Answer:
the attraction water molecules have to other surfaces
Explanation:
Adhesion is defined as the attractive forces between unlike substances, e.g water moving up a capillary tube.
Adhesion is the tendency of dissimilar particles or surfaces to cling to one another(Wikipedia).
So, what we mean by adhesion in this context, is the attraction water molecules to other surfaces.
Carbon dioxide is the atmospheric compound that is
responsible (after it have dissolved in water) for both the general acidity of
normal rainwater and the acidity of much soil. Natural rain water is slightly
acidic because carbon dioxide combined with water in the air forms carbonic
acid, which is a weak acid.
Answer:
15.9 g
Explanation:
(Take the atomic mass of C=12.0, H=1.0, O=16.0)
no. of moles = mass / molar mass
no. of moles of octane used = 11.2 / (12.0x8 + 1x18)
= 0.0982456 mol
Since oxygen is in excess and octane is the limiting reagent, the no. of moles of H2O depends on the no. of moles of octane used.
From the balanced equation, the mole ratio of octane : water = 2:18 = 1: 9,
so this means, one mole of octane produced 9 moles of water.
Using this ratio, we can deduce that (y is the no. of moles of water produced):
y = 0.0982456x9
y= 0.88421 mol
Since mass = no. of moles x molar mass,
mass of water produced = 0.88421 x (1.0x2+16.0)
=15.9 g
Balance the chemical equation for the chemical reaction.
Convert the given information into moles.
Use stoichiometry for each individual reactant to find the mass of product produced.
The reactant that produces a lesser amount of product is the limiting reagent.
The reactant that produces a larger amount of product is the excess reagent.
To find the amount of remaining excess reactant, subtract the mass of excess reagent consumed from the total mass of excess reagent given.
Answer:
2
Step-by-step explanation:
A. Moles before mixing
<em>Beaker I:
</em>
Moles of H⁺ = 0.100 L × 0.03 mol/1 L
= 3 × 10⁻³ mol
<em>Beaker II:
</em>
Beaker II is basic, because [H⁺] < 10⁻⁷ mol·L⁻¹.
H⁺][OH⁻] = 1 × 10⁻¹⁴ Divide each side by [H⁺]
[OH⁻] = (1 × 10⁻¹⁴)/[H⁺]
[OH⁻] = (1 × 10⁻¹⁴)/(1 × 10⁻¹²)
[OH⁻] = 0.01 mol·L⁻¹
Moles of OH⁻ = 0.100 L × 0.01 mol/1 L
= 1 × 10⁻³ mol
B. Moles after mixing
H⁺ + OH⁻ ⟶ H₂O
I/mol: 3 × 10⁻³ 1 × 10⁻³
C/mol: -1 × 10⁻³ -1 × 10⁻³
E/mol: 2 × 10⁻³ 0
You have more moles of acid than base, so the base will be completely neutralized when you mix the solutions.
You will end up with 2 × 10⁻³ mol of H⁺ in 200 mL of solution.
C. pH
[H⁺] = (2 × 10⁻³ mol)/(0.200 L)
= 1 × 10⁻² mol·L⁻¹
pH = -log[H⁺
]
= -log(1 × 10⁻²)
= 2
