The grams of Fe₂O₃ that are formed is 47.68 g
<u><em>calculation</em></u>
Step 1: write the equation for reaction
4 Fe +3O₂ → 2 Fe₂O₃
Step 2: find the moles of Fe
moles = mass÷ molar mass
= 33.4 g÷55.8 g/mol =0.5986 moles
Step 3 : use the mole ratio to determine the moles of Fe₂O₃
That is from equation above Fe:Fe₂O₃ is 4:2 therefore the moles of Fe₂O₃ is = 0.5986 moles x 2/4 =0.2993 moles
Step 4 : find the mass of Fe₂O₃
mass = mass x molar mass
The molar mass of Fe₂O₃ = (55.8 x 2 +(15.9 x3) = 159.3 g/mol
mass is therefore = 0.2993 moles x 159.3 g/mol =47.68 g
Answer:
48.75 g of AgCl
11.60 g of H₂S
Solution:
The Balance Chemical Equation is as follow,
Ag₂S + HCl → AgCl + H₂S
<u>Calculate amount of AgCl produced</u><u>,</u>
According to equation,
247.8 g (1 mol) of Ag₂S produces = 143.32 g (1 mol) of AgCl
So,
84.3 g of Ag₂S will produce = X g of AgCl
Solving for X,
X = (84.3 g × 143.32 g) ÷ 247.8 g
X = 48.75 g of AgCl
<u>Calculate amount of H</u><u>₂</u><u>S produced</u><u>,</u>
According to equation,
247.8 g (1 mol) of Ag₂S produces = 34.1 g (1 mol) of H₂S
So,
84.3 g of Ag₂S will produce = X g of H₂S
Solving for X,
X = (84.3 g × 34.1 g) ÷ 247.8 g
X = 11.60 g of H₂S
Answer:411,000 Grams
Explanation: 411 x 100=411,000
Answer:
There are 5 significant digits in 0.23100.
Explanation:
This is because all non-zero digits are considered significant and zeros after decimal points are considered significant.
Answer:
using higher concentration of the nucleophile
Explanation:
In SN2 reaction, the attack of the nucleophile on the substrate occurs simultaneously as the leaving group departs. The entering group normally attacks through the back side of the molecule. The reaction is concerted and bimolecular. This implies that the concentration of the nucleophile is important in the rate equation for the reaction. Hence increasing the concentration of the nucleophile will increase the rate of SN2 reaction.
