Answer:
5.52cm³ of water will rise and might spill over the edge
Explanation:
Use the change in volume of a liquid with changing temperature equation which is written as
ΔV = β x V₀ x ΔT, where β is the coefficient of expansion, V₀ is the volume being submerged and ΔT is the difference in temperature
ΔV = (69 x 10⁻⁶) x (0.1 x 0.1 x 0.1) x (85 - 5)
ΔV = 5.52 x 10⁻⁶ m³
ΔV = 5.52cm³
Answer:
The answer to your question is below
Explanation:
H₂SO₄ = sulphuric acid
KOH = potassium hydroxide
This is a neutralization reaction
H₂SO₄ + 2 KOH ⇒ K₂SO₄ + 2H₂O
2 ----------- K ----------- 2
1 ----------- S ---------- 1
4 ---------- H --------- 4
6 ------------ O ---------- 6
C. dim light is <span>the answer
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Answer:
V₂ = 568.9 mL
Explanation:
Given data:
Volume of gas = 550 mL
Pressure of a gas = 960 mmHg
Temperature = 200.0°C ( 200+273 = 473 K)
Final volume = ?
Final pressure = 830 mmHg
Final temperature = 150°C (150+273 = 423 K)
Solution:
Formula:
P₁V₁/T₁ = P₂V₂/T₂
P₁ = Initial pressure
V₁ = Initial volume
T₁ = Initial temperature
P₂ = Final pressure
V₂ = Final volume
T₂ = Final temperature
Solution:
V₂ = P₁V₁ T₂/ T₁ P₂
V₂ = 960 mmHg × 550 mL × 423 K / 473 K ×830 mmHg
V₂ = 223344000 mL / 392590
V₂ = 568.9 mL
