What conversion factor will change moles co2 to grams co2? moles h2o to molecules h2o

See explanation

Explanation:

To do this, let's part with an example. Let's suppose we have 3.6 moles of CO₂ and 2.5 moles of H₂O.

To do the conversion of moles to mass, you can either use a conversion factor or the formula which is:

m = n * MM (1)

Where:

n: moles

MM: molar mass of compound.

From the above formula we can have a conversion factor. First, the reported molar mass of CO₂ is 44 g/mol

So with the above formula, we just need to replace and we get the conversion factor:

m = 3.6 moles * 44 g/moles  ---> cancel moles:

m = 158.4 g

And this is how we convert moles to mass.

For the case of moles to molecules, we use the avogadro number which is 6.02x10²³. And we know with this that in 1 mole of any compound we have 6.02x10²³ atoms or molecule, so:

1 mole > 6.02x10²³ molecules

the conversion factor is the above, then to convert the moles of water:

molecules of water = 2.5 moles * 6.02x10²³ molecules/mole

molecules of water = 1.505x10²⁴ molecules

And this is how we convert moles to molecules

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

explanation: