Answer:
molecular weight of C12H22O11 or grams This compound is also known as Lactose or Sucrose or Maltose. The SI base unit for amount of substance is the mole. 1 mole is equal to 1 moles C12H22O11, or 342.29648 grams.
Explanation:
Answer:
5.89 g × 10⁶ μg
Explanation:
Step 1: Convert 5.00 scruples to grains
We will use the conversion factor 1 scruple = 20 grains.
5.00 scruple × 20 grain/1 scruple = 100 grain
Step 2: Convert 100 grains to ounces
We will use the conversion factor 1 oz = 480 grains.
100 grain × 1 oz/480 grain = 0.208 oz
Step 3: Convert 0.208 oz to grams
We will use the conversion factor 1 oz = 28.34 g.
0.208 oz × 28.34 g/1 oz = 5.89 g
Step 4: Convert 5.89 g to micrograms
We will use the conversion factor 1 g = 10⁶ μg.
5.89 g × 10⁶ μg/1 g = 5.89 g × 10⁶ μg
This is false
А
Because they are compounds, they cannot be pure substances.
Because:Compounds are pure substance
Answer:
± 1 or ± 2
Explanation:
Electrovalent bonds are chemical bonds that are established on the premise of transferring electrons between two atoms.
In this bond type, a higly electronegative atom, typically a non-metal receives electrons from an atom with lesser electronegativity, a metal.
To know the number of electrons involved in forming electrovalent bonds, we typically look at the groups of atoms that combines to form the bond.
Metals are found in group I and II on the periodic table. Metals are electropositive and are good electron donors. These metals have 1 and 2 electrons in their valence shell respectively. In like manners, the more electronegative atoms are found in group VI and VII. The elements in these groups are non-metals with high electronegativity and requires just 1 and 2 electrons to complete their octet.
The chemical reaction is:
<span>SO2Cl2 </span>⇔<span> SO2+ Cl2
This shows that the forward reaction the compound is decomposed to smaller compounds and the backward reaction the larger compound is produced. Adding more chlorine to the mixture when in equilibrium, will shift the equilibrium to the left forming the larger compound.</span>
