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2 years ago

Calculate the density of a 50 ml solution that is 5% water and 95% ethanol. show your calculation.

Physics

1 answer:

vichka [17]2 years ago

6 0

The density of the solution from the given compositions of water and ethanol is 799.55 kg/m³.

The given parameters;

Volume of the solution, v = 50 ml
Composition of water, 5%
Composition of ethanol, 95%
Density of water, = 1000 kg/m³
Density of ethanol, = 789 kg/m³

The density of the solution from the given compositions of water and ethanol is obtained by finding the respective amount of water and ethanol composition is the solution.

The density of the solution is calculated as follows;

$\rho = \frac{5(1000 \ kg/m^3) + 95(789 \ kg/m^3)}{100} \\\\\rho = 799.55 \ kg/m^3$

Thus, the density of the solution from the given compositions of water and ethanol is 799.55 kg/m³.

Learn more here:brainly.com/question/17826958

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Answer:

$q=10^{-6}(e^{-4000t}-4e^{-1000t}+3)C$

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#To find the particular solution:

$Q(t)=A,\ Q\prime(t)=0,Q\prime \prime(t)=0\\\\0+0+4000000A=12\\\\A=3\times10^{-6}\\\\Q(t)=3\times10^{-6},\\\\q=q_c+Q(t)\\\\q=c_1e^{-4000t}+c_2e^{-1000t}+3\times10^{-6}\\\\q\prime=-4000c_1e^{-4000t}-1000c_2e^{-1000t}\\q\prime(0)=0\\\\-4000c_1-1000c_2=0\\c_1+c_2+3\times10^{-6}=0\\\\#solving \ simultaneously\\\\c_1=10^{-6},c_2=-4\times10^{-6}\\\\q=10^{-6}e^{-4000t}-4\times10^{-6}e^{-1000t}+3\times10^{-6}\\\\q=10^{-6}(e^{-4000t}-4e^{-1000t}+3)C$

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