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Troyanec [42]
4 years ago
13

Alveolar air (a mixture of nitrogen, oxygen, and carbon dioxide) has a total pressure of 0.998 atm. If the partial pressure of o

xygen gas is 0.198 atm and the partial pressure of nitrogen gas is 0.770 atm, what is the partial pressure of carbon dioxide in millimeters of mercury?
Chemistry
1 answer:
inn [45]4 years ago
6 0

Answer:

The partial pressure of carbon dioxide is 22.8 mmHg

Explanation:

Dalton's Law is a gas law that relates the partial pressures of the gases in a mixture. This law says that the pressure of a gas mixture is equal to the sum of the partial pressures of all the gases present.

In this case:

Ptotal=Pnitrogen + Poxygen + Pcarbondioxide

You know that:

  • Ptotal= 0.998 atm
  • Pnitrogen= 0.770 atm
  • Poxygen= 0.198 atm
  • Pcarbondioxide= ?

Replacing:

0.998 atm=0.770 atm + 0.198 atm + Pcarbondioxide

Solving:

Pcarbondioxide= 0.998 atm - 0.770 atm - 0.198 atm

Pcarbondioxide= 0.03 atm

Now you apply the following rule of three: if 1 atm equals 760 mmHg, 0.03 atm how many mmHg equals?

Pcarbondioxide=\frac{0.03 atm*760 mmHg}{1 atm}

Pcarbondioxide= 22.8 mmHg

<u><em>The partial pressure of carbon dioxide is 22.8 mmHg</em></u>

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A compound with chemical formula na2cx3 has formula mass 106 amu .. what is the atomic mass of element x
maria [59]

The element "X" is "O" (oxygen).

<h3>Calculation:</h3>

Given,

Chemical formula = Na₂CX₃

Formula mass = 106 amu

Molar mass of Na = 23 amu

Molar mass of C = 12 amu

To find,

Element X =?

We will equate the equation as follows,

2(23) + 12 + 3(y) = 106

46 + 12 + 3y =106

58 + 3y = 106

3y = 106 - 58

3y = 48

y = 48/3

y = 16

We know that Oxygen has molecular mass of 16. Therefore the element "X" is "O".

Learn more about molar mass here:

brainly.com/question/22997914

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5 0
2 years ago
How many milliliters of a 3.4 M NaCl solution would be needed to prepare each solution?
Ksenya-84 [330]

Answer:

a. Approximately 1.3\; \rm mL.

b. Approximately 7.2\; \rm mL.

Explanation:

The unit of concentration "\rm M" is equivalent to "\rm mol \cdot L^{-1}", which means "moles per liter."

However, the volume of both solutions were given in mililiters \rm mL. Convert these volumes to liters:

\displaystyle 45\; \rm mL = 45\; \rm mL \times \frac{1\; \rm L}{1000\; \rm mL} = 0.045\; \rm L.

\displaystyle 330\; \rm mL = 330\; \rm mL \times \frac{1\; \rm L}{1000\; \rm mL} = 0.330\; \rm L.

In a solution of volume V where the concentration of a solute is c, there would be c \cdot V (moles of) formula units of this solute.

Calculate the number of moles of \rm NaCl formula units in each of the two solutions:

Solution in a.:

n = c \cdot V = 0.045\; \rm L \times 0.10\; \rm mol \cdot L^{-1} = 0.0045\; \rm mol.

Solution in b.:

n = c \cdot V = 0.330\; \rm L \times 0.074\; \rm mol \cdot L^{-1} = 0.02442\; \rm mol.

What volume of that 3.4\; \rm M (same as 3.4 \; \rm mol \cdot L^{-1}) \rm NaCl solution would contain that many

For the solution in a.:

\displaystyle V = \frac{n}{c} = \frac{0.0045\; \rm mol}{3.4\; \rm mol \cdot L^{-1}} \approx 0.0013\; \rm L.

Convert the unit of that volume to milliliters:

\displaystyle 0.0013\; \rm L = 0.0013\; \rm L \times \frac{1000\; \rm mL}{1\; \rm L} = 1.3\; \rm mL.

Similarly, for the solution in b.:

\displaystyle V = \frac{n}{c} = \frac{0.02442\; \rm mol}{3.4\; \rm mol \cdot L^{-1}} \approx 0.0072\; \rm L.

Convert the unit of that volume to milliliters:

\displaystyle 0.0072\; \rm L = 0.0072\; \rm L \times \frac{1000\; \rm mL}{1\; \rm L} = 7.2\; \rm mL.

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Arlecino [84]

Answer:b

Explanation:

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Which element has 21 neutrons and 20 electrons?
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Scandium has 21 neutrons
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