Answer:
9.51 × 10⁴ kL
Explanation:
Step 1: Given data
Volume of the sample (V): 9.51 × 10⁹ cL
Step 2: Convert "V" to liters
We will use the conversion factor 1 L = 100 cL.
9.51 × 10⁹ cL × (1 L / 100 cL) = 9.51 × 10⁷ L
Step 3: Convert "V" to kL
We will use the conversion factor 1 kL = 1000 L.
9.51 × 10⁷ L × (1 kL / 1000 L) = 9.51 × 10⁴ kL
9.51 × 10⁹ cL is equal to 9.51 × 10⁴ kL.
Answer:
The main use for hydrogen sulfide is in the production of sulfuric acid and elemental sulfur. ... H2S is used to prepare the inorganic sulfides you need to make those products. As a reagent and intermediate, hydrogen sulfide is beneficial because it can prepare other types of reduced sulfur compounds.
Answer:
Explanation:
Given data:
Mass of N₂ = 48.7 g
Moles of N₂ = ?
Molecules of N₂ = ?
Mass of N₂O = ?
Moles of N₂O = ?
Atoms of N₂O = ?
Solution:
Chemical equation:
2N₂ + O₂ → 2N₂O
Number of moles of N₂:
Number of moles = mass/ molar mass
Number of moles = 48.7 g / 28 g/mol
Number of moles = 1.74 mol
Number of molecules of N₂:
1 mole = 6.022 × 10²³ molecules
1.74 mol× 6.022 × 10²³ molecules/1 mol
10.5 × 10²³ molecules
Now we will compare the moles of N₂ with N₂O from balance chemical equation:
N₂ : N₂O
2 : 2
1.74 : 1.74
Moles of N₂O:
1.74 mol
Mass of N₂O:
Mass = number of moles × molar mass
Mass = 1.74 mol × 44 g/mol
Mass = 76.56 g
Sucrose is an example of a non- reducing sugar.
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