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maksim [4K]
3 years ago
14

Zinc has a specific heat capacity of 0.390 J/goC. What is its molar heat capacity? Enter your answer numerically to three signif

icant figures in units of J/moloC.
Chemistry
1 answer:
ch4aika [34]3 years ago
3 0

Answer:

The answer to your questions is  Cm = 25.5 J/mol°C  

Explanation:

Data

Heat capacity = 0.390 J/g°C

Molar heat capacity = ?

Process

1.- Look for the atomic number of Zinc

     Z = 65.4 g/mol

2.- Convert heat capacity to molar heat capacity

       (0.390 J/g°C)(65.4 g/mol)

- Simplify and result

   Cm = 25.5 J/mol°C  

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Answer:

a

Explanation:

5 0
3 years ago
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How many atoms are in 2.70 moles of iron (Fe) atoms?
vredina [299]
The answer is <span>1.63 × 1024 atoms Fe.
</span>

Avogadro's number is the number of units (atoms, molecules) in 1 mole of substance:

<span>6.023 × 10²³ atoms per 1 mole
</span>So, how many atoms are per 2.70 moles:

6.023 × 10²³ atoms : 1 mole = x : 2.70 moles
x = 6.023 × 10²³ atoms * 2.70 moles : 1 mole
x = 16.3 × 10²³ = 1.63 × 10 × 10²³ = 1.63 × 10²⁴ atoms


3 0
3 years ago
What is the wavelength range,in nanometer,for infrared light? In what portion of this range does the earth receive IR from the s
andrezito [222]
  • The wavelength range of Infrared radiation is 700 nanometers to 1 millimeter.
  • The sun emits mainly near-infrared which is mainly composed of wavelength below 4 micrometers.
  • The thermal range of infrared ranges between wavelengths 3.5 and 2.0 micrometers

Explanation:

The wavelength range of Infrared radiation is 700 nanometers to 1 millimeter. This also translates to a frequency range of 430 TeraHertz  to 300 Giga Hertz.

Because the sun is a star and is hot in comparison to earth and other planetary bodies, the bigger  range of infrared radiation it emits is in the near-infrared which is mainly composed of wavelength below 4 micrometers.

The earth's surface produces infrared radiation of the mid-infrared range while cooler substances will produce far-infrared range

The thermal range of infrared ranges between wavelengths 3.5 and 2.0 micrometers and is produced by black bodies.

Learn More:

For more on infrared radiation check out;

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7 0
3 years ago
The radius of a single atom of a generic element X is 169 picometers (pm) and a crystal of X has a unit cell that is body-center
Virty [35]

The volume of the unit cell is 2.67 x 10⁻²⁸ m³.

<h3>What is the volume of a unit cell of a body-centered cubic crystal?</h3>

In a body-centered cubic unit cell, the volume occupied by the particles of the substance is about 68% of the total unit cell.

Assuming that a single atomic a sphere, the volume is:

Volume(atom) = 4/3 x π x r³

Volume(atom) = 4/3 x π x (169 x 10⁻¹²)³

Volume(atom) = 2.02 x 10⁻²⁹ m³

There are a total of 9 atoms in a body-centered unit cell, so the total volume occupied by atoms is:

2.02 x 10⁻²⁹ x 9

= 1.82 x 10⁻²⁸ m³

Volume of cell = (1.15 x 10⁻²⁸ ) / 0.68

Volume of cell = 2.67 x 10⁻²⁸ m³

Therefore, the volume of the unit cell is 2.67 x 10⁻²⁸ m³.

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8 0
2 years ago
The cell potential of the following electrochemical cell depends on the gold concentration in the cathode half-cell: Pt(s)|H2(g,
Masja [62]

<u>Answer:</u> The concentration of Au^{3+} in the solution is 1.87\times 10^{-14}M

<u>Explanation:</u>

The given cell is:

Pt(s)|H_2(g.1atm)|H^+(aq.,1.0M)||Au^{3+}(aq,?M)|Au(s)

Half reactions for the given cell follows:

<u>Oxidation half reaction:</u> H_2(g)\rightarrow 2H^{+}(1.0M)+2e^-;E^o_{H^+/H_2}=0V ( × 3)

<u>Reduction half reaction:</u> Au^{3+}(?M)+3e^-\rightarrow Au(s);E^o_{Au^{3+}/Au}=1.50V ( × 2)

<u>Net reaction:</u> 3H_2(s)+2Au^{3+}(?M)\rightarrow 6H^{+}(1.0M)+2Au(s)

Substance getting oxidized always act as anode and the one getting reduced always act as cathode.

To calculate the E^o_{cell} of the reaction, we use the equation:

E^o_{cell}=E^o_{cathode}-E^o_{anode}

Putting values in above equation, we get:

E^o_{cell}=1.50-0=1.50V

To calculate the concentration of ion for given EMF of the cell, we use the Nernst equation, which is:

E_{cell}=E^o_{cell}-\frac{0.059}{n}\log \frac{[H^{+}]^6}{[Au^{3+}]^2}

where,

E_{cell} = electrode potential of the cell = 1.23 V

E^o_{cell} = standard electrode potential of the cell = +1.50 V

n = number of electrons exchanged = 6

[Au^{3+}]=?M

[H^{+}]=1.0M

Putting values in above equation, we get:

1.23=1.50-\frac{0.059}{6}\times \log(\frac{(1.0)^6}{[Au^{3+}]^2})

[Au^{3+}]=1.87\times 10^{-14}M

Hence, the concentration of Au^{3+} in the solution is 1.87\times 10^{-14}M

7 0
3 years ago
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