Answer:
Concentration of sulfuric acid in the acid rain sample is 0.0034467 mol/L.
Explanation:
Volume of NaOH = 1.7 ml = 0.0017 L
Molarity of NaOH = 0.0811 M
Moles of NaOH = n
n = 0.0001378 mol
According to reaction, 2 mol of NaOH neutralize 1 mol of sulfuric acid.
Then 0.0001378 mol of NaOH will neutralize:
of sulfuric acid.
Concentration of sulfuric acid in the acid rain sample: x
Concentration of sulfuric acid in the acid rain sample is 0.0034467 mol/L.
Answer:
Ne, Ar, and Kr are gases at STP, unreactive, and are generally monatomic.
Explanation:
they are unreactive and monoatomic and thats why have a very low boiling point.
Answer:
T₁ = 39 K
Explanation:
Given data:
Initial pressure = 1023.6 kpa
Final pressure = 8114 kpa
Final temperature = 36°C (36+ 273= 309K)
Initial temperature = ?
Solution:
P₁/T₁ = P₂/T₂
T₁ = P₁×T₂ /P₂
T₁ = 1023.6 kpa × 309 K /8114 kpa
T₁ = 316292.4 K. Kpa /8114 kpa
T₁ = 39 K
Thus original pressure was 39 k.
The ideal gas under STP is 22.4 L/mol. While the gas has a rule of P1V1/T1=P2V2/T2. So the volume under 101 kPa and 273 K is 0.2*22.4=4.48 L.
The answer is C . Watery feel
