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svet-max [94.6K]
3 years ago
15

bottle contains 425 mL of insulin at a concentration of 50.0 mg/mL . What is the total mass of insulin in the bottle?

Chemistry
1 answer:
Ulleksa [173]3 years ago
3 0
Hey there!

C = m
     ------
       V


50.0 mg/mL =      mass
                         ---------------
                             425 mL

mass =   50.0 * 425

mass => 21250 mg
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3 years ago
Argon (Ar) and helium (He) are initially in separate compartments of a container at 25°C. The
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Answer:

(a) V_B=11.68L

(b) x_{He}=0.533

Explanation:

Hello,

In this case, since the both gases behave ideally, with the given information we can compute the moles of He in A:

n_A=\frac{0.082\frac{atm*L}{mol*K}*298K}{1.974 atm*6.00L}=2.063mol

Thus, since the final pressure is 3.60 bar, we can write:

P=x_{Ar}P_A+x_{He}P_B\\\\P=\frac{n_{Ar}}{n_{Ar}+n_{He}} P_A+\frac{n_{He}}{n_{Ar}+n_{He}} P_B\\\\3.60bar=\frac{2.063mol}{2.063mol+n_{He}} *2.00bar+\frac{n_{He}}{2.063mol+n_{He}} *5.00bar

The moles of helium could be computed via solver as:

n_{He}=2.358mol

Or algebraically:

3.60bar=\frac{1}{2.063mol+n_{He}} *(4.0126+5.00*n_{He})\\\\7.314+3.60n_{He}=4.013+5.00*n_{He}\\\\7.314-4.013=5.00*n_{He}-3.60n_{He}\\\\n_{He}=\frac{3.3}{1.4}=2.358mol

In such a way, the volume of the compartment B is:

V_B=\frac{n_{He}RT}{P_B}=\frac{2.358mol*0.082\frac{atm*L}{mol*K}*298.15K}{4.935atm}\\  \\V_B=11.68L

Finally, he mole fraction of He is:

x_{He}=\frac{2.358}{2.358+2.063}\\ \\x_{He}=0.533

Regards.

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