lolovep45
lolovep45 lolovep45
  • 02-04-2020
  • Chemistry
contestada

11g Na,.0900gh, 2.94 and 5.86 g o (NaOHS)
calculate the Imperial formula
please help me

Respuesta :

Eduard22sly
Eduard22sly Eduard22sly
  • 02-04-2020

Answer:

Na5O4H10S

Explanation:

Data obtained from the question include:

Sodium (Na) = 11g

Hydrogen (H) = 0.9g

Oxygen (O) = 5.86g

Sulphur (S) = 2.94g

To obtain the empirical formula, do the following.

First Divide the above by their individual molar mass

Na = 11/23 = 0.478

H = 0.9/1 = 0.9

O = 5.86/16 = 0.366

S = 2.94/32 = 0.092

Next, divide by the smallest number

Na = 0.478/0.092 = 5

H = 0.9/0.092 = 10

O = 0.366/0.092 = 4

S = 0.092/0.092 = 1

Therefore, the empirical formula is Na5O4H10S

Answer Link

Otras preguntas

Energy from hot magma is called ____.
Write365,098 in word form
Tyler wrote a paper describing life in England in the eighteenth _________, from ______ to ______.
why was otto i declared the roman emperor
Round 125,458 to the nearest hundred thousand
Mr. Derbyshire makes a business trip from his house to loganville in 2 hours. One hour later, he returns home in traffic at a rate 20mph less than his rate goin
Energy from hot magma is called ____.
when a falling object reaches a speed called terminal speed, its speed is no longer unceasing. the object id loosing gravitational potential energy nut no kinet
All of the following are population characteristics EXCEPT (A) number of individuals (B) phenotype (C) sex ratio (D) age distribution (E) death rate
this is in my homework paper 5300 divided by 10 to the second