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sweet [91]
2 years ago
14

Using the Fick equation, calculate cardiac output if a person is consuming 1.41 L O2/min, has an arterial O2 content of 190 ml O

2/L, and a venous O2 content of 25 ml O2/L.
Chemistry
1 answer:
Elden [556K]2 years ago
8 0

Answer:

Cardiac output (CO) = 8.54 l / min

Explanation:

Given:

Oxygen consumption (V) = 1.41 l O2/min

Oxygen in arterial (Ca) = 190 ml O2/L = 0.19 l O2/L

Oxygen in venous (Cv) = 25 ml O2/L = 0.025 l O2/L

Find:

Cardiac output (CO)

Computation:

Cardiac output (CO) = V / (Ca - Cv)

Cardiac output (CO) = 1.41 / (0.19 - 0.025)

Cardiac output (CO) = 8.54 l / min

You might be interested in
Based on the bond energies for the reaction below, what is the enthalpy of the reaction?HC≡CH (g) + 5/2 O₂ (g) → 2 CO₂ (g) + H₂O
Anuta_ua [19.1K]

Answer:

1219.5 kj/mol

Explanation:

To reach this result, you must use the formula:

ΔHºrxn = Σn * (BE reactant) - Σn * (BE product)

ΔHºrxn = [1 * (BE C = C) + 2 * (BE C-H) + 5/2 * (BE O = O)] - [4 * (BE C = O) + 2 * (BE O-H).

The BE values are:

BE C = C: 839 kj / mol

BE C-H: 413 Kj / mol

BE O = O: 495 kj / mol

BE C = O = 799 Kj / mol

BE O-H = 463 kj / mol

Now you must replace the values in the above equation, the result of which will be:

ΔHºrxn = [1 * 839 + 2 * (413) + 5/2 * (495)] - [4 * (799) + 2 * (463) = 1219.5 kj/mol

8 0
2 years ago
The name "penicillin" is used for several closely-related antibiotics. A 1.2177 g sample of one of these compounds is burned, pr
masya89 [10]

<u>Answer:</u> The empirical formula for the given organic compound is C_7H_{11}O_2SN

<u>Explanation:</u>

The chemical equation for the combustion of hydrocarbon having carbon, hydrogen and oxygen follows:

S_vN_wC_xH_yO_z+O_2\rightarrow CO_2+H_2O+SO_2+NO

where, 'v', 'w' 'x', 'y' and 'z' are the subscripts of sulfur, nitrogen, carbon, hydrogen and oxygen respectively.

We are given:

Mass of CO_2=0.2829g

Mass of H_2O=0.1159g

Mass of SO_2=0.4503g

Mass of NO = 0.2109 g

We know that:

Molar mass of carbon dioxide = 44 g/mol

Molar mass of water = 18 g/mol

Molar mass of sulfur dioxide = 64 g/mol

Molar mass of nitrogen monoxide = 30 g/mol

  • <u>For calculating the mass of carbon:</u>

In 44 g of carbon dioxide, 12 g of carbon is contained.

So, in 2.1654 g of carbon dioxide, \frac{12}{44}\times 2.1654=0.590g of carbon will be contained.

  • <u>For calculating the mass of hydrogen:</u>

In 18 g of water, 2 g of hydrogen is contained.

So, in 0.6965 g of water, \frac{2}{18}\times 0.6965=0.077g of hydrogen will be contained.

  • <u>For calculating the mass of sulfur:</u>

In 64 g of sulfur dioxide, 32 g of sulfur is contained.

So, in 0.4503 g of sulfur dioxide, \frac{32}{64}\times 0.4503=0.225g of sulfur will be contained.

  • <u>For calculating the mass of nitrogen:</u>

In 30 g of nitrogen monoxide, 14 g of nitrogen is contained.

So, in 0.2109 g of nitrogen monoxide, \frac{14}{30}\times 0.2109=0.098g of nitrogen will be contained.

  • Mass of oxygen in the compound = (1.2177) - (0.590 + 0.077 + 0.225 + 0.098) = 0.2277 g

To formulate the empirical formula, we need to follow some steps:

  • <u>Step 1:</u> Converting the given masses into moles.

Moles of Carbon =\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{0.590g}{12g/mole}=0.049moles

Moles of Hydrogen = \frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{0.077g}{1g/mole}=0.077moles

Moles of Oxygen = \frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{0.2277g}{16g/mole}=0.0142moles

Moles of Sulfur = \frac{\text{Given mass of Sulfur}}{\text{Molar mass of sulfur}}=\frac{0.225g}{32g/mole}=0.007moles

Moles of Nitrogen = \frac{\text{Given mass of nitrogen}}{\text{Molar mass of nitrogen}}=\frac{0.098g}{14g/mole}=0.007moles

  • <u>Step 2:</u> Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 0.007 moles.

For Carbon = \frac{0.049}{0.007}=7

For Hydrogen  = \frac{0.077}{0.007}=11

For Oxygen  = \frac{0.0142}{0.007}=2.03\approx 2

For Sulfur  = \frac{0.007}{0.007}=1

For Nitrogen  = \frac{0.007}{0.007}=1

  • <u>Step 3:</u> Taking the mole ratio as their subscripts.

The ratio of C : H : O : S : N = 7 : 11 : 2 : 1 : 1

Hence, the empirical formula for the given compound is C_7H_{11}O_2S_1N_1=C_7H_{11}O_2SN

4 0
2 years ago
Walls made of paraffin wax, a covalent compound, help keep the temperature in a room steady as night changes into day and day in
Over [174]
The wall would absorb extra heat during the day when the sun is out, then release the heat back into the room when the sun goes down.
8 0
2 years ago
Read 2 more answers
What is the maximum volume of a 0.788 M CaCl2 solution that can be prepared using 85.3 g CaCl2
LekaFEV [45]
Molar mass CaCl₂ = 111.0 g/mol

number of moles:

n = mass of solute / molar mass

n = 85.3 / 111.0

n = 0.7684 moles of CaCl₂

M = n / V

0.788 M = <span>0.7684 / V
</span>
V = 0.7684 / 0.788

V = 0.97512 L

hope this helps!


3 0
2 years ago
A particular first-order reaction has a rate constant of 1.35 × 102 s-1 at 25.0°c. what is the magnitude of k at 75.0°c if ea =
IgorC [24]
According to this formula:
K= A*(e^(-Ea/RT) when we have K =1.35X10^2 & T= 25+273= 298K &R=0.0821
Ea= 85.6 KJ/mol So by subsitution we can get A:
1.35x10^2 = A*(e^(-85.6/0.0821*298))
1.35x10^2 = A * 0.03
A= 4333
by substitution with the new value of T(75+273) = 348K & A to get the new K
∴K= 4333*(e^(-85.6/0.0821*348)
  = 2.16 x10^2
8 0
2 years ago
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