Note: You are calculating mass which is determine the gram(g)
You will have to cancel out the mol
(28.97 g/mol) * mol will give grams by itself
Given the mass 3.33 moles of air
28.97 g/mol * 3.33 mol = 96.47 grams
Solution: 96.5 grams
Answer is: mass fo ammonium chloride is 93.625 grams.
V(NH₄Cl) = 5 L.
c(NH₄Cl) = 0.35 M.
n(NH₄Cl) = V(NH₄Cl) · c(NH₄Cl).
n(NH₄Cl) = 5 L · 0.35 mol/L.
n(NH₄Cl) = 1.75 mol.
M(NH₄Cl) = 14 + 1·4 + 35.5 · g/mol = 53.5 g/mol.
m(NH₄Cl) = n(NH₄Cl) · M(NH₄Cl).
m(NH₄Cl) = 1.75 mol · 53.5 g/mol.
m(NH₄Cl) = 93.625 g.
Answer:

Explanation:
Let recall the crystal structure of CsBr obtains a BCC structure. In a BCC structure, there exist only two atom per cell.
The density d of CsBr in g/cm³ can be calculated by using the formula:

where;
z = 1 mole of CsBr
edge length = 428.7 pm = (4.287 × 10⁻⁸)³ cm
molar mass of CsBr = 212.81 g/mol
avogadro's number = 6.023 × 10²³



Answer:
About 0.0940 M.
Explanation:
Recall that NaOH is a strong base, so it dissociates completely into Na⁺ and OH⁻ ions. Because the acid is monoprotic, we can represent it with HA. Thus, the reaction between HA and NaOH is:

Using the fact that it took 15.00 mL of NaOH to reach the endpoint, determine the number of HA that was reacted with:

Therefore, the molarity of the original solution was:
![\displaystyle \left[ \text{HA}\right] = \frac{0.00188\text{ mol}}{20.00\text{ mL}} \cdot \frac{1000\text{ mL}}{1\text{ L}} = 0.0940\text{ M}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cleft%5B%20%5Ctext%7BHA%7D%5Cright%5D%20%3D%20%5Cfrac%7B0.00188%5Ctext%7B%20mol%7D%7D%7B20.00%5Ctext%7B%20mL%7D%7D%20%5Ccdot%20%5Cfrac%7B1000%5Ctext%7B%20mL%7D%7D%7B1%5Ctext%7B%20L%7D%7D%20%3D%200.0940%5Ctext%7B%20M%7D)
In conclusion, the molarity of the unknown acid is about 0.0940 M.