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lukranit [14]
2 years ago
13

Which statement best describes why specific heat capacity is often more useful than heat capacity for scientists when comparing

two materials?
Specific heat capacity is an intensive property and does not depend on sample size.
Specific heat capacity is an extensive property and does not depend on sample size.
Specific heat capacity is an intensive property and depends on the sample size.
Specific heat capacity is an extensive property and depends on the sample size.
Chemistry
2 answers:
zhuklara [117]2 years ago
7 0
<span>Specific heat capacity is an intensive property and does not depend on sample size.</span>
Misha Larkins [42]2 years ago
6 0

The correct option is this: SPECIFIC HEAT CAPACITY IS AN INTENSIVE PROPERTY AND DOES NOT DEPEND ON SAMPLE SIZE.

Generally, all the properties of matters can be divided into two classes, these are intensive and extensive properties. Intensive properties are those properties that are not determined by the quantity of the material that is present or available. Examples of intensive properties are colour, density and specific heat capacity. For instance, whether you have a bucket of water or a cup of water, the quantity does not matter, the colour of water will always remain the same. Extensive properties in contrast, are those properties that depend on the quantity of material that is available. Examples are mass, heat capacity and volume.

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Carbon dioxide readily absorbs radiation with an energy of 4.67 x 10-20 J. What is the wavelength and frequency of this radiatio
Tanya [424]

Answer:

ν = 7.04 × 10¹³ s⁻¹

λ = 426 nm

It falls in the visible range

Explanation:

The relation between the energy of the radiation and its frequency is given by Planck-Einstein equation:

E = h × ν

where,

E is the energy

h is the Planck constant (6.63 × 10⁻³⁴ J.s)

ν is the frequency

Then, we can find frequency,

\nu = \frac{E}{h}=  \frac{4.67 \times 10^{-20}J  }{6.63 \times 10^{-34}J.s} = 7.04 \times 10^{13} s^{-1}

Frequency and wavelength are related through the following equation:

c = λ × ν

where,

c is the speed of light (3.00 × 10⁸ m/s)

λ is the wavelength

\lambda = \frac{c}{\nu } =\frac{3.00 \times 10^{8} m/s }{7.04 \times 10^{13} s^{-1} } =4.26 \times 10^{-6}m.\frac{10^{9}nm }{1m} = 426 nm

A 426 nm wavelength falls in the visible range (≈380-740 nm)

7 0
2 years ago
If it takes three "breaths" to blow up a balloon to 1.2 l, and each breath supplies the balloon with 0.060 moles of exhaled air,
vesna_86 [32]
<span>It takes 3 breaths to get to 1.2 l. One breath is then (1.2 l) / 3 breaths = .4l/breath. To get to 3.0 l we need the difference from 1.2 l. 3.0-1.2 = 1.8 l. Divide the difference by liters/breath (.4) to get how many needed breaths. (1.8 l)/(.4 l/breath) = 4.5 breaths to get the balloon to 3.0 l. In total there were 3 breaths+ 4.5 breaths = 7.5breaths to get to 3.0 l. To find the total moles multiply 7.5breaths by .060 moles/breath 7.5 breaths*.060moles/breath = .45moles</span>
6 0
2 years ago
Read 2 more answers
The molecular mass of air, at standard pressure and temperature, is approximately 28.97 g/mol. Calculate the mass of 3.33 moles
poizon [28]
<h3>Answer:</h3>

               Mass = 96.47 g

<h3>Solution:</h3>

Data Given:

                  M.Mass  =  28.97 g.mol⁻¹

                  Moles  =  3.33 mol

                  Mass  =  ??

Formula Used:

                  Moles  =  Mass ÷ M.Mass

Solving for Mass,

                  Mass  =  Moles × M.Mass

Putting values,

                  Mass  =  3.33 mol × 28.97 g.mol⁻¹

                  Mass  =  96.4701 g

Rounding to four significant numbers,

                 Mass = 96.47 g


6 0
2 years ago
Read 2 more answers
Describe the changes in properties (from metals to nonmetals or from nonmetals to metals) as we move (a) down a periodic group a
defon

Answer:

Explanation:

The changes in properties from metals to non-metals on a periodic table can be measured and determined by the metallicity or electropositivity of elements.

Metallicity is a measure of the tendency of atoms of an element to lose electrons.

a.

Down a periodic group, metallicity increases.

b.

Across a period from left to right electropositivity or metallicity decreases.

Metals are found in the left part of periodic table and the most reactive metal sits in the lower left corner. Non-metals are towards the right side of the table.

7 0
2 years ago
Lars observes a substance to be a solid and to float in water at room temperature (23°C). Based on the given properties, which s
Triss [41]

Answer:

D. Sulfur Hexafluride

Explanation:

  • above it says the substance floats above water at room temperature and lists some substances and their density at room temp!
  • we know that the density of water is 1.0 so the substance in order for it to float has to be less than 1.0 and the densities for Sulfer Hexa, are all less than 1!!

I hope this helped !!

4 0
2 years ago
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