A melting point of a substance is a point at which the sample or substance start converting in liquid. For most substances, melting and freezing points are approximately equal. For example, the melting point and freezing point of mercury is 234.32 kelvins (−38.83 °C or −37.89 °F). Hope this helped!! :)
hi im breanna
Answer:
The mole is simply a very large number that is used by chemists as a unit of measurement.
Explanation:
The mole is simply a very large number,
6.022
×
10
23
, that has a special property. If I have
6.022
×
10
23
hydrogen atoms, I have a mass of 1 gram of hydrogen atoms . If I have
6.022
×
10
23
H
2
molecules, I have a mass of 2 gram of hydrogen molecules. If I have
6.022
×
10
23
C
atoms, I have (approximately!) 12 grams.
The mole is thus the link between the micro world of atoms and molecules, and the macro world of grams and litres, the which we can easily measure by mass or volume. The masses for a mole of each element are given on the periodic table as the atomic weight. So, if have 12 g of
C
, I know, fairly precisely, how many atoms of carbon I have. Given this quantity, I know how many molecules of
O
2
are required to react with the
C
, which I could measure by mass or by volume.
Answer:
725.15 L
Explanation:
The balanced chemical equation for the reaction between Na₂O₂ and CO₂ is the following:
Na₂O₂ + CO₂ → Na₂CO₃ + 1/2 O₂
From the stoichiometry of the reaction, 1 mol of Na₂O₂ reacts with 1 mol CO₂. So, the stoichiometric ratio is 1 mol Na₂O₂/1 CO₂.
Now, we convert the mass of reactants to moles by using the molecular weight (Mw) of each compound:
Mw (Na₂O₂) = (23 g/mol x 2) + (16 g/mol x 2) = 78 g/mol
moles Na₂O₂ = 96.7 g/(78 g/mol) = 1.24 mol Na₂O₂
Mw (CO₂) = 12 g/mol + (16 g/mol x 2) = 44 g/mol
moles CO₂ = 0.0755 g/(44 g/mol) = 1.71 x 10⁻³ mol CO₂
Now, we calculate the number of moles of CO₂ we need to completely react with the mass of Na₂O₂ we have:
1.24 mol Na₂O₂ x 1 mol CO₂/1 mol Na₂O₂ = 1.24 mol CO₂
In 1 L of respired air we have 1.71 x 10⁻³ mol CO₂ (0.0755 g), so we need the following number of liters to have 1.24 mol of CO₂:
1.24 mol CO₂ x 1 L/1.71 x 10⁻³ mol CO₂ = 725.15 L
