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3 years ago

How does the law of conservation of mass apply to this reaction: C2H4 + O2 → H2O + CO2?

2 answers:

7 0

The correct answer is D. The law of conservation of mass states that in a system matter can neither be created or destroyed, hence atoms of each element on reactant side should equal atoms of each element on product side. In the unbalanced equation:

$C_2H_4 + O_2 ==> H_2O + CO_2$ , the are 2 carbon atoms on reactant side as opposed to 1 carbon atom on product side, there are 4 hydrogen atoms on reactant side as opposed to 2 hydrogen atoms on product side, and there are 2 oxygen atoms on reactant side as opposed to 3 atoms of oxygen on product side. To balance the equation we add a coefficient of 3 on O_2 and on the product side we put a coefficient of 2 on both water $H_2O$ and carbon dioxide $CO_2$ .

$C_2H_4 + 3O_2 ==> 2H_2O + 2CO_2$ .

kicyunya [14]3 years ago

4 0

The answer is a

the equation needs to be balanced. There are fewer oxygen atoms in the equation than hydrogen or a carbon

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2 years ago

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2 years ago

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Two small spheres with mass 10 gm and charge q, are suspended from a point by threads of length L=0.22m. What is the charge on e

Nataliya [291]

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$q=1.95*10^{-7}C$

Explanation:

According to the free-body diagram of the system, we have:

$\sum F_y: Tcos(15^\circ)-mg=0(1)\\\sum F_x: Tsin(15^\circ)-F_e=0(2)$

So, we can solve for T from (1):

$T=\frac{mg}{cos(15^\circ)}(3)$

Replacing (3) in (2):

$(\frac{mg}{cos(15^\circ)})sin(15^\circ)=F_e$

The electric force ( $F_e$ ) is given by the Coulomb's law. Recall that the charge q is the same in both spheres:

$(\frac{mg}{cos(15^\circ)})sin(15^\circ)=\frac{kq^2}{r^2}(4)$

According to pythagoras theorem, the distance of separation (r) of the spheres are given by:

$sin(15^\circ)=\frac{\frac{r}{2}}{L}\\r=2Lsin(15^\circ)(5)$

Finally, we replace (5) in (4) and solving for q:

$mgtan(15^\circ)=\frac{kq^2}{(2Lsin(15^\circ))^2}\\q=\sqrt{\frac{mgtan(15^\circ)(2Lsin(15^\circ))^2}{k}}\\q=\sqrt{\frac{10*10^{-3}kg(9.8\frac{m}{s^2})tan(15^\circ)(2(0.22m)sin(15^\circ))^2}{8.98*10^{9}\frac{N\cdot m^2}{C^2}}}\\q=1.95*10^{-7}C$

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3 years ago