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Aleonysh [2.5K]
3 years ago
10

Which equation represents a combustion reaction? Pb(NO3)2 + 2HCl → PbCl2 + 2HNO3 2SO2 + O2 → 2SO3 2C2H6 + 7O2 → 4CO2 + 6H2O Ca +

2HCl → CaCl2 + H2
Physics
2 answers:
Arturiano [62]3 years ago
5 0

Answer:

2C_2H_6+7O_2\rightarrow4CO_2+6H_2O it is a combustion reaction.

Explanation:

Combustion reaction is the type of reaction in which a hydrocarbon heated up in the presence of oxygen to form water and carbon dioxide.

C_2H_6 is ethane and it is a hydrocarbon. The combustion reaction of ethane is as follows :

2C_2H_6+7O_2\rightarrow4CO_2+6H_2O

So, the correct option is (c).

Naddik [55]3 years ago
3 0

Answer:

2C2H6 + 7O2 → 4CO2 + 6H2O

Explanation:

Combustion is when a hydrocarbon (a compound made of only hydrogen and carbon) reacts with oxygen at high temperature to produce CO2 and H2O gases.

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What force is required to move an object 8 m using 24 j of work?
Anna35 [415]

Answer:

3N

Explanation:

Workdone = F x d

F = force

d = distance 8m

Workdone = 24J

24 = F x 8

Divide both sides by 8

24/8 = F x 8/8

3 = F

F = 3N

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As an airplane leaves the runway and goes into the air, has acceleration occurred?
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Car and a train move together along straight, parallel paths with the same constant cruising speed v0. At t=0 the car driver not
musickatia [10]

Answer:

Part A: t = v_0/a_0

Part B: t = v_0/a_0

Part C: v_0^2/a_0

Explanation:

Part A:

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v = v_0 + at\\0 = v_0 - a_0t\\t = \frac{v_0}{a_0}

Part B:

We will use the same kinematics equation:

v = v_0 + at\\v_0 = 0 + a_0t\\t = \frac{v_0}{a_0}

Part C:

The total time takes is 2t.

So the train moves a distance of

x = v_0(2t) = 2v_0(\frac{v_0}{a_0}) = \frac{2v_0^2}{a_0}

And the car moves a distance in Part A and in Part B:

d_A = v_0t + \frac{1}{2}at^2 = v_0(\frac{v_0}{a_0}) - \frac{1}{2}a_0(\frac{v_0^2}{a_0^2}) = \frac{v_0^2}{a_0} - \frac{v_0^2}{2a_0} = \frac{v_0^2}{2a_0}\\d_B = v_0t + \frac{1}{2}at^2 = 0 + \frac{1}{2}a_0(\frac{v_0^2}{a_0^2}) = \frac{v_0^2}{2a_0}

So the total distance that the car traveled is d = \frac{v_0^2}{a_0}

The difference between the train and the car is

x - d = \frac{2v_0^2}{a_0} - \frac{v_0^2}{a_0} = \frac{v_0^2}{a_0}

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