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elixir [45]
3 years ago
12

The total kinetic energy of a body is known as:

Chemistry
1 answer:
Nutka1998 [239]3 years ago
8 0

Answer: thermal energy

Explanation:

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Why it’s possible to walk through air but not a concrete wall
Sunny_sXe [5.5K]

well you can't walk through the air but you can't walk through a concrete wall because all of the solid atoms holding it together

3 0
4 years ago
A 0.100 mile sample of gas is at a temperature of 85.0 degrees C and a volume of 3.47 L. What is the pressure of the gas( in mm
vladimir1956 [14]

Answer: 64.6 mmHg

Explanation:

Given that:

Volume of gas V = 3.47L

(since 1 liter = 1dm3

3.47L = 3.47dm3)

Temperature T = 85.0°C

Convert Celsius to Kelvin

(85.0°C + 273 = 358K)

Pressure P = ?

Number of moles of gas N = 0.100 mole

Note that Molar gas constant R is a constant with a value of 0.0082 ATM dm3 K-1 mol-1

Then, apply ideal gas equation

pV = nRT

p x 3.47dm3 = 0.10 x (0.0082 atm dm3 K-1 mol-1 x 358K)

p x 3.47dm3 = 0.29 atm dm3

p = (0.29 atm dm3 / 3.47 dm3)

p = 0.085 atm

Recall that pressure of the gas is required in mm hg, so convert 0.085 atm to mm Hg

If 1 atm = 760 mm Hg

0.085atm = 0.085 x 760

= 64.6 mm Hg

Thus, the pressure of the gas is 64.6 mm hg

6 0
3 years ago
What is the mass of 3.00 moles of magnesium chloride, MgCl2? Express your answer with the appropriate units.
xenn [34]
<span>To find the mass of 3.00 moles of magnesium chloride (MgCl2), first record the atomic mass of magnesium (Mg) and chloride (Cl), which are both listed on the periodic table as follows: Mg=24 g/mole Cl=38 g/mole Now, double the Cl mass since there are 2 Cl moles in MgCl2 and then add it to the Mg mass like so: (38 g/mole*2 moles)+24 g/mole=100 g/mole Finally, to calculate the mass of 3.00 moles of MgCl2, convert the combined atomic mass to grams as follows: 3.00 moles * 100 g/mole = 300 g</span>
5 0
4 years ago
The Balmer series, named after Johann Balmer, is a portion of the hydrogen emission spectrum produced from the transitions betwe
horrorfan [7]

Explanation:

The wavelength of the balmer series is calculated using the following steps;

- Find the Principle Quantum Number for the Transition

- Calculate the Term in Brackets

- Multiply by the Rydberg Constant

- Find the Wavelength

The Balmer series in a hydrogen atom relates the possible electron transitions down to the n = 2 position to the wavelength of the emission that scientists observe.

The λ symbol represents the wavelength, and RH is the Rydberg constant for hydrogen, with RH = 1.0968 × 107 m−1

n=7 to n=2

- The principal quantum numbers are 2 and 7.

-  (1/2²) − (1 / n²₂)

For n₂ = 7, you get:

(1/2²) − (1 / n²₂) = (1/2²) − (1 / 7²)

= (1/4) − (1/49)

= 0.2230

- RH = 1.0968 × 107 m−1, to find a value for 1/λ. The formula and the example calculation gives:

1/λ = RH [(1/2²) − (1 / n²₂)]

= 1.0968 × 107 m−1 × 0.2230

= 2445864 m−1

- λ = 1 / 2445864 m−1

= 4.08 × 10−7 m

= 408 nanometers

≈ 410nm

n=6 to n=2

- The principal quantum numbers are 2 and 6.

-  (1/2²) − (1 / n²₂)

For n₂ = 6, you get:

(1/2²) − (1 / n²₂) = (1/2²) − (1 / 6²)

= (1/4) − (1/36)

=  0.2222

- RH = 1.0968 × 107 m−1, to find a value for 1/λ. The formula and the example calculation gives:

1/λ = RH [(1/2²) − (1 / n²₂)]

= 1.0968 × 107 m−1 × 3/16

= 2437090 m−1

- λ = 1 / 2437090 m−1

= 4.10 × 10−7 m

= 410 nanometers

n=5 to n=2

- The principal quantum numbers are 2 and 5.

-  (1/2²) − (1 / n²₂)

For n₂ = 5, you get:

(1/2²) − (1 / n²₂) = (1/2²) − (1 / 5²)

= (1/4) − (1/25)

= 0.21

- RH = 1.0968 × 107 m−1, to find a value for 1/λ. The formula and the example calculation gives:

1/λ = RH [(1/2²) − (1 / n²₂)]

= 1.0968 × 107 m−1 × 0.21

= 2303280 m−1

- λ = 1 / 2303280 m−1

= 4.34 × 10−7 m

= 434 nanometers

n=4 to n=2

- The principal quantum numbers are 2 and 4.

-  (1/2²) − (1 / n²₂)

For n₂ = 4, you get:

(1/2²) − (1 / n²₂) = (1/2²) − (1 / 4²)

= (1/4) − (1/16)

= 0.1875

- RH = 1.0968 × 107 m−1, to find a value for 1/λ. The formula and the example calculation gives:

1/λ = RH [(1/2²) − (1 / n²₂)]

= 1.0968 × 107 m−1 × 0.1875

= 2056500 m−1

- λ = 1 / 2056500 m−1

= 4.86 × 10−7 m

= 486 nanometers

n=3 to n=2

- The principal quantum numbers are 2 and 3.

-  (1/2²) − (1 / n²₂)

For n₂ = 3, you get:

(1/2²) − (1 / n²₂) = (1/2²) − (1 / 3²)

= (1/4) − (1/9)

= 0.13889

- RH = 1.0968 × 107 m−1, to find a value for 1/λ. The formula and the example calculation gives:

1/λ = RH [(1/2²) − (1 / n²₂)]

= 1.0968 × 107 m−1 × 0.13889

= 1523345 m−1

- λ = 1 / 1523345 m−1

= 6.56 × 10−7 m

= 656 nanometers

7 0
3 years ago
Atoms containing the same number of protons and different numbers of neutrons are _____. (1 point)
Vadim26 [7]
I believe that it could be isotopes
3 0
4 years ago
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