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Aneli [31]
3 years ago
10

A sample of an unidentified compound contains 29.84% sodium, 67.49% chromium, and 72.67% oxygen. What is the compound's empirica

l formula? If the molar mass of the molecular formula is 523.96 g/mol, find the molecular formula.
Empirical Formula:
Molecular Formula:​
Chemistry
1 answer:
vladimir1956 [14]3 years ago
5 0

Answer:

i would help i just dont know sorry

Explanation:

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barium chloride (Ba(CIO3)2) breaks down to form barium chloride and oxygen. what is the balanced equation for this equation?
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Answer:

Ba(ClO₃)₂ → BaCl₂ + 3 O₂

Explanation:

When exposed to heat, barium chlorate (Ba(ClO₃)₂ breaks down into an inorganic compound (Barium chloride - BaCl₂) and a molecule (Oxygen - O₂).

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According to the Kinetic Energy Theory, the more energy a particle has, the _______ it moves.
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It is recommended that drinking water contain 1.6 ppm fluo- ride (F) to prevent tooth decay. Consider a cylindrical reservoir wi
Ne4ueva [31]

Answer:

32,127.02 grams of hydrogen hexafluorosilicate will contain this 25,434  grams of F.

Explanation:

Volume of cylindrical reservoir = V

Radius of the cylindrical reservoir = r = d/2

d = diameter of the cylindrical reservoir = d =4.50\times 10^1 m=45 m

r = d/2 = 22.5 m

Depth of the reservoir = h =  10.0 m

V=\pi r^2 h

=3.14\times (22.5 m)^2\times 10.0 m=15,896.25 m^3=15,896,250 L

1 m^3=1000 l

Volume of water cylindrical reservoir : V

Density of water,d = 1 kg/L

Mass of water cylindrical reservoir =  m

m=d\times V=1 kg/L\times 15,896,250 L=15,896,250 kg

1.6 kilogram of fluorine per million kilograms of water. (Given)

Concentration of fluorine in water = 1.6 kg/ 1000,000 kg of water

In 1000,000 kg of water = 1.6 kg of fluorine

Then 15,896,250 kg of water have x mass of fluorine:

\frac{x}{15,896,250 kg\text{kg of water}}=\frac{1.6 kg}{1000,000 \text{kg of water}}

x=\frac{1.6 kg}{1000,000}\times 15,896,250 kg=25.434 kg

15,896,250 kg water of contains mass 25.434 kg of fluorine.

25.434 kg = 25434 g

25,434  grams of fluorine  should be added to give 1.6 ppm.

Percentage of fluorine in hydrogen hexafluorosilicate :

Molar mass hydrogen hexafluorosilicate = 144 g/mol

F\%=\frac{6\times 19 g/mol}{144 g/mol}\times 100=79.16\%

Total mass of hydrogen hexafluorosilicate = m'

79.16\%=\frac{25,434 g}{m'}\times 100

m' = 32,127.02 g

32,127.02 grams of hydrogen hexafluorosilicate will contain this 25,434  grams of F.

5 0
3 years ago
a 32 gram radioactive sample has a half-life of 2 days how much of the original sample remains after 6 days
bixtya [17]
First, we will get the "n", the number of half lives, it is the elapsed time over the half life. In the problem, the time is measured in days, so we have 

6/2 = 3 

to get the ending amount of radioactive sample, we have

32g x (1/2)³ = 4 grams of radioactive sample ⇒ the amount left after 6 days


3 0
3 years ago
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