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seraphim [82]
3 years ago
11

At 600. K how many moles of gas are in a 1.00 L cylinder at atmospheric pressure?

Chemistry
1 answer:
Anton [14]3 years ago
8 0

Answer:

0.02moles

Explanation:

To answer this question, the general gas law equation is used. The General gas law is:

Pv = nRT

Where; P = standard atmospheric pressure (1 atm)

V = volume (L)

n = number of moles

R = Gas law constant

T = Temperature

For this question; volume = 1.00L, atmospheric pressure (P) = 1 atm, R = 0.0821 L-atm / mol K, T = 600K, n = ?

Therefore; Pv = nRT

n = PV/RT

n = 1 × 1/ 0.0821 × 600

n = 1/49.26

n = 0.0203moles

Hence, there are 0.02 moles of gas.

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Basic molecules contain more ...
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Option C. hydroxide ions (OH-).

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According to molecular orbital theory, the regions of the wave function with the highest probability of finding electrons are ar
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According to molecular orbital theory,  regions of  wave function with highest probability of finding electrons are areas with constructive interference.

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6 0
1 year ago
Please help, I lost my previous tutor just when she was explaining the answer
TEA [102]

Answer:

HCO₂/H₂O is not the acid-base conjugate pair.

Explanation:

<em>Acid and conjugate base pairs differ by an H+ ion.</em>

Neither HCO₂ nor H₂O has lost or gained protons.

The conjugate acid of H₂O is H₃O⁺

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5 0
2 years ago
If 842 grams of sodium hydroxide reacts with 750.0 grams of aluminum, how many grams of aluminum hydroxide should theoretically
Phantasy [73]

548.55 grams of aluminum hydroxide should theoretically form.

Explanation:

Balanced equation for the reaction:

3 NaOH + Al ⇒ Al(OH)3 +3 Na

DATA GIVEN:

mass of NaOH = 842 grams, atomic mass =39.9 grams/mole

mass of Al = 750 grams, atomic mass = 26.9 grams/mole

aluminum hydroxide theoretical yield = ?

Moles of NaOH reacted

number of moles = \frac{mass}{atomic mass of 1 mole}

putting the values in the equation

NaOH = \frac{842}{39.9}

           = 21.1 MOLES OF NaOH

Al = \frac{750}{26.9}

   = 27.8 moles

from the equation

 from 3 moles of NaOH 1 mole of Al(OH)3 is produced

21.1 moles of NaOH will react to give x moles of Al(OH)3

\frac{1}{3} = \frac{x}{21.1}

7.03 moles of Al(OH)3 is formed.

and

1 mole of Al(OH)3 is formed from 1 mole of Al in the reaction

so, 27.8 Moles will react to give give 27.8 moles of Al(OH)3 limiting reagent of the given reaction is NaOH

mass of Al(OH)3 =7.03 x 78 (atomic mass of Al(OH)3)

          = 548.55 grams

theoretical  yield from the given data is 548.55 grams

3 0
2 years ago
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