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3 years ago

How many moles of N2O3 contain 2.55 × 1024 oxygen atoms

1 answer:

6 0

1mol ------------ 3×6,02×10²³ oxygen atoms
x ------------ 2,55×10²⁴

x = 2,5×10²⁴ × 1mol / 3×6,02×10²³ ≈ 0,138*10 mol = 1,38mol

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BRAINLIEST!!

Juliette [100K]

Answer:

Cesium

Explanation:

cuz i looked at the chart from highest to lowest

3 0

3 years ago

1. Unas de las formas de producir nitrógeno gaseoso (N2) es mediante la oxidación de metilamina (CH3NH2), tal como se muestra en

Maslowich

Answer:

a) 4CH₃NH₂ + 9O₂ ⇄ 4CO₂ + 10H₂O + 2N₂

b) m = 5,043 g

c) % = 69,4 %

Explanation:

a) La ecuación balanceada es la siguiente:

4CH₃NH₂ + 9O₂ ⇄ 4CO₂ + 10H₂O + 2N₂

En el balanceo, se tiene en la relación estequiométrica que 4 moles de metilamina reacciona con 9 moles de oxígeno para producir 4 moles de dióxido de carbono, 10 moles de agua y 2 moles de nitrógeno.

b) Para determinar la masa de nitrógeno se debe calcular primero el reactivo limitante:

$n_{O_{2}} = \frac{m}{M} = \frac{25,6 g}{31,99 g/mol} = 0,800 moles$

$n_{CH_{3}NH_{2}} = \frac{4}{9}*0,800 moles = 0,356 moles$

De la ecuación anterior se tiene que la cantidad de moles de metilamina necesaria para reaccionar con 0,800 moles de oxígeno es 0,356 moles, y la cantidad de moles iniciales de metilamina es 0,5 moles, por lo tanto el reactivo limitante es el oxígeno.

Ahora, podemos calcular la masa de nitrógeno producida:

$n_{N_{2}} = \frac{2}{9}*n_{O_{2}} = \frac{2}{9}*0,8 moles = 0,18 moles$

$m_{N_{2}} = n_{N_{2}}*M = 0,18 moles*28,014 g/mol = 5,043 g$

Por lo tanto, se pueden producir 5,043 g de nitrógeno.

c) El redimiento de la reacción se puede calcular usando la siguiente fórmula:

$\% = \frac{R_{r}}{R_{T}}*100$

<u>Donde</u>:

$R_{r}$ : es el rendimiento real

$R_{T}$ : es el rendimiento teórico

$\% = \frac{3,5}{5,043}*100 = 69,4$

Entonces, el procentaje de rendimiento de la reacción es 69,4%.

Espero que te sea de utilidad!

5 0

3 years ago

Can anyone do this question.i really need a answer

PilotLPTM [1.2K]

Answer:

17.4

Explanation:

kasi 15.2

16.2

=17.4 thanks me later pa brainliest din

7 0

3 years ago

What mass of oxygen reacts during the incomplete combustion of 18.0 g of propane?

liq [111]

Answer:

45.82 gm O2

Explanation:

From your balanced equation in the first part, you can see that 2 moles propane need 7 moles O2

Find the number of moles of propane in 18 gm

mole wt of propane = ~ 12 * 3 + 8 *1 = 44 gm/mole

18 gm / 44 gm/mole = 9/22 gm/mole propane

we need 7/2 of this as O2 moles 7/2 * 9/22 = 1.432moles of O2

O2 mole weight =~ 32

1.432 moles * 32 gm/mole = 45.82 gm O2

4 0

2 years ago

A 1.25-g sample contains some of the very reactive compound Al(C6H5)3. On treating the compound with aqueous HCl, 0.951 g of C6H

Oksanka [162]

<span>83.9% First, determine the molar masses of Al(C6H5)3 and C6H6. Start by looking up the atomic weights of the involved elements. Atomic weight aluminum = 26.981539 Atomic weight carbon = 12.0107 Atomic weight hydrogen = 1.00794 Molar mass Al(C6H5)3 = 26.981539 + 18 * 12.0107 + 15 * 1.00794 = 258.293239 g/mol Molar mass C6H6 = 6 * 12.0107 + 6 * 1.00794 = 78.11184 g/mol Now determine how many moles of C6H6 was produced Moles C6H6 = 0.951 g / 78.11184 g/mol = 0.012174851 mol Looking at the balanced equation, it indicates that 1 mole of Al(C6H5)3 is required for every 3 moles of C6H6 produced. So given the number of moles of C6H6 you have, determine the number of moles of Al(C6H5)3 that was required. 0.012174851 mol / 3 = 0.004058284 mol Then multiply by the molar mass to get the number of grams that was originally present. 0.004058284 mol * 258.293239 g/mol = 1.048227218 g Finally, the weight percent is simply the mass of the reactant divided by the total mass of the sample. So 1.048227218 g / 1.25 g = 0.838581775 = 83.8581775% And of course, round to 3 significant digits, giving 83.9%</span>

8 0

3 years ago