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lara31 [8.8K]
3 years ago
15

How does the ability to balance a chemical equation relate to that reaction obeying the Law of Conservation of Matter?

Physics
1 answer:
Aliun [14]3 years ago
8 0
Example: 
<span>CH4 + 2 O2 → CO2 + 2 H2O </span>

<span>In the balanced equation above, 1 mole of CH4 will react with 2 moles of O2 to produce 1 mole of CO2 and 1 mole of H2O. </span>
<span>The mass of 1 mole of CH4 = 12 + (4 * 1) = 16 grams </span>
<span>The mass of 2 moles of O2 = 2 * 32 = 64 grams </span>
<span>Total mass of reactants = 16 + 64 = 80 grams </span>

<span>The mass of 1 mole of CO2 = 12 + (2 * 16) = 44 grams </span>
<span>The mass of 2 moles of H2O = 2 * 18 = 36 grams </span>
<span>Total mass of products = 44 + 36 = 80 grams </span>

<span>Before the reaction, there was 80 grams of matter, and after the reaction there is 80 grams of matter! </span>
<span>Matter is conserved!

</span>
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Answer:

Option A

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The speed of a mechanical wave is fastest in the solid medium and the slowest in the gaseous medium. Hence, as the wave traverses from gaseous medium to the solid medium, its speed increases.

Thus, option A is correct

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3 years ago
Engineers at the Space Centre must determine the net force needed for a rockets engine to achieve an acceleration of 70 m/s2. As
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Answer:

3,150,000N

Explanation:

According to Newton's second law;

F = mass * acceleration

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acceleration = 70m/s^2

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An extended period when temperatures are well below average is known as a/an
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Consider a space shuttle which has a mass of about 1.0 x 105 kg and circles the Earth at an altitude of about 200.0 km. Calculat
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Answer:

1.6675×10^-16N

Explanation:

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g = GM/r²

G is the gravitational constant

M is the mass = 1.0 x 10^5 kg

r is the altitude = 200km = 200,000m

Substitute into the formula

g = 6.67×10^-11 × 1.0×10^5/(2×10^5)²

g = 6.67×10^-6/4×10^10

g = 1.6675×10^{-6-10}

g = 1.6675×10^-16N

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2 years ago
A scientist wants to model an internal organ with connective tissue as a mass on a spring. The mass of the organ is 2.0 kg, and
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Answer:

Spring constant, k = 5483.11 N/m

Explanation:

It is given that,

Mass of the organ, m = 2 kg

The natural period of oscillation is, T = 0.12 s

Let k is the spring constant for the spring in the scientist's model. The period of oscillation is given by :

T=2\pi\sqrt{\dfrac{m}{k}}

k=\dfrac{4\pi^2 m}{T^2}

k=\dfrac{4\pi^2 \times 2\ kg}{(0.12\ s)^2}

k = 5483.11 N/m

So, the  spring constant for the spring in the scientist's model is 5483.11 N/m.

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3 years ago
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