Answer:
14.68 moles of He
Explanation:
To do this, just remember Avogadro's Constant or Avogadro's number. This constant tells us how many units ( in this case atoms) there are in a mole of ANY type of substance.
Avogadro's constant is 6.022140857 × 10²³ units per mole.
Now that we know how many atoms there are in 1 mole, we can use this as our conversion factor.
8.84 x 10²⁴ atoms of He → moles of He

So the answer would be:
14.68 moles of He
Answer:
The answer is True.
Explanation:
The sentence above is true it all adds up.
Answer:
0.0238M SbCl3, 1.07M H+, 1.14M Cl-
Explanation:
The total volume of the solution is:
4mL + 5.00mL + 12.0mL = 21mL
As the volume of the SbCl3 is 5.00mL, the dilution factor is:
21mL / 5.00mL = 4.2 times
The concentration of SbCl3 is:
0.10M SbCl3 / 4.2 times = 0.0238M SbCl3
The concentration of H+ = [HCl]:
4.5M / 4.2 times = 1.07M H+
The initial concentration of Cl- is:
3 times SbCl3 + HCl = 0.10M*3 + 4.5M =
<em>3 times SbCl3 because 1 mole of SbCl3 contains 3 moles of Cl-</em>
4.8M Cl- / 4.2 times = 1.14M Cl-
The volume of 0.20 moles of helium at STP is 4.5 liters.
Explanation:
Given:
Number of moles = 0.20 moles
To Find:
The volume of Helium at STP =?
Solution:
According to ideal gas law
PV = nRT
where
P is pressure,
V is volume,
n is the number of moles
R is the gas constant, and
T is temperature in Kelvin.
The question already gives us the values for p and T
,because helium is at STP. This means that temperature is 273.15 K and pressure is 1 atm
.
We also already know the gas constant. In our case, we'll use the value of
0.08206 L atm/K mol since these units fit the units of our given values the best
On substituting these values we get



V = 4.5 Liters
An ion is a charged atom or molecule. It is charged because the number of electrons do not equal the number of protons in the atom or molecule. An atom can acquire a positive charge or a negative charge depending on whether the number of electrons in an atom is greater or less then the number of protons in the atom. An example is Iron (III) , Iron (II) , lithium, and hydrogen.