Question

question 3 (2 points)
how many moles of oxygen are in 100. l at 1.00 atm and 273.15 k?
there are two ways to solve this problem.
this is a missing variable type of problem so we can use the equation 1.
also, since the conditions are stp, we can convert 100. l into moles using 2 as
the conversion factor.
a. missing variable b. change in conditions c. density d. molar volume at stp
e. partial pressure f. 1 mol = 22.4 l g. d = m/v h. pv = nrt
i. p₂v₂/n₂t₂ = p₁v₁/n₁t₁ j. p_total = p₁ + p₂ + …… k. p l. p₁ m. p₂
n. v o. v₁ p. v₂ q. n r. n₁ s. n₂ t. r u. t
v. t₁ w. t₂ x. 1.25 y. 12.5 z. 125

question 4 (2 points)
what is the density of chlorine gas at stp? the molar mass of cl₂ is 70.9 g.
this is a 1 type of problem so we use the equation 2.

Explanation:

Step1: Identify the formula for the first blank

Since it's a missing - variable problem (finding the number of moles \(n\) when \(P\), \(V\), and \(T\) are known), we use the ideal gas law \(PV=nRT\).

Step2: Identify the conversion factor for the second blank

At STP (\(P = 1\ atm\), \(T=273.15\ K\)), the molar volume of a gas is \(22.4\ L/mol\). So, the conversion factor is \(1\ mol = 22.4\ L\).

Answer:

1. H. \(PV = nRT\)
2. F. \(1\ mol=22.4\ L\)