A tank at is filled with of dinitrogen difluoride gas and of carbon dioxide gas. You can assume both gases behave as ideal gases under these conditions. Calculate the mole fraction and partial pressure of each gas, and the total pressure in the tank. Be sure your answers have the correct number of significant digits. dinitrogen difluoride molar fraction: partial pressure: carbon dioxide mole fraction: partial pressure: Total pressure in tank:

For N₂F₂:

Molar fraction = 0.84

Partial pressure = 1.12 atm

For SF₄:

Molar fraction = 0.16

Partial pressure = 0.208 atm

Explanation:

It seems your question is missing the values required to solve the problem. However, an internet search showed me the following values for your question. If the values in your problem are different, your answer will be different as well, however the solving method will remain the same:

" A 5.00L tank at 0.7°C is filled with 16.5g of dinitrogen difluoride gas and 5.00g of sulfur tetrafluoride gas. You can assume both gases behave as ideal gases under these conditions. "

First we calculate the moles of each gas, using their molar mass:

Total mol number = 0.25 + 0.0463 = 0.2963 mol

Now we use PV=nRT to calculate the partial pressure of each gas:

P = ?

V = 5.00 L

T = 0.7 °C ⇒ 0.7 + 273.16 = 273.86 K

For N₂F₂:

For SF₄:

part 1: the half-life of the element is 22 years.

part 2: it will take 132.02 year for the sample to decay from 308.0 g to 4.8125 g.

explanation:

part 1: there are two ways to determine the half life time.

first way: the half life time can be determined directly from its definition as it is the time required for the sample to decay to its half concentration.

from the data the initial concentration was 45.0 g and decayed to its half value (22.5 g) in 22.0 years.

second way: the radio active decay is considered first order reaction so we can use the laws of first order reactions.

k = (1/t) ln (a/a-x), where k is the rate constant of the reaction,

t is the time of the reaction

a is the initial concentration of the reactant

a-x is the remaining concentration at time (t)

a = 45.0 g the initial concentration at t = 0 year.

at t = 22 year, a-x = 22.5 g

then k = (1/22 year) ln (45.0 g/ 22.5 g) = 0.0315 year-1.

the half-life time (t1/2) = ln 2/ k = 0.693/ k =0.693/ 0.0315 = 22 years.

part 2:

using the same law k = (1/t) ln (a/a-x),

using the given data; a = 308.0 g and a-x = 4.8125 g and the calculated k from the part 1 (k = 0.0315 year-1)

t = (1/k) ln (a/a-x) = (1/0.0315) ln (308.0/ 4.8125) = 132.02 years

a c d and f

just finished the assignment