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patriot [66]
1 year ago
8

Arterial blood contains about 0.25 g of oxygen per liter at 37°C and standard atmospheric pressure. Under these conditions, the

Henry's law constant is kH = 3.7 × 10^–2 mol/(L • atm), and the mole fraction of O2 in the atmosphere is 0.209Calculate the solubility (in M) of O2 in the blood of a climber on Mt. Everest, where Patm = 0.35 atm.________ M
Chemistry
1 answer:
Olin [163]1 year ago
6 0

Firstly we need to determine the partial pressure of O2:

\begin{gathered} P_{O_2}=X\times P_T \\ P_{O_2}:partial\text{ }pressure \\ X:mole\text{ }fraction \\ P_T:total\text{ }pressure \\  \\ P_{O_2}=0.209\times0.35\text{ }atm \\ P_{O_2}=0.073\text{ }atm \end{gathered}

We will now use the Henry's Law equation to determine the solubility of the gas:

\begin{gathered} c=K_H\times P_{O_2} \\ c:solubility\text{ }or\text{ }concentration\text{ }of\text{ }the\text{ }gas(M) \\ K_H:Henry^{\prime}sLawconstant=3.7\times10^{-2}M\text{ }atm^{-1} \\ P_{O_2}:partial\text{ }pressure\text{ }of\text{ }the\text{ }gas=0.073atm \\  \\ c=3.7\times10^{-2}M\text{ }atm^{-1}\times0.073 \\ c=2.7\times10^{-3}M \end{gathered}

Answer: Solubility is 2.7x10^-3 M

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The graphics below shows how photosynthesis and cellular respiration are related.
lisov135 [29]

Options found from another source are:

a. oxygen. b. glucose. c. energy stored as ATP. d. carbon dioxide and water

Answer:

c energy stored as ATP

Explanation:

Cellular respiration converts glucose into energy in the form of ATP (c). The answer cannot be oxygen (a), because this is required for this process as a final electron acceptor. In terms of photosynthesis, oxygen is released as a by-product. The answer cannot be glucose (b) because that is our starting point for respiration, and what is synthesised during photosynthesis. The answer cannot be (d) as carbon dioxide and water are released by cellular respiration, and required by photosynthesis

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A cork has a mass of 3 grams and a volume of 16 cms calculate the density
REY [17]

hey mate here is ur answer

solution

mass{m}=3 gram

             =3/1000

             

volume{v}=16cm

                =16/100

               

density=m/v

           =3/1000÷16/100

           =3/160

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6 0
3 years ago
How many moles of NH3 can be produced from 12.0 mol of H2 and excess N2? Express your answer numerically in moles. View Availabl
VladimirAG [237]

Answer:

A) 8.00 mol NH₃

B) 137 g NH₃

C) 2.30 g H₂

D) 1.53 x 10²⁰ molecules NH₃

Explanation:

Let us consider the balanced equation:

N₂(g) + 3 H₂(g) ⇄ 2 NH₃(g)

Part A

3 moles of H₂ form 2 moles of NH₃. So, for 12.0 moles of H₂:

12.0molH_{2}.\frac{2molNH_{3}}{3molH_{2}} =8.00molNH_{3}

Part B:

1 mole of N₂ forms 2 moles of NH₃. And each mole of NH₃ has a mass of 17.0 g (molar mass). So, for 4.04 moles of N₂:

4.04molN_{2}.\frac{2molNH_{3}}{1molN_{2}} .\frac{17.0gNH_{3}}{1molNH_{3}} =137gNH_{3}

Part C:

According to the <em>balanced equation</em> 6.00 g of H₂ form 34.0 g of NH₃. So, for 13.02g of NH₃:

13.02gNH_{3}.\frac{6.00gH_{2}}{34.0gNH_{3}} =2.30gH_{2}

Part D:

6.00 g of H₂ form 2 moles of NH₃. An each mole of NH₃ has 6.02 x 10²³ molecules of NH₃ (Avogadro number). So, for 7.62×10⁻⁴ g of H₂:

7.62 \times 10^{-4} gH_{2}.\frac{2molNH_{3}}{6.00gH_{2}} .\frac{6.02\times 10^{23}moleculesNH_{3}  }{1molNH_{3}}=1.53\times10^{20}moleculesNH_{3}

3 0
3 years ago
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