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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. Calculate; The acceleration of any object due to Earth's gravity is -9.81 m/s. For every

RideAnS [48]

As the time of motion increases, the velocity of the object increases downwards.

The given parameters;

acceleration due to gravity, g = -9.81 m/s²
velocity of the object, v₀ = 9.8 m/s

The final velocity of the object at different time is calculated as follows;

when the time = 1 second;

v = v₀ - gt

v = 9.8 - 9.8(1)

v = 0

when the time = 2 second;

v = v₀ - gt

v = 9.8 - 9.8(2)

v = -9.8 m/s

when the time = 3 second;

v = v₀ - gt

v = 9.8 - 9.8(3)

v = -19.6 m/s

Thus, we can conclude that as the time of motion increases, the velocity of the object increases downwards.

Learn more here:brainly.com/question/16782759

8 0

3 years ago

Question 9 In an RC series circuit, ε = 12.0 V, R = 1.07 MΩ, and C = 2.66 µF. (a) Calculate the time constant. (b) Find the maxi

meriva

Answer:

a.) τ = 2.85 s b.) Q = 3.19 * 10^-5 C c.) t = 1.691 s

Explanation:

So we are told that it is a RC circuit. We are told $Q = C V [1 - e^(-t/RC)]$ = 12.0 V, R = 1.07 MΩ and C = 2.66 µF.

a.) The time constant for RC circuit, τ = RC. Substituting our known values we get:

τ = RC where R = (1.07 * 10 ^ 6)Ω and C = (2.66 * 10 ^ -6) F

τ = (1.07 * 10 ^ 6)Ω * (2.66 * 10 ^ -6) F = 2.8462 s ≈ 2.85 s

τ = 2.85 s

b.) The relationship between capacitance, potential, charge is given:

$Q = CV[1-e^{-t/RC} ]$

The capacitor is fully charge when t approaches infinity, therefore:

$Q = \lim_{t \to \infty} a_n CV[1-e^{-t/RC} ]$

When t approaches infinity, the term e becomes very small (e^-∞ = 0), therefore we can simplify the equation and plug in our values

$Q = (2.66*10^{-6}) F * (12.0)V *[1 - 0] = 3.192 * 10^{-5}$

Q = 3.19 * 10^-5 C

c.) Using the same equation as before, we can substitute Q in and solve for Q:

$(14.3 * 10 ^ 6) C = (2.66*10^{-6})F *(12.0)V*[1-e^{-t/(2.85s)}]\\0.552 = e^{-t/(2.85s)}\\t = -1 * 2.85 * ln(0.552) \\t = 1.69120678 s$

t = 1.691 s

Hope this helps! I'm not sure what the units you want, so convert to the desired units.

6 0

3 years ago

You are observing the radiation from a distant active galaxy and you notice that the amplitude of the signal varies in strength

GREYUIT [131]

Answer:

Period of the signal.

Explanation:

So, this question is all about a concept in physics or astronomy which is called or known as Radiation Astronomy and Galactic Nuclei that are active. This concept talks most about Quasars; a powerful radiating object which derives its power from black holes.

When You take a look at Quasars, we get the to know that the more you think you can see, the more they move away from us.

Thus, when "You are observing the radiation from a distant active galaxy and you notice that the amplitude of the signal varies in strength regularly over a certain period. The maximum possible size for the source of this radiation can now be calculated from the "PERIOD OF THE SIGNAL.

NB: not the amplitude but the period.

7 0

4 years ago

A strip of copper metal is riveted to a strip of aluminum. the two metals are then heated. the coefficient of linear expansion o

olganol [36]

We can base this on the equation of thermal expansion.

ΔL = L₀αΔT
where
ΔL is the expansion of length upon heating
L₀ is the initial length
α is the coefficient of linear expansion
ΔT is the temperature difference

So, if α for Aluminum is greater than α for Copper, then after heating, aluminum would be longer than copper.

4 0

3 years ago

8th grade science help help

Maslowich

Answer: A

Explanation:

7 0

3 years ago