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ahrayia [7]
4 years ago
9

You have a chemical sealed in a glass container filled with air. the glass container has a mass of 200.0 grams and the chemical

has a mass of 750 grams. the total mass of the setup is 950 grams. the chemical is ignited by means of a magnifying glass focusing sunlight onto the reactant. after the chemical has completely burned, what is the total mass of the setup?
Chemistry
1 answer:
valentinak56 [21]4 years ago
8 0
After the chemical has completely burned, the total mass of the set up remains being the same than before: 950g.

The mass is conserved during chemical reactions; this is, the mass of the reactants is equal to the mass of the products, so there is not change in mass, and you can conclude quickly that the mass of the set up at the end is 950, the same of starting.
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3 years ago
Q = smT
Effectus [21]

Answer:

The heat required is 1262.91 joules or 1.26 kilojoules.

Explanation:

The quantity of heat (Q) required to raise the temperature of a substance depends on its Mass (M), specific heat capacity (C) and change in temperature (Φ)

Thus, Q = MCΦ

Since,

Q = ?

Mass of aluminum sample = 33.0g

C = 0.89 J/g℃

Φ = (Final temperature - Initial temperature)

= 68°C - 25°C = 43°C

Then, Q = MCΦ

Q = 33.0g x 0.89 J/g℃ x 43°C

Q = 1262.91 J

Since the heat in joules is 1262.91, obtain heat in kilojoules.

If 1000 joules = 1 kilojoules

1262.91 joules = Z

To get Z, cross multiply

Z x 1000 = 1262.91 x 1

1000Z = 1262.91

Z = 1262.91 / 1000

Z = 1.26291 kilojoules (when placed as 3 significant figures, Z is 1.26 kilojoules)

Thus, the heat required is 1262.91 joules or 1.26 kilojoules.

4 0
3 years ago
A balloon contains 2.0 L of air at 101.5 kPa . You squeeze the balloon to a volume of 0.25 L.
8_murik_8 [283]
<h3>Answer:</h3>

812 kPa

<h3>Explanation:</h3>
  • According to Boyle's law pressure and volume of a fixed mass are inversely proportional at constant absolute temperature.
  • Mathematically, P\alpha \frac{1}{V}

At varying pressure and volume;

P1V1=P2V2

In this case;

Initial volume, V1 = 2.0 L

Initial pressure, P1 = 101.5 kPa

Final volume, V1 = 0.25 L

We are required to determine the new pressure;

P2=\frac{P1V1}{V2}

Replacing the known variables with the values;

P2=\frac{(101.5)(2.0L)}{0.25L}

           = 812 kPa

Thus, the pressure of air inside the balloon after squeezing is 812 kPa

8 0
3 years ago
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