Im pretty sure the anservis a
53 x 1000 = 53000
53000/ 44
1204.545 moles of CO2
Answer:
108.43 grams KNO₃
Explanation:
To solve this problem we use the formula:
Where
- ΔT is the temperature difference (14.5 K)
- Kf is the cryoscopic constant (1.86 K·m⁻¹)
- b is the molality of the solution (moles KNO₃ per kg of water)
- and<em> i</em> is the van't Hoff factor (2 for KNO₃)
We <u>solve for b</u>:
- 14.5 K = 1.86 K·m⁻¹ * b * 2
Using the given volume of water and its density (aprx. 1 g/mL) we <u>calculate the necessary moles of KNO₃</u>:
- 275 mL water ≅ 275 g water
- moles KNO₃ = molality * kg water = 3.90 * 0.275
- moles KNO₃ = 1.0725 moles KNO₃
Finally we <u>convert KNO₃ moles to grams</u>, using its molecular weight:
- 1.0725 moles KNO₃ * 101.103 g/mol = 108.43 grams KNO₃
Explanation:
The steps given in the question are incorrect.
Step 1 should be convert atoms to moles (n). Step 2 should be convert moles (n) to mass (m).
Step 1
Use dimensional analysis to convert the number of atoms to moles.
1 mole atoms = 6.022 × 10²³ atoms
n(Ag) = 2.3 × 10²⁴ Ag atoms × (1 mol Ag/6.022 × 10²³ Ag atoms) = 3.8193 mol Ag
Step 2
Convert the moles of Ag to mass.
mass (m) = moles (n) × molar mass (M)
n(Ag) = 3.8193 mol Ag
M(Ag) = atomic weight on the periodic table in g/mol = 107.868 g Ag/mol Ag
m(Ag) = 3.8193 mol × 107.868 g/mol = 412 g Ag = 410 g Ag rounded to two significant figures
The mass of 2.3 × 10²⁴ Ag atoms is approximately 410 g.