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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

You might be interested in

A 5 cm spring is suspended with a mass of 3.8589 g attached to it which extends the spring by 1.5747 cm. The same spring is plac

lana66690 [7]

Answer:

charges of the beads is 1.173 × $10^{-15}$ C

Explanation:

given data

mass = 3.8589 g = 0.003859 kg

spring length = 5 cm = 0.05 m

extend spring x = 1.5747 cm = 0.15747 m

spring's extension = 0.0116 m

to find out

charges of the beads

solution

we know that force is

force = mass × g

force = 0.003859 × 9.8

force = 0.03782 N

so we know force for mass

force = -kx

so k = force / x

put here force and x value

k = -0.03782 / 0.1575

k = -0.24 N/m

and

force for spring's extension

force = -kx

force = -0.24 ( 0.0116) = 0.002784 N

so here

total length L = 0.05 + 0.0116 = 0.0616

so charges of the beads = force × L² / ke

charges of the beads = 0.002784 × (0.0616)² / (9 × $10^{9}$ )

so charges of the beads = 1.173 × $10^{-15}$ C

3 0

3 years ago

Explain how the extension of a spring is determined

fgiga [73]

Answer:

For a given spring the extension is directly proportional to the force applied For example if the force is doubled, the extension doubles When an elastic object is stretched beyond its limit of proportionality the object does not return to its original length when the force is removed

Explanation:

6 0

3 years ago

A girl stands on the edge of a merry-go-round of radius 1.71 m. If the merry go round uniformly accerlerates from rest to 20 rpm

Mashutka [201]

Answer:

$a = 0.53 m/s^2$

Explanation:

initially the merry go round is at rest

after 6.73 s the merry go round will accelerates to 20 rpm

so final angular speed is given as

$\omega = 2\pi f$

$\omega = 2\pi ( \frac{20}{60})$

$\omega = 2.10 rad/s$

so final tangential speed is given as

$v = r\omega$

$v = 1.71 (2.10) = 3.58 m/s$

now average acceleration of the girl is given as

$a = \frac{v_f - v_i}{\Delta t}$

$a = \frac{3.58 - 0}{6.73}$

$a = 0.53 m/s^2$

8 0

3 years ago

Baseball player A bunts the ball by hitting it in such a way that it acquires an initial velocity of 1.3 m/s parallel to the gro

pashok25 [27]

Answer:

Explanation:

cSep 20, 2010

well, since player b is obviously inadequate at athletics, it shows that player b is a woman, and because of this, she would not be able to hit the ball. The magnitude of the initial velocity would therefore be zero.

Anonymous

Sep 20, 2010

First you need to solve for time by using

d=(1/2)(a)(t^2)+(vi)t

1m=(1/2)(9.8)t^2 vertical initial velocity is 0m/s

t=.45 sec

Then you find the horizontal distance traveled by using

v=d/t

1.3m/s=d/.54sec

d=.585m

Then you need to find the time of player B by using

d=(1/2)(a)(t^2)+(vi)t

1.8m=(1/2)(9.8)(t^2) vertical initial velocity is 0

t=.61 sec

Finally to find player Bs initial horizontal velocity you use the horizontal equation

v=d/t

v=.585m/.61 sec

so v=.959m/s

5 0

3 years ago

Read 2 more answers

A proton travels with a speed of 5.02×10⁶ m/s in a direction that makes an angle of 60.0° with the direction of a magnetic ficld

Triss [41]

The magnitude of the magnetic force on the proton is 1.25× 10—¹³ N.

Speed of the proton = 5.02 × 10 ⁶ m /a

Angel of between the velocity and the magnetic force = 60 °

The magnitude of magnetic field B = 0.180 T

The magnitude of the magnetic force on the proton is,

$F = q(v \times B)$

$F = qvB \: sin \: θ$

$F = 1.6 \times 10 ^{ - 19} \times 5.02 \times 10 ^{6} \times 0.180 \times \: sin \: 60°$

$= 1.25 \times 10 ^{ - 13} \: N$

Therefore, the magnitude of the magnetic force on the proton is 1.25× 10—¹³ N.

To know more about magnetic force, refer to the below link:

brainly.com/question/23096032

#SPJ4

4 0

1 year ago