FLOOR. Most of the molecules that dont sick to something fall on the floor so most would be on the FLOOR.
Answer:
S/.486 es el valor del anillo
Explanation:
Para hallar el precio del anillo se deben encontrar las moles de oro que contiene este.
Si el anillo es de 90g y solo el 59.1% contiene oro, la cantidad de oro en gramos es:
90g × 59.1% = 53.19g Oro en el anillo
Ahora, para convertir los gramos de oro a moles se debe usar la masa atómica del oro (197g/mol), así:
53.19g × (1mol / 197g) = <em><u>0.27 moles de oro contiene el anillo</u></em>.
Ya que cada mol de oro cuesta S/.1800, 0.27 moles de oro (Y por lo tanto, el anillo) costarán:
0.27mol × (S/.1800 / 1mol oro) =
<h3>S/.486 es el valor del anillo</h3>
Answer:
2Cu2^+ + 2I^- ----> 2Cu^+ + I2
Explanation:
The reaction performed in the experiment is;
2 Cu(NO3)2 + 4 KI → 2 CuI (s) + 4 KNO3 + I2
The iodide ions reduces Cu^2+ to Cu^+ which is insoluble in water hence the precipitate. This is so because iodine is a good oxidizing agent seeing that it requires one electron to fill its outermost shell. Potassium on the other hand is a good reducing agent since it easily looses its one electron.
The oxidation - reduction equation is as follows;
2Cu2^+ + 2e ----> 2Cu^+ reduction half equation
2I^- ----> I2 + 2e. Oxidation half equation
Balanced redox reaction equation;
2Cu2^+ + 2I^- ----> 2Cu^+ + I2
Answer:
Li2S> Na2S> K2S> CsS
Explanation:
The lattice energy of ionic species depends on the relative sizes of ions in the ionic compounds. As the size of ions increases, the lattice energy decreases and vice versa.
When the size of the anions are the same, the lattice energy now depends on the relative sizes of the cations. Therefore, since all the compounds are sulphides and the order of magnitude of ionic sizes is: Li^+ < Na^+ < K^+ < Cs^+.
Therefore, the order of decrease in lattice energy is; Li2S> Na2S> K2S> CsS
equilibrium
Heat transfers from a body with high temperature to a body with low temperature until both bodies are in the same temperature.
