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kap26 [50]
3 years ago
5

The partial pressure of O2 in your lungs varies from 25 mm Hg to 40 mm Hg. What mass of O2 can dissolve in 1.0L OF water at 25.0

oC when the O2 partial pressure is 40. mm Hg?
Chemistry
1 answer:
Inga [223]3 years ago
7 0

The partial pressure of O2 in your lungs varies from 25 mm Hg to 40 mm Hg. What mass of O2 can dissolve in 1.0L OF water at 25.0oC when the O2 partial pressure is 40. mm Hg? Since the normal Water solubility of oxygen at 25oC and pressure = 1 bar is at 40 mg/L water, the partial pressure is only40 mmg Hg

Mass of 02 dissolve = (40 mmHg/760 mmHg)*( 40 mg/L) (1 L) =2.10 mg O2 dissolved

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A mixture of NaBrO 3 , NaBrO3, NaHCO 3 , NaHCO3, Na 2 CO 3 , Na2CO3, and NaBr NaBr was heated, producing H 2 O , H2O, CO 2 , CO2
nata0808 [166]

Answer:

Composition of initial mixture is:

9.02g of NaBrO₃

15.84g of  Na₂CO₃

17.06g of NaHCO₃

82.58g NaBr

Explanation:

For the reactions:

2NaBrO₃(s) ⟶ 2NaBr(s) + 3O₂(g)

2NaHCO₃(s) ⟶ Na₂O(s) + H₂O(g) + 2CO₂(g)

Na₂CO₃(s) ⟶ Na₂O(s)+CO₂(g)

All H₂O(g) comes from NaHCO₃. Thus, initial moles and mass of NaHCO₃ are:

1.83g H₂O ₓ (1 mol H₂O / 18.02g) ₓ (2 mol NaHCO₃ / 1 mol H₂O) = <em>0.203moles NaHCO₃</em> ₓ (84g / 1mol NaHCO₃) =

<em>17.06g of NaHCO₃</em>

CO₂ comes from NaHCO₃ and Na₂CO₃.

15.51g of CO₂ are:

15.51g CO₂ ₓ (1mol / 44.01g) =<em> 0.352moles of CO₂</em>

As 2 moles of NaHCO₃ produce 2 moles of CO₂, moles of CO₂ that comes from NaHCO₃ are 0.203moles NaHCO₃. Moles of CO₂ that comes from Na₂CO₃ are:

0.352mol CO₂ - 0.203mol CO₂ = <em>0.149mol CO₂</em>

<em />

These moles of CO₂ are produced from:

0.149mol CO₂ ₓ (1 mol Na₂CO₃ / 1 mol CO₂) ₓ (106g / 1mol Na₂CO₃) =

<em>15.84g of  </em>Na₂CO₃

And all O₂ comes from NaBrO₃. Initial mass of NaBrO₃ is:

2.87g O₂ ₓ (1 mol O₂ / 32g) ₓ (2 mol NaBrO₃ / 3 mol O₂) ₓ (150.9g / 1mol NaBrO₃) =

<em>9.02g of </em>NaBrO₃

If initial mass of the mixture was 124.5g, mass of NaBr was:

124.5g - 9.02g of NaBrO₃ - 15.84g of  Na₂CO₃ - 17.06g of NaHCO₃ =

<em>82.58g NaBr</em>

<em />

<em>Composition of initial mixture is:</em>

<em>9.02g of NaBrO₃</em>

<em>15.84g of  Na₂CO₃</em>

<em>17.06g of NaHCO₃</em>

<em>82.58g NaBr</em>

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