Number of Atoms in Methanol are calculated as,
Write chemical formula of Methanol,
CH₃-OH
Find out Subscripts:
There is only one subscript (₃) written after Hydrogen, it means there are 3 Hydrogen atoms attached to Carbon atoms.
Add all atoms in molecule:
Carbons = 1
Hydrogens = 4
Oxygens = 1
Total Atoms ______
6 Atoms
Answer:
PubChem CID 6393
Structure
Find Similar Structures
Chemical Safety
Compressed Gas
Laboratory Chemical Safety Summary (LCSS) Datasheet
Molecular Formula CF4
Molecular Weight
88.004 g/mol
Explanation:
The molarity of the solution of H₃PO₄ needed to neutralize the KOH solution is 0.35 M
<h3>Balanced equation </h3>
H₃PO₄ + 3KOH —> K₃PO₄ + 3H₂O
From the balanced equation above,
- The mole ratio of the acid, H₃PO₄ (nA) = 1
- The mole ratio of the base, KOH (nB) = 3
<h3>How to determine the molarity of H₃PO₄ </h3>
- Volume of acid, H₃PO₄ (Va) = 10.2 mL
- Molarity of base, Ca(OH)₂ (Mb) = 0.2 M
- Volume of base, Ca(OH)₂ (Vb) = 53.5 mL
- Molarity of acid, H₃PO₄ (Ma) =?
MaVa / MbVb = nA / nB
(Ma × 10.2) / (0.2 × 53.5) = 1 / 3
(Ma × 10.2) / 10.7 = 1 / 3
Cross multiply
Ma × 10.2 × 3 = 10.7
Ma × 30.6 = 10.7
Divide both side by 30.6
Ma = 10.7 / 30.6
Ma = 0.35 M
Learn more about titration:
brainly.com/question/14356286
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Answer:
Radiation is energy that comes from a source and travels through space at the speed of light.
Answer:
Total worth of gold in the ocean = $5,840,000,000,000,000
Explanation:
As stated in the question above, 4.0 x 10^-10 g of gold was present in 2.1mL of ocean water.
Therefore, In 1 L of ocean water there will be,
(4.0 x 10^-10)/0.0021
= 1.9045 x 10^-7 g of gold per Liter of ocean water.
So in 1.5 x 10^-21 L of ocean water, there will be
(1.9045 x 10^-7) * (1.5 x 10^-21)
= 2.857 x 10^14 g of gold in the ocean.
1 gram of gold costs $20.44, that is 20.44 dollars/gram. The total cost of the gold present in the ocean is
20.44 * (2.857 x 10^14)
= $5,840,000,000,000,000
