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dolphi86 [110]
3 years ago
5

If a reaction starts with 30 grams how many should it end with ?

Physics
1 answer:
zysi [14]3 years ago
4 0

30grams

Explanation:

If a reaction starts with 30grams then the reaction should end with 30grams.

This in conformity with the law of conservation of mass.

  • The law states that "in an isolated system, mass is neither created nor destroyed during chemical transformation".
  • Mass is the quantity of matter contained in a substance.
  • In chemical reactions, the mass of reactants must always be the same with the mass of the product baring any loss.
  • In an isolated system, there is no exchange of energy and mass.
  • Chemical systems are usually treated as isolated systems in which mass is conserved.

Learn more:

Chemical laws brainly.com/question/5896850

#learnwithBrainly

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Ludmilka [50]

Answer:

121.0 W

Explanation:

We use the equation for rate of heat transfer during radiation.

Q/t = σεA(T₂⁴ - T₁⁴)

Since temperature of surroundings = T₁ = -20.0°C = 273 +(-20) = 253 K, and temperature of skier's clothes = T₂ = 5.50°C = 273 + 5.50 = 278.5 K.

Surface area of skier , A = 1.60 m², emissivity of skier's clothes,  ε = 0.70 and σ =  5.67 × 10⁻⁸ W/m²K⁴ .

Therefore, the rate of heat transfer by radiation Q/t is

Q/t = σεA(T₂⁴ - T₁⁴) = (5.67 × 10⁻⁸ W/m²K⁴ ) × 0.70 × 1.60 m² × (278.5⁴ - 253⁴) = 6.3054 × (1918750544.0625) × 10⁻⁸ W = 1.2098 × 10² W = 120.98 W ≅ 121.0 W

7 0
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If element X has 17 protons, how many electrons does it have?
OleMash [197]
18 electrons because protons minus the atomic mass so it would be 35-17 which gives you 18.
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An object travels with velocity v = 4.0 meters/second and it makes an angle of 60.0° with the positive direction of the y-axis.
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Saturn moves in an orbit around the Sun with radius 10 AU. How many degrees does it move on the Celestial in one year? (Hint: Ca
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Answer:

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Explanation:

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Where r is the distance of the planet to the sun, G is the gravitational constant and m is the mass of the sun.

Now, we don't actually need to solve the values of the constants, since we now that the distance from the sun to Saturn is 10 times the distance from the sun to the earth. We now this because 1 AU is the distance from the earth to the sun.

Now, we divide the expression used to calculate the orbital period of Saturn by the expression used to calculate the orbital period of the earth. Notice that the constants will cancel and we will get the rate of orbital periods in terms of the distances to the sun:

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Knowing that the orbital period of the earth is 1 year, the orbital period of Saturn will be \sqrt{10^3}} years, or 31.62 years.

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