94.20 g/3.16722 mL = 29.74 g/mL
The ratio of mass to volume is equal to the substance's density. Thus, 29.74 g/mL is the density of whatever substance it may be. Density does not change for incompressible matter like solid and some liquids. Although, it may be temperature dependent.
The percent A by mass for substance AB =<u> 75%</u>
<h3>Further explanation</h3>
Proust states the Comparative Law that compounds are formed from elements with the same Mass Comparison, so that compounds have a fixed composition of elements
Empirical formula is the mole ratio of compounds forming elements.
From Substance AB₂ is 60.0% A by mass.
Let's say that AB₂ mass = 100 gram, then
mass A = 60 gram
mass B = 40 gram : 2 (coefficient in compound AB₂ = 2) = 20 gram
In compound AB:
Total mass = mass A + mass B
Total mass = 60 + 20 grams = 80 grams
Then the percentage of compound A = (60: 80) = 75%
<h3>Learn more</h3>
Grams of KO₂ needed to form O₂
brainly.com/question/2823257
Keywords : percent mass, substance
#LearnwithBrainly
1) Balanced chemical reaction:
2H2 + O2 -> 2H20
Sotoichiometry: 2 moles H2: 1 mol O2 : 2 moles H2O
2) Reactant quantities converted to moles
H2: 5.00 g / 2 g/mol = 2.5 mol
O2: 50.0 g / 32 g/mol = 1.5625 mol
Limitant reactant: H2 (because as per the stoichiometry it will be consumed with 1.25 mol of O2).
3) Products
H2 totally consumed -> 0 mol at the end
O2 = 1.25 mol consumed -> 1.5625 mol - 1.25 mol = 0.3125 mol at the end
H2O: 2.5 mol H2 produces 2.5 mol H2O -> 2.5 mol at the end.
Total number of moles: 0.3125mol + 2.5 mol = 2.8125 mol
4) Pressure
Use pV = nRT
n = 2.8125
V= 9 liters
R = 0.082 atm*lit/K*mol
T = 35 C + 273.15 = 308.15K
p = nRT/V = 7.9 atm
The chemical formula for ammonia is NH3. So first, you need to find the molar mass of ammonia (how many grams in one mole).
N=14g
H3=3g
So one mole of NH3 is 17 grams, you can divide 82.9 grams by 17 grams to find the number of molecules. The answer should be 4.876 moles (molecules) of ammonia. Hope this helps!
Answer:
9.1
Explanation:
Step 1: Calculate the basic dissociation constant of propionate ion (Kb)
Sodium propionate is a strong electrolyte that dissociates according to the following equation.
NaC₃H₅O₂ ⇒ Na⁺ + C₃H₅O₂⁻
Propionate is the conjugate base of propionic acid according to the following equation.
C₃H₅O₂⁻ + H₂O ⇄ HC₃H₅O₂ + OH⁻
We can calculate Kb for propionate using the following expression.
Ka × Kb = Kw
Kb = Kw/Ka = 1.0 × 10⁻¹⁴/1.3 × 10⁻⁵ = 7.7 × 10⁻¹⁰
Step 2: Calculate the concentration of OH⁻
The concentration of the base (Cb) is 0.24 M. We can calculate [OH⁻] using the following expression.
[OH⁻] = √(Kb × Cb) = √(7.7 × 10⁻¹⁰ × 0.24) = 1.4 × 10⁻⁵ M
Step 3: Calculate the concentration of H⁺
We will use the following expression.
Kw = [H⁺] × [OH⁻]
[H⁺] = Kw/[OH⁻] = 1.0 × 10⁻¹⁴/1.4 × 10⁻⁵ = 7.1 × 10⁻¹⁰ M
Step 4: Calculate the pH of the solution
We will use the definition of pH.
pH = -log [H⁺] = -log 7.1 × 10⁻¹⁰ = 9.1
