Answer:
3.51× 10²³ formula units
Explanation:
Given data:
Mass of CaO = 32.7 g
Number of formula units = ?
Solution:
First of all we will calculate the number of moles.
Number of moles = mass/molar mass
Number of moles = 32.7 g/ 56.1 g/mol
Number of moles = 0.583 mol
Number of formula units:
1 mole = 6.022 × 10²³ formula units
0.583 mol × 6.022 × 10²³ formula units / 1 mol
3.51× 10²³ formula units
The number 6.022 × 10²³ is called Avogadro number.
Answer: -
Acetic acid
Explanation: -
The intermolecular force of attraction depends on the strength or extent of Hydrogen bonding present in a substance.
Benzene and chloroform does not have hydrogen bonding being non polar molecule.
Water has hydrogen bonding being polar.
However acetic acid being most polar has the maximum hydrogen bonding.
Thus acetic acid has the strongest intermolecular forces of attraction.
Mg + 2H(NO)3 ⇒ Mg(NO3)2 + H2
If there weren't any predators, the population of prey would increase and they would starve due to too many mouths to feed.
This is a straightforward dilution calculation that can be done using the equation
where <em>M</em>₁ and <em>M</em>₂ are the initial and final (or undiluted and diluted) molar concentrations of the solution, respectively, and <em>V</em>₁ and <em>V</em>₂ are the initial and final (or undiluted and diluted) volumes of the solution, respectively.
Here, we have the initial concentration (<em>M</em>₁) and the initial (<em>V</em>₁) and final (<em>V</em>₂) volumes, and we want to find the final concentration (<em>M</em>₂), or the concentration of the solution after dilution. So, we can rearrange our equation to solve for <em>M</em>₂:

Substituting in our values, we get
![\[M_2=\frac{\left ( 50 \text{ mL} \right )\left ( 0.235 \text{ M} \right )}{\left ( 200.0 \text{ mL} \right )}= 0.05875 \text{ M}\].](https://tex.z-dn.net/?f=%5C%5BM_2%3D%5Cfrac%7B%5Cleft%20%28%2050%20%5Ctext%7B%20mL%7D%20%5Cright%20%29%5Cleft%20%28%200.235%20%5Ctext%7B%20M%7D%20%5Cright%20%29%7D%7B%5Cleft%20%28%20200.0%20%5Ctext%7B%20mL%7D%20%5Cright%20%29%7D%3D%200.05875%20%5Ctext%7B%20M%7D%5C%5D.)
So the concentration of the diluted solution is 0.05875 M. You can round that value if necessary according to the appropriate number of sig figs. Note that we don't have to convert our volumes from mL to L since their conversion factors would cancel out anyway; what's important is the ratio of the volumes, which would be the same whether they're presented in milliliters or liters.