Answer is: volume of CO₂ is 0,113 dm³.
Ideal gas law = pV = nRT.
p = 850 PSI = 5860543,6992 Pa.
Psi <span>is the abbreviation of pound per square inch.
T = 21</span>°C = 294,15 K.
n = 0,273 mol.
R = 8,314 J/K·mol.
V = nRT ÷ p
V = 0,273 mol · 8,314 J/K·mol · 294,15 K ÷ 5860543,6992 Pa.
V = 0,00011 m³ = 0,113 dm³.
Answer:
We need 78.9 mL of the 19.0 M NaOH solution
Explanation:
Step 1: Data given
Molarity of the original NaOH solution = 19.0 M
Molarity of the NaOH solution we want to prepare = 3.0 M
Volume of the NaOH solution we want to prepare = 500 mL = 0.500 L
Step 2: Calculate volume of the 19.0 M NaOH solution needed
C1*V1 = C2*V2
⇒with C1 = the concentration of the original NaOH solution = 19.0 M
⇒with V1 = the volume of the original NaOH solution = TO BE DETERMINED
⇒with C2 = the concentration of the NaOH solution we want to prepare = 3.0 M
⇒with V2 = the volume of the NaOH solution we want to prepare = 500 mL = 0.500 L
19.0 M * V2 = 3.0 M * 0.500 L
V2 = (3.0 M * 0.500L) / 19.0 M
V2 = 0.0789 L
We need 0.0789 L
This is 0.0789 * 10^3 mL = 78.9 mL
We need 78.9 mL of the 19.0 M NaOH solution
Answer:
false thought ia ion of neon = clarity active
Explanation:
First, we have to get the initial [C6H8O6] = mass/molar mass
when the molar mass of C6H8O6 = 176.12 g/mol
∴[C6H8O6] = 0.25 g / 176.12 g/mol
= 0.00142 M
when
C6H8O6 ⇄ H+ + C6H7O6-
intial 0.00142 M 0 0
change -X +X +X
Equ (0.00142-X) X X
so, Ka = [H+][C6H7O6-] / [C6H8O6]
by substitution:
8 x 10^-5 = X * X / (0.00142-X) by solving this equation for X
∴ X = 0.000299
∴[H+] = 0.000299
∴PH = -㏒[H+]
= -㏒ 0.000299
= 3.52
B. the frogs are a limiting factor for the gnats
the frogs limit the reproduction of the gnats, and therefore with less frogs the gnat population can increase
