This problem is being solved using Ideal Gas Equation.
PV = nRT
Data Given:
Initial Temperature = T₁ = 27 °C = 300 K
Initial Pressure = P₁ = constant
Initial Volume = V₁ = 8 L
Final Temperature = T₂ = 78 °C = 351 K
Final Pressure = P₂ = constant
Final Volume = V₂ = ?
As,
Gas constant R and Pressures are constant, so, Ideal gas equation can be written as,
V₁ / T₁ = V₂ / T₂
Solving for V₂,
V₂ = (V₁ × T₂) ÷ T₁
Putting Values,
V₂ = (8 L × 351 K) ÷ 300 K
V₂ = 9.38 L
Answer:
Motion is defined as a change of position. How we perceive motion depends on our frame of reference. Frame of reference refers to something that is not moving with respect to an observer that can be used to detect motion.
Explanation:
Answer:
carbon
Explanation:
Carbon forms compounds that make up about 18 percent of all the matter in living things. The processes by which organisms consume carbon and return it to their surroundings constitute the carbon cycle.
Answer:
305 litres of NO gas will be produced from 916 L of NO₂
Explanation:
Given the balanced equation of the chemical reaction as follows:
3 NO₂ (g) + H₂O( l) —— 2 HNO₃ (l) + NO (g)
Under standard conditions, 3 moles of No₂ will react with 1 mole of water to produce 1 mole of NO gas.
Molar volume of a gas at STP is 22.4 L
Number of moles of NO₂ gas present in 916 L = 916/22.4 = 40.893 moles of NO₂ gas
From the mole ratio of NO₂ to NO in the equation of reaction,
Number of moles of NO that will be produced = 1/3 × 40.893 moles = 13.631 moles of NO gas
Volume of 13.631 moles of NO gas = 13.631 × 22.4
Volume of NO gas produced = 305.334L
Therefore, Volume of NO gas produced from the reaction of 916 L of NO₂ with water = 305 L
Answer:
2.5L [NaCl] concentrate needs to be 4.8 Molar solution before dilution to prep 10L of 1.2M KNO₃ solution.
Explanation:
Generally, moles of solute in solution before dilution must equal moles of solute after dilution.
By definition Molarity = moles solute/volume of solution in Liters
=> moles solute = Molarity x Volume (L)
Apply moles before dilution = moles after dilution ...
=> (Molarity X Volume)before dilution = (Molarity X Volume)after dilution
=> (M)(2.5L)before = (1.2M)(10.0L)after
=> Molarity of 2.5L concentrate = (1.2M)(10.0L)/(2.5L) = 4.8 Molar concentrate
