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san4es73 [151]
3 years ago
11

Write a balanced equation for the decomposition of ammonium nitrate to form molecular nitrogen, molecular oxygen, and water. Exp

lain your answer.
Physics
1 answer:
VARVARA [1.3K]3 years ago
8 0

The unbalanced reaction is

<em>a</em> NH₄NO₃   ⇒   <em>b</em> N₂ + <em>c</em> O₂ + <em>d</em> H₂O

where <em>a</em>, <em>b</em>, <em>c</em>, and <em>d</em> are unknown constants.

Count how many times each element appears on either side of the reaction.

• reactants:

N = 2<em>a</em>, H = 4<em>a</em>, O = 3<em>a</em>

• products

N = 2<em>b</em>, O = 2<em>c</em> + <em>d</em>, H = 2<em>d</em>

<em />

Now we solve the system of equations,

2<em>a</em> = 2<em>b</em> … … … [1]

4<em>a</em> = 2<em>d</em> … … … [2]

3<em>a</em> = 2<em>c</em> + <em>d</em> … … … [3]

<em />

From [1] we immediately have

<em>a</em> = <em>b</em>

In [2], we get

2<em>a</em> = <em>d</em>

and substituting for <em>d</em> in [3] gives

3<em>a</em> = 2<em>c</em> + 2<em>a</em>

<em>a</em> = 2<em>c</em>

<em />

Let <em>c</em> = 1; then

• <em>a</em> = 2×1 = 2

• <em>b</em> = 2

• <em>d</em> = 2×2 = 4

So, the balanced reaction is

2 NH₄NO₃   ⇒   2 N₂ + O₂ + 4 H₂O

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Explanation:

The critical velocity is that velocity of liquid flow, up to which its flow is streamlined (laminar)& above which its flow becomes turbulent. It's denoted by Vc & it depends upon: Coefficient of viscosity of liquid (η) Density of liquid. Radius of the tube.

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You would want it to be greater than D. friction force

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true or false? short-wave heat radiation is given off by the earth when the earth absorbs long-wave radiation.
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Answer:

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Explanation:

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A block of mass m = 2.0 kg lies on a rough ramp that is inclined at an angle θ = 20oto the horizontal. A force F of magnitude 5.
Marina86 [1]

Answer:

a) 0.64 b) 2.17m/s^2 c) 8.668joules

Explanation:

The block was on the ramp, the ramp was inclined at 20degree. A force of 5N was acting horizontal to the but not parallel to the ramp,

Frictional force = horizontal component of the weight of the block along the ramp + the applied force since the block was just about move

Frictional force = mgsin20o + 5N = 6.71+5N = 11.71

The force of normal = the vertical component of the weight of the block =mgcos20o = 18.44

Coefficient of static friction = 11.71/18.44= 0.64

Remember that g = acceleration due to gravity (9.81m/s^2) and m = mass (2kg)

b) coefficient of kinetic friction = frictional force/ normal force

Fr = 0.4* mgcos 20o = 7.375N

F due to motion = ma = total force - frictional force

Ma = 11.71 - 7.375 = 4.335

a= 4.335/2(mass of the block) = 2.17m/s^2

C) work done = net force *distance = 4.335*2= 8.67Joules

8 0
3 years ago
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