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Vedmedyk [2.9K]
3 years ago
12

Calculate the molarity of a solution made by adding 35.0 mL of concentrated perchloric acid (70.5 % by mass, density 1.67 g/mL)

to some water in a volumetric flask, then adding water to the mark to make exactly 2000 mL of solution. (It is important to add concentrated acid or base to water, rather than the other way, to minimize splashing and maximize safety.)
Chemistry
2 answers:
Kay [80]3 years ago
7 0

Explanation:njeqkfvmdsc,a.xzjr

Nata [24]3 years ago
5 0
Molarity is equal to the number of moles of solute per liter solution.

number of moles solute: (molar mass of perchloric acid is 100.45 g/mol)
35.0 mL * (<span>1.67 g/mL) *0.705 * (mol/100.45 g) = 0.41 moles
liter solution = 2000 ml * (1 L/1000 ml) = 2 L

molarity is equal to 0.41 moles/ 2 L  =0.205 M</span>
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Answer:

n_{P_2O_5}^{max}=1.4molP_2O_5

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In this case, since the reaction between phosphorous and oxygen to form diphosphorous pentoxide is:

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Thus, since phosphorous is in excess and oxygen and diphosphorous pentoxide are in a 5/2:1 mole ratio, we can compute the maximum moles of product as shown below:

n_{P_2O_5}^{max}=112 gO_2*\frac{1molO_2}{32.00gO_2}*\frac{1molP_2O_5}{5/2molO_2}\\\\  n_{P_2O_5}^{max}=1.4molP_2O_5

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At room temperature (20°C} and pressure, the density of air is 1.189 g/L. An object will float in air if its density is less tha
Alekssandra [29.7K]

Explanation:

Density =\frac{Mass}[Volume}

Density of the air ,d= 1.189 g/L

(a) Density of the evacuated ball

Mass of the ball ,m = 0.12 g

Volume of the ball =V=560 cm^3=560 ml=0.560 L

D =\frac{0.12 g}{0.560 L}=0.214 g/L

D<d, teh evacuated ball will flaot in air.

(b) Density of the evacuated ball D = 0.214 g/L

Density of carbon dioxide gas = d_1=1.830 g/L

Mass of the carbon dioxide gas :

1.830 g/L\times 0.560 L=1.0248 g

Total density of filled ball with carbon dioxide gas:

\frac{0.12 g+1.0248 g}{0.560 L}==2.044 g/L

The ball filled with carbon dioxide will not float in the air because total density of filled ball is greater than the density of an air.

(c) Density of the evacuated ball D = 0.214 g/L

Density of hydrogen gas = d_2=0.0899 g/L

Mass of the hydrogen gas :

1.830 g/L\times 0.560 L=0.050344 g

Total density of filled ball with hydrogen gas:

\frac{0.12 g+0.050344 g}{0.560 L}==0.3041 g/L

The ball filled with hydrogen will float in the air because total density of filled ball is lessor than the density of an air.

(d) Density of the evacuated ball D = 0.214 g/L

Density of oxygen gas = d_3=1.330 g/L

Mass of the oxygen gas :

1.330 g/L\times 0.560 L=1.7448 g

Total density of filled ball with oxygen gas:

\frac{0.12 g+1.7448 g}{0.560 L}=1.5442 g/L

The ball filled with oxygen will not float in the air because total density of filled ball is greater than the density of an air.

(e) Density of the evacuated ball D = 0.214 g/L

Density of nitrogen gas = d_4=1.165 g/L

Mass of the nitrogen gas :

1.165 g/L\times 0.560 L=0.6524 g

Total density of filled ball with nitrogen gas:

\frac{0.12 g+0.6524 g}{0.560 L}==1.3792 g/L

The ball filled with nitrogen will not float in the air because total density of filled ball is greater than the density of an air.

f) Mass must be added to sink the ball = m

Density of ball > Density of the air ; to sink the ball.

\frac{0.12g +m}{0.560L}>1.189 g/L

m > 0.54584 g

For any case weight added to ball to make it sink in an air should be grater than the value of 0.54584 grams.

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