Answer:
36.66%
Explanation:
Step 1: Given data
- Mass of the sample: 2.875 g
Step 2: Calculate the mass of salt
The mass of the sample is equal to the sum of the masses of the components.
m(sample) = m(iron) + m(sand) + m(salt)
m(salt) = m(sample) - m(iron) - m(sand)
m(salt) = 2.875 g - 0.660 g - 1.161 g
m(salt) = 1.054 g
Step 3: Calculate the percent of salt in the sample
We will use the following expression.
%(salt) = m(salt) / m(sample) × 100%
%(salt) = 1.054 g / 2.875 g × 100% = 36.66%
The combustion reaction of octane is as follow,
C₈H₁₈ + 25/2 O₂ → 8 CO₂ + 9 H₂O
According to balance equation,
8 moles of CO₂ are released when = 114.23 g (1 mole) Octane is reacted
So,
6.20 moles of CO₂ will release when = X g of Octane is reacted
Solving for X,
X = (114.23 g × 6.20 mol) ÷ 8 mol
X = 88.52 g of Octane
Result:
88.52 g of Octane is needed to release 6.20 mol CO₂.
They are atoms. a molecule is made up of atoms.
Answer:
B
Explanation:
The main aim of lab rules is to minimize the no of accidents
Acid base reactions also known as neutralization reactions are reactions that occur between acids and bases thus the name. Anytime this happens the product will niether an acid or base it will be neutralized to a neutral pH as in that of pure water pH 7.
Precipitation reactions occur when cations and anions in aqueous solution combine to form an insoluble ionic solid called a precipitate. Whether or not such a reaction occurs can be determined by using the solubility rules for common ionic solids.
