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nlexa [21]
3 years ago
15

A compound contains potassium, nitrogen, and oxygen. the experimental analysis gave values of 45.942 % potassium and 16.458 % ni

trogen, by weight, the remainder is oxygen. what is the empirical formula of the compound?
Chemistry
2 answers:
9966 [12]3 years ago
7 0
It should be KNO3 to the best of my understanding
Tanya [424]3 years ago
4 0

Answer:

KNO₂

Explanation:

To get the empirical formula of a compound, we follow a series of steps.

Step 1: Given data

%K = 45.942%

%N = 16.458%

%O = 100.000% - 45.942% - 16.458% = 37.600%

Step 2: Divide each percentage by the atomic mass of the element

K: 45.942 / 39.098 = 1.1750

N: 16.458 / 14.007 = 1.1750

O: 37.600 / 15.999 = 2.3501

Step 3: We divide all the coefficients by the smallest number

K: 1.1750 / 1.1750 = 1

N: 1.1750 / 1.1750 = 1

O: 2.3501 / 1.1750 = 2

The empirical formula is KNO₂.

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3 years ago
The process by which hot and cold air are transferred in the atmosphere is
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hope that helps :)

3 0
3 years ago
Which of the following is a strong base?
malfutka [58]
C. NaOH ammmonia is also an base but not as strong as NaOH
3 0
2 years ago
PLEASE HELP DUE SOON
jonny [76]

Answer:

This is all true if the atom has to be neutral.

Also what does V mean?

Helium: one shell with 2 neutrons and 2 protons in the center, with 2 electrons in the first shell.

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Nitrogen: two shells with 7 neutrons and 7 protons in the center, with 2 electrons in the first shell, and 5 electrons in the second shell.

Flourine: two shells with 9 protons and 10 neutrons in the center, with 2 electrons in the first shell, and 7 electrons in the second shell.

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3 0
2 years ago
Se hace reaccionar 4,00 g de aluminio y 42,00 g de bromo, según la reacción: Al(s)+Br2(l)⟶AlBr3(s) Calcular las moles de AlBr3(s
NeX [460]

Answer:

0.145 moles de AlBr3.

Explanation:

¡Hola!

En este caso, al considerar la reacción química dada:

Al(s)+Br2(l)⟶AlBr3(s)

Es claro que primero debemos balancearla como se muestra a continuación:

2Al(s)+3Br2(l)⟶2AlBr3(s)

Así, calculamos las moles del producto AlBr3 por medio de las masas de ambos reactivos, con el fin de decidir el resultado correcto:

n_{AlBr_3}^{por\ Al}=4.00gAl*\frac{1molAl}{27gAl} *\frac{2molAlBr_3}{2molAl}=0.145mol AlBr_3\\\\n_{AlBr_3}^{por\ Br_2}=42.00gr*\frac{1molr}{160g Br_2} *\frac{2molAlBr_3}{3molBr_2}=0.175mol AlBr_3

Así, inferimos que el valor correcto es 0.145 moles de AlBr3, dado que viene del reactivo límite que es el aluminio.

¡Saludos!

3 0
2 years ago
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