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arsen [322]
3 years ago
6

A representation of a chemical reaction that uses symbols to show the relationship between the reactants and the products is cal

led a __________.
a.
symbolic statement
b.
chemical equation
c.
mole ratio
d.
chemical proportion


Please select the best answer from the choices provided
Physics
2 answers:
VARVARA [1.3K]3 years ago
8 0

Answer:

(B) chemical equation

Explanation:

Chemical equation is a representation of how reactant form products using chemical symbols.

The reactant are usually in the left hand side of the equation, while the products are kept at the right hand side of the equation. A long arrow is placed in-between them to indicate the direction of the reaction. Some chemical reactions requires number to balance the amount of reactant and products in the equation since law of conservation of matter must be strictly adhered to. Eg

NaOH + HCl ———》 NaCl + H2O

mezya [45]3 years ago
5 0

Answer:

a.

it's a symbolic statement. as it shows only sybols

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, b)    f = 0.851 Hz

, c)  v = 1,069 m / s

, d)  x = 0

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         k = F / x

          k = 24.0 / 0.200

          k = 120 N / m

b) the angular velocity of the simple harmonic movement is

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        w = 5,345 rad / s

Angular velocity and frequency are related.

       w = 2π f

        f = w / 2π

        f = 5.345 / 2π

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        x = A cos (wt + Ф)

As the body is released without initial velocity, Ф = 0

        x = 0.2 cos wt

Speed ​​is

       v = dx / dt

       v = -A w sin wt

The speed is maximum for sin wt = ±1

       v = A w

       v = 0.200 5.345

       v = 1,069 m / s

d) when the function sin wt = -1 the function cos wt = 0, whereby the position for maximum speed is

       x = A cos wt = 0

       x = 0

e) the acceleration is

       a = d²x / dt² = dv / dt

       a = - Aw² cos wt

The acceleration is maximum when cos wt = ± 1

       a = A w²

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        a = 5.71 m / s²

f) the position for this acceleration is

       x = A cos wt

       x = A

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h) the position is

         x = 1/3 A

Let's calculate the time to reach this point

         x = A cos wt

        1/3 A = A cos 5.345t

         t = 1 / w cos⁻¹(1/3)

The angles are in radians

t = 1.23 / 5,345

t = 0.2301 s

Speed ​​is

v = -A w sin wt

v = -0.2 5.345 sin (5.345 0.2301)

v = -1.01 m / s

i) acceleration

a = -A w² sin wt

a = - 0.2 5.345² cos (5.345 0.2301)

      a = -1.91 m / s²

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