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maxonik [38]
4 years ago
15

In the reaction C6H12 + ________ O2----> 6H20 + 6CO2, what coefficient should be placed in front of 02 to balance the reactio

n?
Physics
2 answers:
Alborosie4 years ago
7 0
All you need to do is count the number of Oxygen the Products:
In 6H2O, you have 6 oxygen, and in 6CO2, you have 6 Dioxygen so 12 Oxygen. So a total of 12 + 6 = 18 atoms of Oxygen, or 9 atoms of dioxygen.
So the coefficient that should be placed in front of O2 to balance the reaction is 9:
C6H12 + 9O2 ---> 6H2O + 6CO2

Hope this Helps! :)
astraxan [27]4 years ago
6 0

Answer : The coefficient placed in front of O_2 to balance the reaction should be, 9

Explanation :

Balanced chemical reaction : It is defined as the number of atoms of the individual elements present on reactant side in the reaction must be equal to the product side.

The given unbalanced chemical reaction is,

C_6H_{12}+O_2\rightarrow 6H_2O+6CO_2

This reaction is an unbalanced chemical reaction because in this reaction the number oxygen atoms are not balanced.

In order to balance the chemical reaction, we put the coefficient '9' before the oxygen molecule.

Thus, the balanced chemical reaction will be,

C_6H_{12}+9O_2\rightarrow 6H_2O+6CO_2

Hence, the coefficient placed in front of O_2 to balance the reaction should be, 9

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Explanation:

The problem doesn't specify that the units have to be g/mL, so you can calculate the density in kg/L without converting the mass or volume.

Just make sure that either way, you write the units.

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If I push a box at a constant rate is there friction force acting on it?
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Yes, the friction is acting in the opposite direction you are pushing.
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3 years ago
You are driving your 1700 kg car at 21 m/s down a hill with a 5.0∘ slope when a deer suddenly jumps out onto the roadway. You sl
podryga [215]

Answer:

d = 27.7 m

Explanation:

Here the car is driving on the inclined plane

So here we can say that work done by the gravity and work done by friction is equal to change in kinetic energy of the system

So here we can write it as

mgsin\theta \times d - F_f \times d = \frac{1}{2}mv_f^2 - \frac{1}{2}mv_i^2

now we have

m = 1700 kg

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F_f = 1.5 \times 10^4 N

(1700)(9.81)sin5 (d) - (1.5\times 10^4)d = 0 - \frac{1}{2}(1700)(21^2)

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6 0
3 years ago
A piece of driftwood moves up and down as water waves pass beneath it. However, it does not move toward the shore with the waves
KIM [24]

Answer:A piece of driftwood moves up and down as water waves pass beneath it. However, it does not move toward the shore with the waves. What does this demonstrate about the propagation of waves through a medium?

A) Waves transmit energy but not matter as they progress through a medium.

B) Waves transmit matter but not energy as they progress through a medium.

C) Waves do not transmit matter or energy as they progress through a medium.

D) Waves transmit energy as well as matter as they progress through a medium.

Explanation:

A piece of driftwood moves up and down as water waves pass beneath it. However, it does not move toward the shore with the waves. What does this demonstrate about the propagation of waves through a medium?

A) Waves transmit energy but not matter as they progress through a medium.

B) Waves transmit matter but not energy as they progress through a medium.

C) Waves do not transmit matter or energy as they progress through a medium.

D) Waves transmit energy as well as matter as they progress through a medium.

4 0
3 years ago
Q1 is located at the origin, Q2 is located at x = 2.50 cm and Q3 is located at x = 3.50 cm. Q1 has a charge of +4.92μC and Q3 ha
Inessa05 [86]

Answer:

+1.11\mu C

Explanation:

A charge located at a point will experience a zero electrostatic force if the resultant electric field on it due to any other charge(s) is zero.

Q_1 is located at the origin. The net force on it will only be zero if the resultant electric field intensity due to Q_2 and Q_3 at the origin is equal to zero. Therefore we can perform this solution without necessarily needing the value of Q_1.

Let the electric field intensity due to Q_2 be +E_2 and that due to Q_3 be -E_3 since the charge is negative. Hence at the origin;

+E_2-E_3=0..................(1)

From equation (1) above, we obtain the following;

E_2=E_3.................(2)

From Coulomb's law the following relationship holds;

+E_2=\frac{kQ_2}{r_2^2}\\  

-E_3=\frac{kQ_3}{r_3^2}

where r_2 is the distance of Q_2 from the origin, r_3 is the distance of Q_3 from the origin and k is the electrostatic constant.

It therefore means that from equation (2) we can write the following;

\frac{kQ_2}{r_2^2}=\frac{kQ_3}{r_3^2}.................(3)

k can cancel out from both side of equation (3), so that we finally obtain the following;

\frac{Q_2}{r_2^2}=\frac{Q_3}{r_3^2}................(4)

Given;

Q_2=?\\r_2=2.5cm=0.025m\\Q_3=-2.18\mu C=-2.18* 10^{-6}C\\r_3=3.5cm=0.035m

Substituting these values into equation (4); we obtain the following;

\frac{Q_2}{0.025^2}=\frac{2.18*10^{-6}}{0.035^2}\\\\hence;\\\\Q_2=\frac{0.025^2*2.18*10^{-6}}{0.035^2}\\

Q_2=\frac{0.00136*10^{-6}}{0.00123}=1.11*10^{-6}C\\\\Q_3=+1.11\mu C

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