Answer:
Mass = 51 g
Explanation:
Given data:
Mass of nitrogen = 41.93 g
Mass of ammonia formed = ?
Solution:
Chemical equation:
N₂ + 3H₂ → 2NH₃
Number of moles of nitrogen:
Number of moles = mass/molar mass
Number of moles = 41.93 g/ 28 g/mol
Number of moles = 1.5 mol
now we will compare the moles of nitrogen and ammonia.
N₂ : NH₃
1 : 2
1.5 : 2/1×1.5 = 3 mol
Mass of ammonia formed:
Mass = number of moles × molar mass
Mass = 3 mol × 17 g/mol
Mass = 51 g
The anode is the electrode where the oxidation occurs.
Cathode is the electrode where the reducction occurs.
Equations:
Mn(2+) + 2e- ---> Mn(s) Eo = - 1.18 V
2Fe(3+) + 2e- ----> 2 Fe(2+) 2Eo = + 1.54 V
The electrons flow from the electrode with the lower Eo to the electrode with the higher Eo yielding to a positive voltage.
Eo = 1.54 V - (- 1.18) = 1.54 + 1.18 = 2.72
Answer: 2.72 V
Earth spins at 23.5
A day is one rotation
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>
2.2311 moles of gas are there in a 50. 0 l container at 22. 0 °c and 825 torrs.
<h3>What is an ideal gas?</h3>
An Ideal gas is a hypothetical gas whose molecules occupy negligible space and have no interactions, and which consequently obeys the gas laws exactly.
Assuming the gas is ideal, we can solve this problem by using the following equation:
PV = nRT
Where:
P = 825 torr ⇒ 825 / 760 = 1.08 atm
V = 50 L
n = ?
R = 0.082 atm·L·mol⁻¹·K⁻¹
T = 22 °C ⇒ 22 + 273.16 = 295.16 K
We input the data:
1.08 atm x 50 L = n x 0.082 atm·L·mol⁻¹·K⁻¹ x 295.16 K
And solve for n:
24.20312
n = 2.2311 mol
Hence, 2.2311 moles of gas are there in a 50. 0 L container at 22. 0 °c and 825 torrs.
Learn more about ideal gas here:
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