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kirza4 [7]
3 years ago
7

A person accidentally swallows a drop of liquid oxygen, O2(l), which has a density of 1.149 g/mL. Assuming the drop has a volume

of 0.053 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:
nasty-shy [4]3 years ago
8 0

Answer:

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.

Anna [14]3 years ago
4 0
You need to divid 02(1) in to the first number that’s there
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What is the concentration of a solution containing 1.11 g sugar (sucrose, C12H22O11, MW = 342.3 g/mol, d = 1.587 g/cm3) in 432 m
djyliett [7]

Answer:

0.0075 M

0.0060 m

Explanation:

Our strategy here is to use the definition of molarity and molality to solve this question.

The molarity is the number of moles of solute, sucrose in this case, per liter of solution.

The molality is the number of moles of solute per kilogram of solvent.

So the molarity of the  solution is

M = moles of solute/ V solution

As we see we need the volume of solution since we are only given the volume of solvent, but this will be easy to compute since we have the density of  sucrose.

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Volume of solution = Vol Sucrose + Vol glycerine

d = m/V ⇒ Vsucrose = m / d = 1.11 g/ 1.587 g/cm³ = 0.70 cm³

Vol solution = 432 mL + 0.70 mL = 432.7 mL  (1cm³  = 1 mL)

Vol solution = 432.7 mL x 1 L / 1000 mL = 0.4327 L

⇒ M = 3.2 x 10⁻³  mol / 0.4327 L = 0.0075  M

For the molarity what we need is to first calculate the kilograms of glycerine from the given density:

d = m/v ⇒ m = d x v = 1.261 g/cm³ x  432 cm³ = 544.75 g

Converting to Kg:

544.75 g x 1 Kg/ 1000 g = 0.544 kg

Now the molality is

m = mol sucrose/ kg solvent = 3.24 x 10⁻³ mol / 0.544 Kg = 0.0060 m

Note: In the calculation for  volume of solution we could have approximated it to that of just glycerine, but since the density of sucrose was given we calculated the total volume of solution to be more rigorous.

8 0
3 years ago
Chemistry 10th grade PLEASE really need these answers :(
oksian1 [2.3K]

Answer:

I hope this helped you.

Explanation:

A: 1

B: 3

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