The initial mass of sodium hydroxide is 3.3 g (answer C)
<u><em>calculation</em></u>
Step 1 : find the moles of iron (ii) hydroxide ( Fe(OH)₂
moles = mass÷ molar mass
from periodic table the molar mass of Fe(OH)₂ = 56 + [16 +1]2 = 90 g/mol
moles is therefore = 3.70 g÷ 90 g/mol = 0.041 moles
Step 2: use the mole ratio to calculate the moles of sodium hydroxide (NaOH)
from given equation NaOH : Fe(OH)₂ is 2 :1
therefore the moles of NaOH = 0.041 x 2 = 0.082 moles
Step 3: find mass of NaOH
mass = moles x molar mass
from the periodic table the molar mass of NaOH = 23 +16 +1 = 40 g/mol
mass = 0.082 moles x 40 g/mol = 3.3 g ( answer C)
A catalyst is a chemical substance that hastens the chemical reaction. This does not participates in the creating the product(s) but allows it to be formed easily. With this, it is now known that the rate of the reaction becomes relatively higher compared to the uncatalyzed reactions.
Therefore, the answer to this item is the rate of the reaction becomes faster.
<u>Answer:</u> The chemical equation is written below.
<u>Explanation:</u>
Every balanced chemical equation follows law of conservation of mass.
This law states that mass can neither be created nor be destroyed but it can only be transformed from one form to another form. This also means that total number of individual atoms on reactant side must be equal to the total number of individual atoms on the product side.
The chemical equation for the reaction of elemental boron and oxygen gas follows:

By Stoichiometry of the reaction:
4 moles of elemental boron reacts with 3 moles of oxygen gas to produce 2 moles of diboron trioxide.
The chemical equation for the reaction of diboron trioxide and water follows:

By Stoichiometry of the reaction:
1 mole of diboron trixoide reacts with 3 moles of water to produce 2 moles of boric acid.
Hence, the chemical equations are written above.
10.0gNaCl/2.0Lsolution= 5.0g/L
Answer:
a. The second run will be faster.
d. The second run has twice the surface area.
Explanation:
The rate of a reaction is proportional to the surface area of a catalyst. Given the volume (V) of a sphere, we can find its surface area (A) using the following expression.

The area of the 10.0 cm³-sphere is:

The area of each 1.25 cm³-sphere is:

The total area of the 8 1.25cm³-spheres is 8 × 5.61 cm² = 44.9 cm²
The ratio of 8 1.25cm³-sphere to 10.0 cm³-sphere is 44.9 cm²/22.4 cm² = 2.00
Since the surface area is doubled, the second run will be faster.