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Tanzania [10]
2 years ago
6

QUESTION 5

Chemistry
1 answer:
mrs_skeptik [129]2 years ago
7 0
Answer is b. Hope it help
You might be interested in
A 14.3-cm3 sample of tin has a mass of 0.104 kg.
VladimirAG [237]
This is a true statement if it is density you are looking for... Density problem.....

Density is the ratio of the mass of an object to its volume.
D = m / V
D = 104g / 14.3 cm³ = 7.27 g/cm³ .............. to three significant digits

The conventions for the units of density is that grams per cubic centimeter (g/cm³) are usually used for solids, but will work for anything. Grams per milliliter (g/mL) are usually used for liquids and grams per liter (g/L) are for gases. Therefore, by convention, the units for tin (a solid) should be in grams per cubic centimeter.

Since 1 mL is equivalent to 1 cm³, then the density could be expressed as 7.27 g/mL.

The accepted value for the density of tin is 7.31 g/cm³
7 0
3 years ago
What is the temperature of 0.500 moles of helium that occupies a volume of 15.0 L at a pressure of 1.6 atm?
denis23 [38]

Answer:

600K

Explanation:

PV=nRT

T=PV/nR

= 1.6atm* 15.0L/ 0.5mol*0.0821LatmK^-1mol^-1

=600K

6 0
3 years ago
An electron charge e mass m and a positron charge e mass m revolve around their common center of mass under the influence of the
Katen [24]

Answer: v = 2π2 Kme2 Z / nh

Explanation:

The formula for velocity of an electron in the nth orbit is given as,

v = 2π2 Kme2 Z / nh

v = velocity

K = 1/(4πε0)

m= mass of an electron

e = Charge on an electron

Z= atomic number

h= Planck’s constant

n is a positive integer.

3 0
3 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
2 years ago
The energy diagram illustrated in the model corresponds to what student B
KatRina [158]

Answer: Exothermic reaction

Explanation:

Exothermic reactions lose energy, or ΔH, when they react.

4 0
3 years ago
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