Answer:
The empirical formula simply states the ratios of the elements that make up the molecule, while molecular formulas specify the amounts of each element exactly, not just giving the ratio. Let's take ethane (C2H6) as an example.
Answer:
The veins (blue) take oxygen-poor blood back to the heart. Arteries begin with the aorta, the large artery leaving the heart. They carry oxygen-rich blood away from the heart to all of the body's tissues.
Explanation:
hope this helps
<span>Kp is the equilibrium pressure constant calculated from the partial pressures of a reaction equation.
Kp =[pCF4]*[p CO2] / [p COF2]^2 = 2.2 x 10^6
When the mole fraction gets doubled we have
Kp = [pCO2]^2*[pCF4]^2 / [pCOF2]^4
Kp = [[pCF4]*[p CO2] / [p COF2]^2] * 2
Kp = (2.2 * 10^6) * 2
Kp = 4.8 * 10^12</span>
Answer:
342
g/mol
Explanation:
Just get out that Periodic Table of the Elements, find the atomic masses for each atom, and add them up!
Credit to "SCooke" (User)
Answer:
mix 1 mL of the stock solution of MgCl2, 300 mL of the stock solution of NaCl and 699 mL of water.
Explanation:
We need to determine the volume necessary of both stock solutions. When a dilution is made, a certain volume of the stock solution is collected and then more solvent is added to it until the volume is completed. So, the number of moles of the solute in both solutions are the same, and it is the concentration (C) multiplied by the volume (V). If 1 is the stock solution and 2 the diluted solution:
C1*V1 = C2*V2
So, in this case, the two solutions will be mixed, and then the volume will be completed with the solvent. So, for MgCl2:
C1 = 1.0 M
C2 = 1.0 mM = 0.001 M
V2 = 1.0 L = 1000 mL
1*V1 = 0.001*1000
V1 = 1 mL of the stock solution of MgCl2
For NaCl:
C1 = 0.50 M
C2 = 0.15 M
V2 = 1000 mL
0.50*V1 = 0.15*1000
V1 = 300 mL
So, it will be necessary 1 mL of the stock solution of MgCl2, 300 mL of the stock solution of NaCl and 699 mL of water.