A sample of commercial perchloric acid is 70.0% HClO4 by mass; its density is 1.664 g/mL. How many milliliters of this concentrated HClO4 would be required to prepare 500. mL of 1.50 M HClO4 solution?

The volume of the conc. HClO4 to prepare 500 ml of 1.50 M solution is 64.6 ml.

Based on the given information,

•The mass of HClO4 is 70% by mass, the density of HClO4 is 1.664 g/ml or 1664 g/L.

•It is known that the molar mass of perchloric acid is 100.465 g/mol.  

Now the mass of HClO4 is,

Now the moles can be calculated as,

Now the molarity is,

Now, M1 is 1.50 M, M2 is 11.596 M, V1 is 0.500 L. There is a need to find V2, which can be done by using the dilution law, that is,

Now putting the values we get,

Thus, 64.6 ml of the concentrated HClO4 is need to prepare 500 ml of 1.50 M HClO4 solution.

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We need 64.7 mL of this concentrated HClO4 solution

Explanation:

Step 1: Data given

Percentage of HClO4 solutions = 70.0 %

Density of HClO4 = 1.664 g/mL

Volume of prepared HClO4 solution  =  500.0 mL

Molarity of the prepared HClO solution = 1.50 M

Step 2: Calculate mass HClO4

Let's assume we have 1 L = 1000 mL of solution

Density = mass / volume

Mass = density * volume

Mass = 1.664 g/mL * 1000 mL

Mass = 1664 grams

For a 70.0 M HClO4 solution the mass HClO4 is:

1664 g * 0.70 = 1165 grams HClO4

Step 3: Calculate molarity

1165 g HClO4 / 100.5 g  = 11.59 moles HClO4 per L = 11.59 M  

Step 4: Calculate the volume needed

C1V1 = C2V2  

⇒with C1 = the molarity of the 70.0 % HClO4 = 11.59 M

⇒with V1 = the volume of the 70.0 % HClO4 = TO BE DETERMINED

⇒with C2 = the molarity of the prepared HClO4 solution = 1.50 M

⇒with V2 = the volume of the prepared HClO4 solution = 500.0 mL = 0.500 L

11.59 * V2 = 1.50 M * 0.500

V2 = (1.50 * 0.500) / 11.59

V2 = 0.0647 L = 64.7 mL

We need 64.7 mL of this concentrated HClO4 solution

first let us lay out the formula for potential

and kinetic energy:

pe = mgh

ke = (mv^2)/2

where: m=mass, v=velocity, g=gravitational

acceleration, h=height

calculating for mass using the known values of

pe:

340j = m (9.81 m/s^2) (50ft) (1m/3.28ft)

m = 2.27 kg

the law of conservation of energy states that

energy is neither created nor destroyed. therefore, the change in kinetic

energy from 50 ft to 35 ft would be equal to the change in potential energy

from 50 ft to 35 ft. however it would be opposite in signs since one is losing

while other is gaining.

ke(35ft) – ke(50ft) = - [ pe(35ft) – pe(50ft]

ke(50ft) = 0 since the brick is initially at rest

ke(35ft) = 340j - 2.27kg (9.81 m/s^2) (35ft)

(1m/3.28 ft)

ke(35ft)

= 102.38 j

alkaline metals are lithium, sodium, rubidium, potassium, cesium and francium.

alkali metals have electronegativity from 0.7 to 1, lowest in periodic table of elements, which means that alkali metals (i group in periodic table) has positive oxidation number in compounds.

electronegativity is a chemical property that describes the tendency of an atom to attract a shared pair of electrons towards itself.

alkaline metals (far left in main group) have lowest ionizations energy and easy remove valence electrons,

for example, potassium is the alkali metal and has a single valence electron in the outer electron shell.

electron configuration of potassium is: ₁₉k 1s²2s²2p⁶3s²3p⁶4s¹.

they can not exist alone in nature, because they are very reactive.