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dolphi86 [110]
3 years ago
9

A person accidentally swallows three drops of liquid oxygen, , which has a density of 1.149 g/ml. assuming the drop has a volume

of 0.050 ml, what volume of gas will be produced in the person's stomach at body temperature (37 °c) and a pressure of 1.0 atm?

Chemistry
2 answers:
DerKrebs [107]3 years ago
5 0
<span>134 ml First, let's determine how many moles of oxygen we have. Atomic weight oxygen = 15.999 Molar mass O2 = 2*15.999 = 31.998 g/mol We have 3 drops at 0.050 ml each for a total volume of 3*0.050ml = 0.150 ml Since the density is 1.149 g/mol, we have 1.149 g/ml * 0.150 ml = 0.17235 g of O2 Divide the number of grams by the molar mass to get the number of moles 0.17235 g / 31.998 g/mol = 0.005386274 mol Now we can use the ideal gas law. The equation PV = nRT where P = pressure (1.0 atm) V = volume n = number of moles (0.005386274 mol) R = ideal gas constant (0.082057338 L*atm/(K*mol) ) T = Absolute temperature ( 30 + 273.15 = 303.15 K) Now take the formula and solve for V, then substitute the known values and solve. PV = nRT V = nRT/P V = 0.005386274 mol * 0.082057338 L*atm/(K*mol) * 303.15 K / 1.0 atm V = 0.000441983 L*atm/(K*) * 303.15 K / 1.0 atm V = 0.133987239 L*atm / 1.0 atm V = 0.133987239 L So the volume (rounded to 3 significant figures) will be 134 ml.</span>
DiKsa [7]3 years ago
5 0

volume of gas = 0.1374L

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How many grams of water are produced when 4.50 L of
MA_775_DIABLO [31]

The answer for the following problem has been mentioned below.

  • <em><u>Therefore the mass of the water is 5.802 grams.</u></em>

Explanation:

Given:

volume of oxygen (V) = 4.50 L

Temperature (T) = 425 K

pressure of oxygen (P) = 2.50 atm

Gram molecular mass of oxygen (M) = 16.0 grams

To calculate:

mass of water (m)

We know;

According to the ideal gas equation;

     P × V = n × R × T

As we know;

no of moles = \frac{m}{M}

m represents the mass of oxygen (m)

M represents the Gram molecular mass (M)

According to above mentioned equation;

           P × V = n × R × T

P represents the pressure of the oxygen

V represents the volume of the oxygen

n represents the no of moles of the oxygen

R represents the universal gas constant

where,

the value of R is 0.0821 L atm/K moles

Substituting the values in the above equation;

                  2.50 × 4.50 = \frac{m}{16.0} × 0.0821 × 425

                   11.25 =  \frac{m}{16.0} × 34.8925

                  180 = m × 34.8925

                  m = \frac{180}{34.8925}

                  m = 5.158 grams

Therefore the mass of the of oxygen is 5.158 grams

Now;

As we know;

           \frac{m_{1} }{M_{1} } = \frac{m_{2} }{M_{2} }

where;

m_{1} represents the mass of the oxygen

M_{1} represents the gram molecular mass of the oxygen

m_{2} represents the mass of the water

M_{2} represents the gram molecular mass of water

    From the above given formula,

      \frac{5.158}{16.0} = \frac{m_{2} }{18}

where;

Gram molecular weight of water = 18.0 u

    m_{2} = 5.802 grams

<em><u>Therefore the mass of the water is 5.802 grams.</u></em>

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