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3 years ago

What is the mass of a 10.0 ml sample of urine with a specific gravity of 1.04?

1 answer:

5 0

The formula of specific gravity is:

Specific gravity = $\frac{density of an object}{density of water}$

To get this, you can derive a formula for mass. If you transpose the density of water to the other side of the equation (by specific gravity) you will multiply it with specific gravity. Now the density of water is 1 g/mL so you will retain the value of the specific gravity. That will just leave you with this formula:

Specific gravity = density of an object

Density of an object can be computed using the formula:

Density = mass/volume now you can fuse the two formulas to get this:

Specific gravity = mass/volume

From there we can derive the formula for mass by transposing volume to the other side of the equation:

Volume x specific gravity = mass

Lets's use your given in our new equation:

10.0 mL x 1.04 = mass
10.4 = mass

The mass of a 10.0 mL sample of urine with a specific gravity of 1.04 is 10.4g.

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The radioactive isotope of lead, Pb-209, decays at a rate proportional to the amount present at time t and has a half-life of 3.

Marina86 [1]

Answer:

Therefore it will take 7.66 hours for 80% of the lead decay.

Explanation:

The differential equation for decay is

$\frac{dA}{dt}= kA$

$\Rightarrow \frac{dA}{A}=kdt$

Integrating both sides

ln A= kt+c₁

$\Rightarrow A= e^{kt+c_1}$

$\Rightarrow A=Ce^{kt}$ [ $e^{c_1}=C$ ]

The initial condition is A(0)= A₀,

$\therefore A_0=Ce^{0.k}$

$\Rightarrow C=A_0$

$\therefore A=A_0e^{kt}$ .........(1)

Given that the $Pb_{209}$ has half life of 3.3 hours.

For half life $A=\frac12 A_0$ putting this in equation (1)

$\frac12A_0=A_0e^{k\times3.3}$

$\Rightarrow ln(\frac12)= 3.3k$ [taking ln both sides, $ln \ e^a=a$ ]

$\Rightarrow k=\frac{ln \frac12}{3.3}$

⇒k= - 0.21

Now A₀= 1 gram, 80%=0.8

and A= (1-0.8)A₀ = (0.2×1) gram = 0.2 gram

Now putting the value of k,A and A₀in the equation (1)

$\therefore 0.2=1e^{(-0.21)\times t}$

$\Rightarrow e^{-0.21t}=0.2$

$\Rightarrow -0.21t= ln(0. 2)$

$\Rightarrow t= \frac{ln (0.2)}{-0.21}$

⇒ t≈7.66

Therefore it will take 7.66 hours for 80% of the lead decay.

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