d. Fe(s) and Al(s)
<h3>Further explanation</h3>
In the redox reaction, it is also known
Reducing agents are substances that experience oxidation
Oxidizing agents are substances that experience reduction
The metal activity series is expressed in voltaic series
<em>Li-K-Ba-Ca-Na-Mg-Al-Mn- (H2O) -Zn-Cr-Fe-Cd-Co-Ni-Sn-Pb- (H) -Cu-Hg-Ag-Pt-Au </em>
The more to the left, the metal is more reactive (easily release electrons) and the stronger reducing agent
The more to the right, the metal is less reactive (harder to release electrons) and the stronger oxidizing agent
So that the metal located on the left can push the metal on the right in the redox reaction
The electrodes which are easier to reduce than hydrogen (H), have E cells = +
The electrodes which are easier to oxidize than hydrogen have a sign E cell = -
So the above metals or metal ions will reduce Pb²⁺ (aq) will be located to the left of the Pb in the voltaic series or which have a more negative E cell value (greater reduction power)
The metal : d. Fe(s) and Al(s)
Answer:
67.8%
Explanation:
La reacción de descomposición del CaCO₃ es:
CaCO₃ → CO₂ + CaO
<em>Donde 1 mol de CaCO₃ al descomponerse produce 1 mol de CO₂ y 1 mol de CaO.</em>
Usando la ley general de los gases, las moles de dioxido de carbono son:
PV = nRT.
<em>Donde P es presión (1atm), V es volumen (20L), n son moles de gas, R es la constante de los gases (0.082atmL/molK) y T es temperatura absoluta (15 + 273.15 = 288.15K). </em>Reemplazando los valores en la ecuación:
PV / RT = n
1atmₓ20L / 0.082atmL/molKₓ288.15K = 0.846 moles
Como 1 mol de CO₂ es producido desde 1 mol de CaCO₃, las moles iniciales de CaCO₃ son 0.846moles.
La masa molar de CaCO₃ es 100.087g/mol. Así, la masa de 0.846moles de CaCO₃ es:
0.846moles ₓ (100.087g / mol) = <em>84.7g de CaCO₃</em>
Así, la pureza del marmol es:
(84.7g de CaCO₃ / 125g) ₓ 100<em> = </em>
<h3>67.8%</h3>
Here we go, 4 attachments, you made me work LOL
1. If you find shielding that blocks gamma radiation,then it will most likely also block the other two types. (as the hardest radiation)
2. When nuclear radiation enters the tube of a Geiger counter, it the excited atoms of the gas contained in the tube. (this type <span>deposits all energy into a single atom)</span>
3 The particles that make up protons and neutrons and are thought to be basic units of matter are <span>quarks.</span>
Answer:
0.041 L = 41.3 mL
Explanation:
This problem we will solve by considering the stoichiometry of the reaction and the definition of molarity.
Number of moles in .800 L solution:
0.800 L x 0.0240 M = 0.800 L x .0240 mol/L = 0.0192 mol Fe³⁺
to form the precipitate Fe(OH)₃ we will need 3 times .0192
mol NaOH required = 0.057
given the concentration of 1.38 mol M NaOH we can calculate how many milliliters of NaOH will contain 0.057 mol:
1.L/1.38 mol NaOH x 0.057 mol NaOH = 0.041 L
0.041 L x 1000 mL/1L = 41.3 mL
