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3 years ago

6

How does a balanced chemical equation demonstrate the Law of Conservation of Mass?

1 answer:

lbvjy [14]3 years ago

6 0

Answer:

It shows the same number of atoms of each element on both sides of the equation

A balanced equation demonstrates the conservation of mass by having the same number of each type of atom on both sides of the arrow.

Explanation:

Matter cannot be created or destroyed in chemical reactions. This is the law of conservation of mass. In every chemical reaction, the same mass of matter must end up in the products as started in the reactants. Balanced chemical equations show that mass is conserved in chemical reactions.

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1)Pitch is measure of the ____________ of sound.

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Answer:

1) wave length

2) parallel reflection

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A sealed test tube traps 25.0 cm3 of air at a pressure of 1.00 atm and temperature of 18°C. The test tube’s stopper has a diamet

puteri [66]

Answer:

180° C

Explanation:

First we start by finding the area of the stopper.

A = πd²/4, where d = 1.5 cm = 0.015 m

A = 3.142 * 0.015² * ¼

A = 1.767*10^-4 m²

Next we find the force on the stopper

F = (P - P•)A, where

F = 10 N

P = pressure inside the tube,

P• = 1 atm

10 = (P - 101325) * 1.767*10^-4

P - 101325 = 10/1.767*10^-4

P - 101325 = 56593

P = 56593 + 101325

P = 157918 Pascal

Now, remember, in an ideal gas,

P1V1/T1 = P2V2/T2, where V is constant, then we have

P1/T1 = P2/T2, and when we substitute the values, we have

101325/(273 + 18) = 157918/ T2

101325/291 = 157918/ T2

T2 = (157918 * 291)/101325

T2 = 453 K

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4 years ago

(a) The Sun orbits the Milky Way galaxy once each 2.60 x 108y , with a roughly circular orbit averaging 3.00 x 104 light years i

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Answer:

Part a)

$a_c = 1.67 \times 10^{-10} m/s^2$

Part b)

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Explanation:

Time period of sun is given as

$T = 2.60 \times 10^8 years$

$T = 2.60 \times 10^8 (365 \times 24 \times 3600) s$

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Now the radius of the orbit of sun is given as

$R = 3.00 \times 10^4 Ly$

$R = 3.00 \times 10^4 (3\times 10^8)(365 \times 24 \times 3600)m$

$R = 2.84 \times 10^20 m$

Part a)

centripetal acceleration is given as

$a_c = \omega^2 R$

$a_c = \frac{4\pi^2}{T^2} R$

$a_c = \frac{4\pi^2}{(8.2\times 10^{15})^2}(2.84 \times 10^{20})$

$a_c = 1.67 \times 10^{-10} m/s^2$

Part b)

orbital speed is given as

$v = \frac{2\pi R}{T}$

$v = \frac{2\pi (2.84 \times 10^{20})}{8.2 \times 10^{15}}$

$v = 2.18 \times 10^5 m/s$

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