Answer:
0.047 %
Explanation:
Step 1: Given data
- Partial pressure of ozone (pO₃): 0.33 torr
- Total pressure of air (P): 695 torr
Step 2: Calculate the %v/v of ozone in the air
Air is a mixture of gases. We can find the %v/v of ozone (a component) in the air (mixture) using the following expression.
<em>%v/v = pO₃/P × 100%</em>
%v/v = 0.33 torr/695 torr × 100%
%v/v = 0.047 %
Answer:
one in a 2s orbital
Explanation:
Because of the peak near the nucleus in the 2s curve there is a higher probability of finding a 2s within 4 Å of the nucleus. In a multi-electron atom an electron in a 2s orbital will have a lower energy than one in a 2p orbital
First, let us find the corresponding amount of moles H₂ assuming ideal gas behavior.
PV = nRT
Solving for n,
n = PV/RT
n = (6.46 atm)(0.579 L)/(0.0821 L-atm/mol-K)(45 + 273 K)
n = 0.143 mol H₂
The stoichiometric calculations is as follows (MW for XeF₆ = 245.28 g/mol)
Mass XeF₆ = (0.143 mol H₂)(1 mol XeF₆/3 mol H₂)(245.28 g/mol) = <em>11.69 g</em>
<span>There
are a number of ways to express concentration of a solution. This includes
molarity. Molarity is expressed as the number of moles of solute per volume of
the solution. We calculate the mass of the solute by first determining the number of moles needed. And by using the molar mass, we can convert it to units of mass.
Moles </span>(nh4)3po4 = 0.250 L (0.150 M) = 0.0375 moles (nh4)3po4
Mass = 0.0375 mol (nh4)3po4 (149.0867 g / mol) = 5.59 g (nh4)3po4
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