Robert Boyle, the 17th century British chemist, first noticed that the volume of a given amount of gas is inversely proportional to its pressure when kept at a constant temperature. When working with ideal gases we use PV = nRT, but remember n, R, and T are all constant. Therefore we have:
PV(before) = PV(after)
P(0.5650) = (715.1)(1.204)
The answer is: 1.5 moles of oxygen are present.
V(O₂) = 33.6 L; volume of oxygen.
p(O₂) = 1.0 atm; pressure of oxygen.
T = 0°C; temperature.
Vm = 22.4 L/mol; molar volume at STP (Standard Temperature and Pressure).
At STP one mole of gas occupies 22.4 liters of volume.
n(O₂) = V(O₂) ÷ Vm.
n(O₂) = 33.6 L ÷ 22.4 L/mol.
n(O₂) = 1.50 mol; amount of oxygen.
Aqueous solutions of barium nitrate and potassium phosphate are mixed.
What is the precipitate and how many molecules are formed?
Barium nitrate has a chemical symbol of Ba(NO3)2 and potassium phosphate
has a chemical symbol K2PO4. The reaction between these two is a double
replacement reaction yielding barium phosphate and potassium nitrate.
The chemical equation representing the reaction is,
Ba(NO3)2 + K2PO4 à KNO3 +
BaPO4
4) is correct
This is because water is polar and it will mix with a polar solvent. A good rule for remembering the behavior of non-polar and polar compounds when it comes to being miscible is that "like dissolves like."
Answer:
The volume of the sample is 17.4L
Explanation:
The reaction that occurs requires the same amount of CO and NO. As the moles added of both reactants are the same you don't have any limiting reactant. The only thing we need is the reaction where 4 moles of gases (2mol CO + 2mol NO) produce 3 moles of gases (2mol CO2 + 1mol N2). The moles produced are:
0.1800mol + 0.1800mol reactants =
0.3600mol reactant * (3mol products / 4mol reactants) = 0.2700 moles products.
Using Avogadro's law (States the moles of a gas are directly proportional to its pressure under constant temperature and pressure) we can find the volume of the products:
V1n2 = V2n1
<em>Where V is volume and n moles of 1, initial state and 2, final state of the gas</em>
Replacing:
V1 = 23.2L
n2 = 0.2700 moles
V2 = ??
n1 = 0.3600 moles
23.2L*0.2700mol = V2*0.3600moles
17.4L = V2
<h3>The volume of the sample is 17.4L</h3>
