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irakobra [83]
3 years ago
7

The setup in the diagram is left outside during the day and night . Bubbles are continuously produced regardless of the presence

of sunlight.what can you predict of the composition in the bubbles
A. The bubbles are always O2
B. The bubbles are always carbon dioxide CO2
C. During the day the bubbles are CO2 and in the night O2
D. During the day they are O2 and in the night CO2
Chemistry
1 answer:
Angelina_Jolie [31]3 years ago
6 0

Answer:

A

Explanation:

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I think c biological processes

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- A jackrabbit's powerful legs are an example of a
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I think it is b.

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3 years ago
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
Help on part "c": The forensic technician at a crime scene has just prepared a luminol stock solution by adding 19.0g of luminol
Contact [7]

1. The molarity of the stock solution of luminol = 1,431 M

2. 0.12 moles of luminol is present in 2.00 L of the diluted spray

3. The volume of the stock solution (Part A) would contain the number of moles present in the diluted solution (Part B): 83.86 ml

<h3>Further explanation</h3>

Stoichiometry in Chemistry studies about chemical reactions mainly emphasizing quantitative, such as the calculation of volume, mass, amount, which is related to the number of ions, molecules, elements, etc.

In stoichiometry included :

  • 1. Relative atomic mass
  • 2. Relative molecular mass

is the relative atomic mass of the molecule

  • 3. mole

1 mole is the number of particles contained in a substance with the same number of atoms in 12 gr C-12

1 mole = 6.02.10²³ particles

While the number of moles can also be obtained by dividing the mass (in grams) by the relative mass of the element or the relative mass of the molecule

\large{\boxed{\bold{mol\:=\:\frac{grams}{ relative\:mass} }}}

Luminol (C₈H₇N₃O₂) is a substance used to detect traces of blood in the scene of a crime because it reacts with iron in the blood

a. a luminol stock solution by adding 19.0g of luminol into a total volume of 75.0mL of H2O.

So the molarity is

  • 1. Luminol mole

- the relative molecular mass of Luminol

= 8. C + 7.H + 3.N + 2.16

= 8.12 + 7.1 + 3.14 + 2.16

= 177 grams / mol

so the mole:

mol = gram / relative molecular mass

mole=\frac{19}{177}

mole = 0.1073

2. Molarity (M)

M = mole / volume

M\:=\:{\frac{ 0.1703 }{75.10^{-3} L}

M = 1,431

  • b. luminol concentration in a spray bottle 6.00 × 10⁻² M. So that in 2 L of solution, the number of moles is:

mole = M x volume

mole = 6.10⁻² x 2

mole = 0.12

  • c. Molarity of the stock solution (Part A) = 1,431 M

the number of moles present in the diluted solution (Part B) = 0.12

So the volume of the stock solution (Part A) would contain the number of moles present in the diluted solution (Part B) is:

volume = mol / M

volume\:=\:\frac{0.12}{1.431}

volume = 0.08386 L = 83.86 ml

<h3>Learn more</h3>

moles of water you can produce

brainly.com/question/1405182

the number of each atom present in the compound's formula

brainly.com/question/5303004

the ratio of hydrogen atoms (H) to oxygen atoms (O) in 2 l of water

brainly.com/question/10861183

Keywords: mole, volume, molarity, Luminol, the relative molecular mass

6 0
3 years ago
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