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kirza4 [7]
4 years ago
13

Suppose you seal 30 g of ag2o(s) in a 1.00 liter evacuated vessel, and heat it to 450 k. what pressure (of o2) would develop in

the vessel at equilibrium?
Chemistry
1 answer:
kodGreya [7K]4 years ago
3 0
<span>We would use the ideal gas equation. PV = NRT
 No of moles of ag2o = 30/ 231.735 = 0.12944
 The chemical reaction is given by
  2 Ag2O = 4 Ag + O2
  2* 0.12944 of Ag2O requires 0.12944 of O2
 So we have P * 1 = 0.258 * 8.2 * 450
  P = 952.02 atm</span>
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If you made 6 moles of NO2 how many grams of N2 did you use?
mars1129 [50]

Answer:

3

Explanation

N is 14, so multiply it by 6 (because 6 moles). Since it's N2, divide it by 2, then divide it by 14 (N in grams).

4 0
3 years ago
please help I don’t understand the difference between ammonium and ammonium ion. I possibly did 9.a) correct but part b is a bit
ozzi

You may find the electron diagram of ammonia and the ammonium ion in the attached picture.

Explanation:

Ammonia NH₃ have covalent bonds with the hydrogen atoms in which for each bond one electron comes from nitrogen atom (blue dot) and the other from the hydrogen atom (red dot). You may see that the nitrogen remains with the lone pair of electrons that does not participate in bonding.

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ammonium ion

brainly.com/question/10874844

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5 0
3 years ago
What is the ph (to nearest 0.01 ph unit) of a solution prepared by mixing 58.0 ml of 0.0179 m naoh and 60.0 ml of 0.0294 m ba(oh
Nina [5.8K]

Answer: -

12.59

Explanation: -

Strength of NaOH = 0.0179 M

Volume of NaOH = 58.0 mL = 58.0/1000 = 0.058 L

Number of moles = 0.0179 M x 0.058 L

= 1.04 x 10⁻³ mol

Mol of [OH⁻] given by NaOH = 1.04 x 10⁻³ mol

Strength of Ba(OH)₂ = 0.0294 M

Volume of Ba(OH)₂ = 60.0 mL = 60.0/1000 = 0.060 L

Number of moles = 0.0294 M x 0.060 L

= 1.76 x 10⁻³ mol

Mol of [OH⁻] given by Ba(OH)₂ =2 x 1.76 x 10⁻³ mol

Total [OH⁻] = 1.04 x 10⁻³ mol + 2 x 1.76 x 10⁻³ mol

= 4.56 x 10⁻³ mol

Total volume of the mixture = 58.0 + 60.0

= 118.0 mL

118.0 mL of the solution has 4.56 x 10⁻³ mol [OH⁻]

1000 mL of the solution has \frac{4.56 x 10-3 mol x 1000mL }{118 mL}

= 0.0386 mol

Using the relation

pOH = - log [OH-]

= - log 0.0386

= 1.41

Using the relation

pH + pOH = 14

pH = 14 - 1.41

= 12.59

5 0
3 years ago
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