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romanna [79]
2 years ago
12

1. One of my goals is to get 8 hours of sleep a night. If I sleep 7,259,102 second, how many hours of sleep is that (w/ proper s

ig figs)?
2. Did I meet my goal?
3. How many seconds are in 8.0000 hours?
Chemistry
1 answer:
soldier1979 [14.2K]2 years ago
7 0

Answer:

2016.417222 hours

Explanation: you def made your goal

28800 seconds

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All of the following are equal to Avogadro's number EXCEPT ____. a. the number of atoms of bromine in 1 mol Br2 b. the number of
Colt1911 [192]

All the following are equal to Avogadro's number EXCEPT a. the number of atoms of bromine in 1 mol Br₂.

1 mol Br₂ contains Avogadro’s number of molecules of Br₂.

However, each molecule contains two atoms of Br, so there are

<em>2 × Avogadro’s number of Br atoms </em>in 1 mol Br₂.

8 0
2 years ago
If the molar absorptivity constant for the red dye solution is 5.56×104 M-1cm-1, calculate the molarity of the red dye solution
Shtirlitz [24]

Explanation:

a) Using Beer-Lambert's law :

Formula used :

A=\epsilon \times c\times l

where,

A = absorbance of solution = 0.945

c = concentration of solution = ?

l = length of the cell = 1.20 cm

\epsilon = molar absorptivity of this solution =5.56\times 10^4 M^{-1} cm^{-1}

0.945=5.56\times 10^4 M^{-1} cm^{-1}\times 1.20 \times c

c=1.4163\times 10^{-5} M=14.16 \mu M

(1\mu M=10^{-6} M)

14.16 μM is the molarity of the red dye solution at the optimal wavelength 519nm and absorbance value 0.945.

b) c=1.4163\times 10^{-5} mol/L

1 L of solution contains 1.4163\times 10^{-5} moles of red dye.

Mass of 1.4163\times 10^{-5} moles of red dye:

1.4163\times 10^{-5}\times 879.86g/mol=0.01246 g

(w/v)\%=\frac{\text{Mass of solute (g)}}{\text{Volume of solvent (mL)}}\times 100

red(w/v)\%=\frac{0.01246 g}{1000 mL}\times 100=0.001246\%

c) In order to dilute red dye solution by 5 times, we will need to add 1 L of water to solution of given concentration.

Concentration of red dye solution = c=1.4163\times 10^{-5} M

Concentration of red solution after dilution = c'

c=c'\times 5

1.4163\times 10^{-5} M=c'\times 5

c'=2.83\times 10^{-6} M

The final concentration of the diluted solution is 2.83\times 10^{-6} M

8 0
3 years ago
The faster the object is moving the more portential energy it has?
Mazyrski [523]

Answer:

sure

Explanation:

8 0
3 years ago
Ammonia and oxygen react to form nitrogen and water.
Nata [24]

Answer:

A. 19.2 g of O2.

B. 3.79 g of N2.

C. 54 g of H2O.

Explanation:

The balanced equation for the reaction is given below:

4NH3(g) + 3O2(g) → 2N2+ 6H2O(g)

Next, we shall determine the masses of NH3 and O2 that reacted and the masses of N2 and H2O produced from the balanced equation.

This is illustrated below:

Molar mass of NH3 = 14 + (3x1) = 17 g/mol

Mass of NH3 from the balanced equation = 4 x 17 = 68 g

Molar mass of O2 = 16x2 = 32 g/mol

Mass of O2 from the balanced equation = 3 x 32 = 96 g

Molar mass of N2 = 2x14 = 28 g/mol

Mass of N2 from the balanced equation = 2 x 28 = 56 g

Molar mass of H2O = (2x1) + 16 = 18 g/mol

Mass of H2O from the balanced equation = 6 x 18 = 108 g

Summary:

From the balanced equation above,

68 g of NH3 reacted with 96 g of O2 to produce 56 g of N2 and 108 g of H2O.

A. Determination of the mass of O2 needed to react with 13.6 g of NH3.

This is illustrated below:

From the balanced equation above,

68 g of NH3 reacted with 96 g of O2.

Therefore, 13.6 g of NH3 will react with = (13.6 x 96)/68 = 19.2 g of O2.

Therefore, 19.2 g of O2 are needed for the reaction.

B. Determination of the mass of N2 produced when 6.50 g of O2 react.

This is illustrated below:

From the balanced equation above,

96 g of O2 reacted to produce 56 g of N2.

Therefore, 6.5 g of O2 will react to produce = (6.5 x 56)/96 = 3.79 g of N2.

Therefore, 3.79 g of N2 were produced from the reaction.

C. Determination of the mass of H2O formed from the reaction of 34 g of NH3.

This is illustrated below:

From the balanced equation above,

68 g of NH3 reacted to 108 g of H2O.

Therefore, 34 g of NH3 will react to produce = (34 x 108)/68 = 54 g of H2O.

Therefore, 54 g of H2O were obtained from the reaction.

4 0
2 years ago
The half life of potassium 40 is approximately 1.25 billion years. An igneous rock is found to contain about 1/4 of its original
vodomira [7]

Answer:

2.5 billion years

Explanation:

4 0
2 years ago
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