Answer:
empirical formula = C3H7
molecular formula = C6H14
3.0 × 10¹¹ RBC's (or) 3E11 RBC's
Solution:
Step 1: Convert mm³ into L;
As,
1 mm³ = 1.0 × 10⁻⁶ Liters
So,
0.1 mm³ = X Liters
Solving for X,
X = (0.1 mm³ × 1.0 × 10⁻⁶ Liters) ÷ 1 mm³
X = 1.0 × 10⁻⁷ Liters
Step 2: Calculate No. of RBC's in 5 Liter Blood:
As given
1.0 × 10⁻⁷ Liters Blood contains = 6000 RBC's
So,
5.0 Liters of Blood will contain = X RBC's
Solving for X,
X = (5.0 Liters × 6000 RBC's) ÷ 1.0 × 10⁻⁷ Liters
X = 3.0 × 10¹¹ RBC's
Or,
X = 3E11 RBC's
The first answer is B and the second answer is B
Answer:
The molar mass of lysine using the ideal gas equation for this problem is 146.25 g/mole.
Explanation:
The ideal gas equation PV = nRT, was derived from the ABC laws (Avogadros, Boyles and Charles laws). We need to obtain the value for the number of moles n.
The parameters of this equation are:
P = 1.918 atm
V = 750.0mL = 0.75L
n = ?
R = 0.0821
T = 25 degree celcius = 25 + 273 = 298 degree kelvin.
From this formular, n = (PV)/(RT)
n = (1.918 X 0.75)/(0.0821 X 298 )
n = 0.0588
n, no of mole = mass/molar mass
0.0588 = 8.6/MM
MM = 8.6/0.0588
MM = 146.25g/mole.
I'm not sure how many sign fig's you are required to have.
However I think the final answer would be 0.05 Moles, because of the .5g, that is considered 1 sign fig.
