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irina1246 [14]
3 years ago
15

Una piedra es lanzada horizontalmente desde la azote de un edificio de 30 metros de altura, con una velocidad de 12 m/s. Hallar

el tiempo que la piedra cae al piso y a qué distancia de la base del edificio llego la piedra.
Chemistry
1 answer:
drek231 [11]3 years ago
3 0

Answer: La piedra tarda 2.47 segundos en caer al suelo, y cae a 29.64 metros de la base del edificio.

Explanation:

Ok, tenemos 2 sistemas de ecuaciones para este problema.

Primero, el vertical:

La aceleración es la aceleración gravitatoria, entonces:

a = -9.8m/s^2

para la velocidad podemos integrar sobre el tiempo y obtenemos

v = (-9.8m/s^2)*t + v0

donde v0 es la velocidad inicial, pero la velocidad inicial es solo horizontal, entonces v0 = 0.

Para la posición integramos de vuelta:

p = (1/2)*(-9.8m/s^2)*t^2 + p0

donde p0 es la posición inicial, en este caso, 30m

p = (-4.9m/s^2)*t^2 + 30m

la piedra va a llegar al piso cuando la posición vertical sea igual a 0m

p = 0m =  (-4.9m/s^2)*t^2 + 30m

de aca podemos despejar el tiempo que la piedra tarda en llegar al suelo.

t = √(30/4.9) segundos = 2.47 s.

Ahora, las ecuaciones para el movimiento horizontal son:

Aceleración nula, pues no hay ninguna fuerza actuando en la dirección horizontal.

a = 0

para la velocidad, integramos sobre el tiempo:

v = 0*t + v0 = v0

donde v0 es la velocidad inicial, en este caso, es 12m/s

v = 12m/s

para la posición integramos de vuelta:

p = 12m/s*t + p0

en este caso podemos asumir que estamos inicialmente en el punto x = x0, asi que la posición inicial es p0 = x0.

p = 12m/s*t + x0

entonces, si queremos calcular la distancia entre la base del edificio y el punto donde cae la piedra, tenemos que calcular:

D = p(2.47s) - p(0s)

D = 12m/s*2.47s + x0 - (12m/s*0s + x0)

D = 29.64m

La piedra tarda 2.47 segundos en caer al suelo, y cae a 29.64 metros de la base del edificio.

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Conduct metric Titration of H_2(SO_4) and Ba(OH)_2 Write an equation (including states of matter) for the reaction between H_2(S
meriva

Answer:

a) H₂SO₄ + Ba(OH)₂ ⇄ BaSO₄(s) + 2 H₂O(l)

b) H₂SO₄, H⁺, HSO₄⁻, SO₄²⁻. H₂O, H⁺, OH⁻.

c) H⁺, HSO₄⁻, SO₄²⁻

d) As the titration takes place, reaction [1] proceeds to the right. The conductivity of the solution decreases because the amount of H⁺, HSO₄⁻, SO₄²⁻ decreases. The formed solid is barium sulfate BaSO₄. Since BaSO₄ is very insoluble, the main responsible for conductivity are still H⁺, HSO₄⁻ and SO₄²⁻,

e) At the equivalence point equivalent amounts of H₂SO₄ and Ba(OH)₂ react. The conducting species are Ba²⁺, SO₄²⁻, H⁺ and OH⁻.

f) After the equivalence point there is an excess of Ba(OH)₂. The ions Ba²⁺ and OH⁻ are responsible for the increase in the conductivity, being the major conducting species.

Explanation:

a) Write an equation (including states of matter) for the reaction between H₂SO₄ and Ba(OH)₂.

The <em>balanced equation</em> is:

H₂SO₄ + Ba(OH)₂ ⇄ BaSO₄(s) + 2 H₂O(l)   [1]

b) At the very start of the titration, before any titrant has been added to the beaker, what is present in the solution?

In the beginning there is H₂SO₄ and the ions that come from its <em>dissociation reactions</em>: H⁺, HSO₄⁻, SO₄²⁻. There is also H₂O and a very small amount of H⁺ and OH⁻ coming from its <em>ionization</em>.

H₂SO₄(aq) ⇄ H⁺(aq) + HSO₄⁻(aq)

HSO₄⁻(aq) ⇄ H⁺(aq) + SO₄²⁻(aq)

H₂O(l)  ⇄ H⁺(aq) + OH⁻(aq)

c) What is the conducting species in this initial solution?

The main responsible for conductivity are the <em>ions</em> coming from H₂SO₄: H⁺, HSO₄⁻, SO₄²⁻.

d) Describe what happens as titrant is added to the beaker. Why does the conductivity of the solution decrease? What is the identity of the solid formed? What is the conducting species present in the beaker?

As the titration takes place, reaction [1] proceeds to the right. The conductivity of the solution decreases because the amount of H⁺, HSO₄⁻, SO₄²⁻ decreases. The formed solid is barium sulfate BaSO₄. Since BaSO₄ is very insoluble, the main responsible for conductivity are still H⁺, HSO₄⁻ and SO₄²⁻,

e) What happens when the conductivity value reaches its minimum value (which is designated as the equivalence point for this type of titration)? What is the conducting species in the beaker?

At the <em>equivalence point</em> equivalent amounts of H₂SO₄ and Ba(OH)₂ react. Only BaSO₄ and H₂O are present, and since they are <em>weak electrolytes</em>, there is a small amount of ions to conduct electricity. The conducting species are Ba²⁺ and SO₄²⁻ coming from BaSO₄ and H⁺ and OH⁻ coming from H₂O.

f) Describe what happens at additional titrant is added past the equivalence point. Why does the conductivity of the solution increase? What is the conducting species present in the beaker?

After the equivalence point there is an excess of Ba(OH)₂. The ions Ba²⁺ and OH⁻ are responsible for the increase in the conductivity, being the major conducting species.

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A sample that contains only SrCO3 and BaCO3 weighs 0.800 g. When it is dissolved in excess acid, 0.211 g car- bon dioxide is lib
MrRissso [65]

Answer:

53.9%

Explanation:

1 mole of BaCO₃ yields  1 mole of CO₂,

1 mole of SrCO₃ yields    1 mole of CO₂

m₁ = mass of BaCO₃

m₂ =  mass SrCO₃

molar mass of SrCO₃  = 147.63 g/mol

molar mass of  BaCO₃ = 197.34 g/mol

molar mass of CO₂ = 44.01 g/mol

mole of CO₂ in 0.211 g = 0.211 g / 44.01 = 0.00479

mole of BaCO₃ = m₁ / 197.34

mole of SrCO₃  = m₂ / 147.63

mole of BaCO₃ + mole of SrCO₃  = 0.00479

(m₁ / 197.34) + (m₂ / 147.63) = 0.00479

147.63 m₁ + 197.34 m₂ = 139.55

m₁ + m₂ = 0.8

m₁ = 0.8 - m₂

147.63 (0.8 - m₂) + 197.34 m₂ = 139.55

118.104 - 147.63 m₂ + 197.34 m₂ = 139.55

49.71 m₂ = 139.55 - 118.104 = 21.446

m₂ = 21.446 / 49.71 = 0.431

the percentage of m₂ ( SrCO₃ ) in the sample = 0.431 / 0.8 = 0.539 × 100 = 53.9%

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Answer:

No

Explanation:

The conclusions from Thomoson's claims would be invalid if his experiment could not be replicated.

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Hence, <u>Thomoson's conclusion would be invalid if his experiment could not be replicated. </u>

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