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Ann [662]
3 years ago
9

According to the law of conservation of energy, the energy released by the system must be transferred to and absorbed by the sur

roundings
Chemistry
1 answer:
Andrew [12]3 years ago
5 0
True! 
 Energy released by system is absorbed by surroundings.
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At room temperature (20°C} and pressure, the density of air is 1.189 g/L. An object will float in air if its density is less tha
Alekssandra [29.7K]

Explanation:

Density =\frac{Mass}[Volume}

Density of the air ,d= 1.189 g/L

(a) Density of the evacuated ball

Mass of the ball ,m = 0.12 g

Volume of the ball =V=560 cm^3=560 ml=0.560 L

D =\frac{0.12 g}{0.560 L}=0.214 g/L

D<d, teh evacuated ball will flaot in air.

(b) Density of the evacuated ball D = 0.214 g/L

Density of carbon dioxide gas = d_1=1.830 g/L

Mass of the carbon dioxide gas :

1.830 g/L\times 0.560 L=1.0248 g

Total density of filled ball with carbon dioxide gas:

\frac{0.12 g+1.0248 g}{0.560 L}==2.044 g/L

The ball filled with carbon dioxide will not float in the air because total density of filled ball is greater than the density of an air.

(c) Density of the evacuated ball D = 0.214 g/L

Density of hydrogen gas = d_2=0.0899 g/L

Mass of the hydrogen gas :

1.830 g/L\times 0.560 L=0.050344 g

Total density of filled ball with hydrogen gas:

\frac{0.12 g+0.050344 g}{0.560 L}==0.3041 g/L

The ball filled with hydrogen will float in the air because total density of filled ball is lessor than the density of an air.

(d) Density of the evacuated ball D = 0.214 g/L

Density of oxygen gas = d_3=1.330 g/L

Mass of the oxygen gas :

1.330 g/L\times 0.560 L=1.7448 g

Total density of filled ball with oxygen gas:

\frac{0.12 g+1.7448 g}{0.560 L}=1.5442 g/L

The ball filled with oxygen will not float in the air because total density of filled ball is greater than the density of an air.

(e) Density of the evacuated ball D = 0.214 g/L

Density of nitrogen gas = d_4=1.165 g/L

Mass of the nitrogen gas :

1.165 g/L\times 0.560 L=0.6524 g

Total density of filled ball with nitrogen gas:

\frac{0.12 g+0.6524 g}{0.560 L}==1.3792 g/L

The ball filled with nitrogen will not float in the air because total density of filled ball is greater than the density of an air.

f) Mass must be added to sink the ball = m

Density of ball > Density of the air ; to sink the ball.

\frac{0.12g +m}{0.560L}>1.189 g/L

m > 0.54584 g

For any case weight added to ball to make it sink in an air should be grater than the value of 0.54584 grams.

5 0
3 years ago
The combustion of propane may be described by the chemical equation C 3 H 8 ( g ) + 5 O 2 ( g ) ⟶ 3 CO 2 ( g ) + 4 H 2 O ( g ) C
Kipish [7]

Answer: 72 grams of O_2(g) are needed to completely burn 19.7 g C_3H_8(g)

Explanation:

According to avogadro's law, 1 mole of every substance weighs equal to molecular mass and contains avogadro's number 6.023\times 10^{23} of particles.

To calculate the number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}

Putting in the values we get:

\text{Number of moles}=\frac{19.7g}{44g/mol}=0.45moles

C_3H_8(g)+5O_2(g)\rightarrow 3CO_2(g)+4H_2O(g)

According to stoichiometry:

1 mole of C_3H_8 requires 5 moles of oxygen

0.45 moles of C_3H_8 require= \frac{5}{1}\times 0.45=2.25 moles of oxygen

Mass of O_2=moles\times {\text {Molar mass}}=2.25\times 32=72g

72 grams of O_2(g) are needed to completely burn 19.7 g C_3H_8(g)

7 0
3 years ago
What is the solute and solvent in roundup
olga nikolaevna [1]

Answer:

solute is that we disolve in solvent

solvent is in which we dissolve solute

8 0
3 years ago
Difference between sodium ion and sodium atom.<br>(give in points not in paragraph)​
Natali5045456 [20]

Answer:

Explanation:

SODIUM ATOM;

SODIUM ATOM IS NEUTRAL

SODIUM ION;

IT IS A CHARGED SPECIE WITH A CHARGE OF +1

SODIUM ATOM:

THE NUMBER OF PROTONS AND ELECTRONS ARE SAME ie:11

SODIUM ION:

NUMBER OF PROTONS AND ELECTRONS ARE NOT SAME ie. ELETRON: 10, PROTONS:11

HOPE IT WILL HELP:)

3 0
3 years ago
Name a physical property of gold (Au)
Lana71 [14]
Gold is a soft metal. Both malleable and ductile
7 0
3 years ago
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