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babunello [35]
2 years ago
8

A 1.0 kg ball is thrown into the air and initial velocity of 30 mi./s how high into the air did the ball travel

Chemistry
1 answer:
laila [671]2 years ago
6 0

Answer:

45.9m

Explanation:

Given parameters:

Mass of the ball  = 1kg

Initial velocity  = 30m/s

Unknown:

Height the ball will travel  = ?

Solution:

At the maximum height, the final velocity of the ball  will be 0;

  Using ;

   V²   = U²   - 2gh

V is the final velocity

U is the initial velocity

g is the acceleration due to gravity

h is the height

     0² = 30²   - (2 x 9.8 x h)

    0 = 900 - 19.6h

      h  = 45.9m

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[CO] = 1 mol / 2L = 0.5 M

[
According to the equation:

and by using the ICE table:

             CO(g) + H2O(g) ↔   CO2(g) + H2(g)

initial     0.5            0.5                    0          0

change  -X              -X                   +X         +X
     
Equ       (0.5-X)       (0.5-X)                     X            X

when Kc = X^2 * (0.5-X)^2

by substitution:

1.845 = X^2 * (0.5-X)^2  by solving for X 

∴X = 0.26

∴ [CO2] = X = 0.26
4 0
3 years ago
When 32 grams of aluminum react, the actual yield is 105.5 grams, what is the percent yield?
user100 [1]

Answer:

329.7%

Explanation:

Percent Yield = Actual Yield/ Theoretical Yield x 100%

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5 0
3 years ago
Mass has to do with how many particles
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The law of definite proportions agrees with Dalton atomic theory.

What is Dalton atomic theory?

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different elements have different masses and chemical properties.

The law of definite proportions also known as proust's law ,state that a chemical compound contain the same proportion of elements by mass.this law is one of the stoichiometry .

Thus ,

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A gamma ray primarily consists of pure energy and no mass. True False
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3 years ago
If 0.50 mol/L of butane is added to the original equilibrium mixture and the system shifts to a new equilibrium position, what i
Ostrovityanka [42]

Consider the isomerization of butane with equilibrium constant is 2.5 .The system is originally at equilibrium with :

[butane]=1.0 M , [isobutane]=2.5 M

If 0.50 mol/L of butane is added to the original equilibrium mixture and the system shifts to a new equilibrium position, what is the equilibrium concentration of each gas?

Answer:

The equilibrium concentration of each gas:

[Butane] = 1.14 M

[isobutane] = 2.86 M

Explanation:

Butane  ⇄  Isobutane

At equilibrium

1.0 M               2.5 M

After addition of 0.50 M of butane:

(1.0 + 0.50) M               -

After equilibrium reestablishes:

(1.50-x)M            (2.5+x)

The equilibrium expression will wriiten as:

K_c=\frac{[Isobutane]}{[Butane]}

2.5=\frac{2.5+x}{(1.50-x)}

x = 0.36 M

The equilibrium concentration of each gas:

[Butane]= (1.50-x) = 1.50 M - 0.36M = 1.14 M

[isobutane]= (2.5+x) = 2.50 M + 0.36 M = 2.86 M

3 0
3 years ago
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