Using stoichiometry:
5.5 L of blood x (1000 mL/1L) x (15 g/100 mL) x (1 kg/1000 g) = 0.825 kg
Zero is the value of water potential of pure water at atmospheric pressure.
4.8 g/cm3 with sig figs since it's mass/volume you divide 76 grams by 16 cm3
By definition, one mole (one gram molecular weight) of any substance, contains Avogadro’s number of particles; atoms if you are discussing an element, or molecules if a compound. Avogadro’s number has been determined by several methods, all of the accepted values lie within a range of +-1% about the value of 6.022045 x 10^23/gm. That is a large number, in this case approximately; 602,204,500,000,000,000,000,000 molecules of glucose.
From the web :v
You must use 1880 mL of O₂ to react with 4.03 g Mg.
A_r: 24.305
2Mg + O₂ ⟶ 2MgO
<em>Moles of Mg</em> = 4.03 g Mg × (1 mol Mg/24.305 g Mg) = 0.1658 mol Mg
<em>Moles of O₂</em> = 0.1658 mol Mg × (1 mol O₂/2 mol Mg) = 0.082 90 mol O₂
STP is 25 °C and 1 bar. At STP, 1 mol of an ideal gas has a volume of <em>22.71 L</em>.
<em>Volume of O₂</em> = 0.082 90 mol O₂ × (22.71 L O₂/1 mol O₂) = 1.88 L = 1880 mL
