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Lemur [1.5K]
3 years ago
7

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

Physics
2 answers:
zhannawk [14.2K]3 years ago
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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18. Write conversion factors (as ratios) for the number of:
Romashka [77]

Answer:

Conversion tables show:

1 m = 1.09361 yds

1 Lit = .26418 gal = 1.05672 qt     or 1 qt = .944632 Lit

1 lb = .45359 kg = 2.2046 Lbs / Kg

So X yds = X  m * 1.09361 yds / m = 1.09361 * X yds

Likewise X Lit  = X  qt  / 1.05672 qt/ Lit = X / 1.05672 Lit = .94632  X Lit

So X  Lbs =  X kg * 2.2046 Lbs / Kg = 2.2046 Lbs

6 0
3 years ago
Velocity is the slope of the acceleration vs. time graph.<br> A.) True<br> B.)False<br> (apex)
Mekhanik [1.2K]

Velocity vs. time graph shows the acceleration as a slope whereas displacement vs. time graph shows the velocity as a slope. So, the given statement is false.

Answer: Option B

<u>Explanation:</u>

To understand the acceleration graphically, consider the x axis of the graph as the run and the y axis as the velocity rise. Now, as we all know that,

            \text {Acceleration}=\frac{\text {Change in velocity }}{\text {Time interval}}

We can estimate this through the graph. let's draw the motion of an object with time if it's velocity is changing in every second by 4 m/s. Now if we draw this on graph, we will see that there is a slope between the two corresponding values of time and velocity. This slope defines the acceleration for the object with time.

Now, in the same way, if we draw a distance and time graph respective to the y and x axis; we'll get a slope which defines the velocity of the object i.e. change in distance with time.

Hence, with a velocity vs time graph, we get the slope for acceleration whereas with the distance and time graph, we get the slope for velocity. So both the cases, we see there is no velocity slope on an acceleration and time graph. Hence the statement is false.

8 0
3 years ago
Read 2 more answers
A train moves uniform velocity of 36km/h for 15 sec.find distance
vesna_86 [32]

Answer:

S = V × t

Explanation:

first convert 36km/h to m/s

36km/h=36*5/18=180/18=10m/s

d =s*t

=10*10=100m

5 0
3 years ago
Two small nonconducting spheres have a total charge of 93.0 μC . Part A
Travka [436]

Answer:

charge on each

Q1 = 2.06 ×10^{-5}  C

Q2 = 7.23 × 10^{-5} C

when force were attractive

Q1 = 1.07 × 10^{-4} C

Q2 = -1.39 × 10^{-5} C

Explanation:

given data

total charge = 93.0 μC

apart distance r  = 1.14 m

force exerted  F = 10.3 N

to find out

What is the charge on each and What if the force were attractive

solution

we know that force is repulsive mean both sphere have same charge

so total charge on two non conducting sphere is

Q1 + Q2 = 93.0 μC  = 93 ×10^{-6} C

and

According to Coulomb's law force between two sphere is

Force F = \frac{K Q1 Q2 }{r^2}      .........1

Q1Q2 = \frac{F*r^2}{k}

here F is force and r is apart distance and k is 9 × 10^{9} N-m²/C² put all value we get

Q1Q2 = \frac{ 10.3*1.14^2}{9*10^9}

Q1Q2 = 1.49 ×  10^{-9} C²

and

we have  Q2 = 93 ×10^{-6} C - Q1

put here value

Q1²  - 93 ×10^{-6}  Q1 + 1.49 ×  10^{-9} = 0

solve we get

Q1 = 2.06 ×10^{-5}  C

and

Q1Q2 = 1.49 ×  10^{-9}

2.06 ×10^{-5}  Q2 = 1.49 ×  10^{-9}

Q2 = 7.23 × 10^{-5} C

and

if force is attractive we get here

Q1Q2 = - 1.49 ×  10^{-9} C²

then

Q1²  - 93 ×10^{-6}  Q1 - 1.49 ×  10^{-9} = 0

we get here

Q1 = 1.07 × 10^{-4} C

and

Q1Q2 = - 1.49 ×  10^{-9}

2.06 ×10^{-5}  Q2 = - 1.49 ×  10^{-9}

Q2 = -1.39 × 10^{-5} C

5 0
3 years ago
How do you add this vector graphically
arlik [135]

(-3,3)+(2,3)=(1,6)

this is the answer :)

8 0
3 years ago
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