing the mass (in grams) by the relative mass of the element or the relative mass of the molecule
Luminol (C₈H₇N₃O₂) is a substance used to detect traces of blood in the scene of a crime because it reacts with iron in the blood
a. a luminol stock solution by adding 19.0g of luminol into a total volume of 75.0mL of H2O.
So the molarity is
- 1. Luminol mole
- 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
so the mole:
mol = gram / relative molecular mass
mole = 0.1073
2. Molarity (M)
M = mole / volume
M = 1,431
- b. luminol concentration in a spray bottle 6.00 × 10⁻² M. So that in 2 L of solution, the number of moles is:
mole = M x volume
mole = 6.10⁻² x 2
mole = 0.12
- c. Molarity of the stock solution (Part A) = 1,431 M
the number of moles present in the diluted solution (Part B) = 0.12
So the volume of the stock solution (Part A) would contain the number of moles present in the diluted solution (Part B) is:
volume = mol / M
volume = 0.08386 L = 83.86 ml
<h3>Learn more</h3>moles of water you can produce
the number of each atom present in the compound's formula
the ratio of hydrogen atoms (H) to oxygen atoms (O) in 2 l of water
Keywords: mole, volume, molarity, Luminol, the relative molecular mass
