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2 years ago

If a recycling center collects 3245 aluminum cans and there are 22 aluminum cans in 1 lb what volume in liters

Chemistry

1 answer:

Bad White [126]2 years ago

3 0

The volume in liters of 3245 aluminum cans is 24.8 L.

A recycling center collects 3245 aluminum cans and there are 22 aluminum cans in 1 lb. The mass of 3245 aluminum cans is:

$3245 cans \times \frac{1lb}{22cans} = 147.5 lb$

To convert mass to volume, we need the density of aluminum (5.95 lb/L). The volume corresponding to 147.5 lb of aluminum is:

$147.5 lb \times \frac{1L}{5.95 lb} = 24.8 L$

The volume in liters of 3245 aluminum cans is 24.8 L.

You can learn more about density here: brainly.com/question/1841285

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Which sentences describe decomposers in a food chain?

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Answer:

hey are the final link in the energy flow in a food chain or a food web. They are fungi and animals that feed on dead organic mattet.

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2 years ago

Which highly reactive gas was probably absent from the Earth's primitive atmosphere? O2 (oxygen gas) water vapor methane carbon

rosijanka [135]

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3 years ago

Why is the air in a jet aircraft flying at high altitudes pressurized? *

saw5 [17]

Answer:

Aircraft cabins are therefore pressurized to maintained a similar pressure as that experienced at sea level to ensure normal breathing of passengers.

Explanation:

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3 years ago

Calculate the pH for each of the following cases in the titration of 25.0 mL of 0.150 M sodium hydroxide with 0.150 M HBr(aq). (

Taya2010 [7]

Answer:

a) pH = 13.176

b) pH = 13

c) pH = 12.574

d) pH = 7.0

e) pH = 1.46

f) pH = 1.21

Explanation:

HBr + NaOH ↔ NaBr + H2O

∴ equivalent point:

⇒ mol acid = mol base

⇒ (Va)*(0.150mol/L) = (0.025L)*(0.150mol/L)

⇒ Va = 0.025 L

a) before addition acid:

NaOH → Na+ + OH-

⇒ C NaOH = 0.150 M

⇒ [ OH- ] = 0.150 M

⇒ pOH = - Log ( 0.150 )

⇒ pOH = 0.824

⇒ pH = 14 - pOH

⇒ pH = 13.176

b) after addition 5mL HBr:

⇒ C NaOH = (( 0.025)*(0.150) - (0.005)*(0.150)) / (0.025 + 0.005) = 0.1 M

⇒ C HBr = (0.005)*(0.150) / ( 0.03 ) = 0.025 M

⇒ [ OH- ] = 0.1 M

⇒ pOH = 1

⇒ pH = 13

c) after addition 15mL HBr:

⇒ C NaOH = ((0.025)*(0.150) - (0.015)*(0.150 ))/(0.04) = 0.0375 M

⇒ C HBr = ((0.015)*(0.150))/(0.04) = 0.0563 M

⇒ [ OH- ] = 0.0375 M

⇒ pOH = 1.426

⇒ pH = 12.574

d) after addition 25mL HBr:

equivalent point:

⇒ [ OH- ] = [ H3O+ ]

⇒ Kw = 1 E-14 = [ H3O+ ] * [ OH- ] = [ H3O+ ]²

⇒ [ H3O+ ] = 1 E-7

⇒ pH = 7.0

d) after addition 40mL HBr:

⇒ C HBr = ((0.04)*(0.150) - (0.025)*(0.150)) / (0.04 + 0.025) = 0.035 M

⇒ [ H3O+ ] = 0.035 M

⇒ pH = 1.46

d) after addition 60mL HBr:

⇒ C HBr = ((0.06)*(0.150) - (0.025)*(0.150)) / (0.06+0.025) = 0.062 M

⇒ [ H3O+ ] = 0.062 M

⇒ pH = 1.21

8 0

3 years ago

Read 2 more answers

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kotykmax [81]

Answer:

HCl -> H+ + Cl- (monoprotic acid)

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2 years ago