The empirical formula of metal iodide : CoI₃(Cobalt(III) Iodide)
<h3>Further explanation</h3>
13.02 g sample of Cobalt , then mol Co(MW=58.933 g/mol) :
Mass of metal iodide formed : 97.12 g, so mass of Iodine :
Then mol iodine (MW=126.9045 g/mol) :
mol ratio of Cobalt and Iodine in the compound :
Inorganic molecules are composed of other elements. They can contain hydrogen or carbon, but if they have both, they are organic.
Answer:
3AgCl + Na₃PO₄ —> 3NaCl + Ag₃PO₄
The coefficients are 3, 1, 3, 1
Explanation:
From the question given above, the following data were:
Silver chloride reacts with sodium phosphate to yield sodium chloride and silver phosphate. This can be written as follow:
AgCl + Na₃PO₄ —> NaCl + Ag₃PO₄
The above equation can be balanced as follow:
AgCl + Na₃PO₄ —> NaCl + Ag₃PO₄
There are 3 atoms of Na on the left side and 1 atom on the right side. It can be balance by putting 3 in front of NaCl as shown below:
AgCl + Na₃PO₄ —> 3NaCl + Ag₃PO₄
There are 3 atoms of Cl on the right side and 1 atom on the left. It can be balance by putting 3 in front of AgCl as shown below:
3AgCl + Na₃PO₄ —> 3NaCl + Ag₃PO₄
Thus, the equation is balanced.
The coefficients are 3, 1, 3, 1
Answer:
c is not a true statement
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
