Street: 1. Cars 2. People 3. Wind in the trees 4. Birds 5. Car horn
School: 1. School bell 2. Students laughing 3. Teacher talking 4. Footsteps 5. Lockers
Church: 1. Priest talking 2. People talking 3. Church music 4. Kids crying 5. Benches creaking
I'd say it's partially dissociating into products
Answer:
V₂ = 116126.75 cm³
Explanation:
Given data:
Radius of balloon = 15 cm
Initial pressure = 2 atm
Initial temperature = 35 °C (35 +273 = 308K)
Final temperature = -20°C (-20+273 = 253 K)
Final pressure = 0.3 atm
Final volume = ?
Formula:
P₁V₁/T₁ = P₂V₂/T₂
P₁ = Initial pressure
V₁ = Initial volume
T₁ = Initial temperature
P₂ = Final pressure
V₂ = Final volume
T₂ = Final temperature
Solution:
Initial volume of balloon:
V = 4/3πr³
V = 4/3×22/7×(15cm)³
V = 14137.17 cm³
V₂ = P₁V₁ T₂/ T₁ P₂
V₂ = 2 atm × 14137.17 cm³ × 253 K / 308 K × 0.3 atm
V₂ = 7153408.02 atm .cm³. K / 61.6 K.atm
V₂ = 116126.75 cm³
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
Answer:
SrO = Sr + O - Chemical Equation Balancer.
Explanation:
