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4 years ago

Will an empty balloon have precisely the same apparent weight on a scale as a balloon filled with air? Explain.

2 answers:

4 0

The notion an empty balloon have precisely the same apparent weight on a scale as a balloon filled with air depends on the diameter of the balloon. The weight of the balloon filled with air is equal to the mass of the balloon and the mass of the air inside. The mass of air inside is equal to the density of air multiplied by the volume of the balloon. If the balloon is large, then the two masses are equal whereas if not, the mass of air inside the inflation is neglible

pentagon [3]4 years ago

3 0

They would not, but only for this reason: In the the full balloon, slightly more molecules of air are pressing downwards on the same position. The difference lies in the weight of the volume of the inflated balloon that does not lie in space above the deflated balloon.
An air filled balloon is heavier than the empty one.

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A chunk of tin weighing 18.5 grams and originally at 97.38 °C is dropped into an insulated cup containing 75.7 grams of water at

balandron [24]

Answer:

The final temperature of the water is 22.44°C.

Explanation:

Heat lost by tin will be equal to heat gained by the water

$-q_1=q_2$

Mass of tin = $m_1=18.5 g$

Specific heat capacity of tin = $c_1=0.21 J/g^oC$

Initial temperature of the tin = $T_1=97.38^oC$

Final temperature = $T_2$ =T

$q_1=m_1c_1\times (T-T_1)$

Mass of water= $m_2=75.7 g$

Specific heat capacity of water= $c_2=4.184 J/g^oC$

Initial temperature of the water = $T_3=21.52^oC$

Final temperature of water = $T_2$ =T

$q_2=m_2c_2\times (T-T_3)$

$-q_1=q_2$ (Law of Conservation of energy)

$-(m_1\times c_1\times (T-T_1))=m_2\times c_2\times (T-T_3)$

On substituting all values:

$-(18.5 g\times 0.21 J/g^oC\times (T-97.38^oC))=75.7 g\times 4.184 J/g^oC\times (T-21.52 ^oC)$

we get, T = 22.44°C

The final temperature of the water is 22.44°C.

3 0

3 years ago

What is the percent composition of carbon in heptane, C7H16?

JulsSmile [24]

Answer:

Option d, 84%

Explanation:

1) Determine the molar mass of the heptane

element # of atoms atomic mass total mass

C 7 12.011 g/mol 7×12.011 g/mol = 84.077 g/mol

H 16 1.008 g/mol 16×1.008 g/mol = 16.128 g/mol

-------------------

Molar mass = 100.205 g/mol

2) Percent of carbon, %

% = (mass of element / molar mass )×100

= (84.077 g/mol / 100.205 g/mol )×100 = 83.83 %

Round to two significant figures: 84% ← answer

4 0

3 years ago

If you mix equal concentrations of reactants and products, will the reaction proceed to the right or the left? HF(aq)+NO3−(aq)⇌H

kaheart [24]

Answer:

hypompast

Explanation:

4 0

3 years ago

Calculate the number of milliliters of 0.753 M KOH required to precipitate all of the Pb2+ ions in 135 mL of 0.775 M Pb(NO3)2 so

MrRa [10]

Answer:

278 ml of KOH.

Explanation:

The equation for the reaction is: Pb(NO3)2(aq) + 2KOH(aq) --> Pb(OH)2(s) + 2KNO3(aq)

Pb(NO3)2:

Half ionic equation:

Pb(NO3)2(aq) --> Pb2+ + 2NO3^-

Pb(OH)2 --> Pb2+ + 2OH-

Volume = 135 mL

Molar concentration = 0.775 M

Number of moles = molar concentration * volume

= 0.775 * 0.135

= 0.105 mol of Pb(NO3)2

Since 1 mole of Pb(NO3)2 reacted with 2 moles of KOH to give 1 mole of Pb2+ (Pb(OH)2).

By stoichiometry,

Number of moles of KOH = 0.105 * 2

= 0.21 mol

Molar concentration = number of moles/volume

Volume = 0.21/0.753

= 0.278 l

To ml, 0.278 l * 1000 ml/1l

= 278 ml of KOH.

3 0

3 years ago

A chemistry student needs of glycerol for an experiment. By consulting the CRC Handbook of Chemistry and Physics, the student di

Pachacha [2.7K]

Answer:

The mass of glycerol that the student should weigh out depends on the volume that is needed. For example, the density of glycerol is 1.26 g/mL. Then, if 500 mL of glycerol is needed, 630 g should be weighed.

Explanation:

Density is the amount of mass per unit of volume. Density relates how many grams 1 milliliter of a substance weighs. Thus, if the density of glycerol is 1.26 g/mL means that 1.26 grams of glycerol occupy 1 mL of volume, or, in other words, 1 mL of glycerol weighs 1.26 grams.

Therefore, if 500 mL of glycerol is required to use for an experiment, 630 grams need to be weighed:

1 mL glycerol________ 1.26 g

500 mL glycerol______ x= 500mL * 1.26 g / 1 mL = 630 g

3 0

3 years ago