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4 years ago

Why do larger gases such as Neon produce more color bands (line spectra) than smaller gases like Hydrogen?

1 answer:

7 0

Larger gases produces more spectral lines than the smaller gases because they have more orbitals in their atoms.
Hydrogen has only one orbital in which an electron orbits. At the excited state, that is, when the electron gains energy, the number of energy level it can transcend is very few. For larger elements, they have more orbitals and when excited, they can move from the ground state to other energy levels at which they produce various unique spectral lines.

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Write the chemical equation for the following reaction: the elements carbon and oxygen combine to yield the compound carbon diox

Marta_Voda [28]

Product of this reaction depends on the amount of oxygen. i.) If oxygen is excess: Carbon, as graphite, burns to form gaseous carbon (IV) oxide (carbon dioxide), CO2 Equation : C + O2 ----> CO2. This reaction is also called combustion reaction. The expensive way to produce CO2 is to burn Diamond in the presence of air at 600-800°C. Hope it helped you. :D

8 0

3 years ago

The sun emits electromagnetic waves at many frequencies, but mostly in the infrared, visible, and ultraviolet

meriva

The correct answer is radiant, radiant that get transferred into chemical energy

3 0

4 years ago

Read 2 more answers

Plz help plz help plz help meed

Leviafan [203]

Answer:

5106.38 Ω

Explanation:

From the question given above, the following data were obtained:

Current (I) = 0.0235 A

Voltage (V) = 120 V

Resistor (R) =?

From ohm's law,

V = IR

Where:

V => is the voltage.

I => is the current

R => is the resistor

With the above formula, we can obtain the size of the resistor needed as follow:

Current (I) = 0.0235 A

Voltage (V) = 120 V

Resistor (R) =?

V = IR

120 = 0.0235 × R

Divide both side by 0.0235

R = 120 / 0.0235

R = 5106.38 Ω

Thus, the size of the resistor needed is 5106.38 Ω

3 0

3 years ago

Consider the following two substances andtheir vapor pressures at 298 \rm K.Substance Vapor pressure (\rm mmHg) A 275 B 459Based

Dafna11 [192]

Explanation:

The pressure exerted by vapors or gas on the surface of a liquid is known as vapor pressure.

This means that weaker is the intermolecular forces present in a substance more easily it can form vapors. As a result, it will have high vapor pressure.

As substance B has high vapor pressure which means that it has weak intermolecular forces.

Also, stronger is the intermolecular forces present in a substance more will be its boiling point. Hence, more energy or temperature is required to break the bonds. Hence, substance A has higher boiling point and high heat of vaporization.

When surrounding pressure is less than or equal to its vapor pressure then substance B boils into the gas phase. Hence, substance B will be a gas at 300 mm Hg.

Therefore, we can conclude that characteristics of the two substances will be as follows.

(a) Substance B - has weaker intermoclcular

(b) Substance A - has a higher boiling point

(d) Substance A - has a higher heat of vaporization

8 0

3 years ago

You need to prepare 0.2 liters of a 0.1 M solution of a phosphate buffer with a pH of 7.2. If you have solid K2HPO4 (FW=136.09)

Lera25 [3.4K]

Answer:

To prepare 0,2 L of a 0,1M solution of a phosphate buffer to a pH of 7,2 you need to add 1,05 mL of 6M HCl, 2,722 g of K₂HPO₄ and complete 0,2 L with water.

Explanation:

The acid equilibrium of phosphate buffer for a pH of 7,2 is:

H₂PO₄⁻ ⇄ HPO₄²⁻+ H⁺ pka = 6,86

Using Henderson-Hasselbalch formula:

pH = pka + log₁₀ $\frac{[HPO_{4}^{2-}]}{[H_{2}PO_{4}^-]}$

7,2 = 6,86 + log₁₀ $\frac{[HPO_{4}^{2-}]}{[H_{2}PO_{4}^-]}$

2,18776 = $\frac{[HPO_{4}^{2-}]}{[H_{2}PO_{4}^-]}$ (1)

You need to add 0,2L× 0,1M = 0,02 moles of phosphate buffer, that means:

0,02 moles = H₂PO₄⁻ + HPO₄²⁻ (2)

Replacing (2) in (1):

H₂PO₄⁻: 6,2739x10⁻³ moles

Thus,

HPO₄²⁻: 0,013726 moles

K₂HPO₄ reacts with HCl thus:

K₂HPO₄ + HCl → KH₂PO₄ + KCl

Thus, you need to add 0,02 moles of K₂HPO₄ that will react with 6,2739x10⁻³ moles of HCl To produce the necessary moles for the buffer:

0,02 moles of K₂HPO₄× $\frac{136,09 g}{1 mole}$ = 2,722 g

6,2739x10⁻³ moles of HCl÷ 6 M = 1,05x10⁻³ L ≡ 1,05 mL of 6M HCl

Thus, to prepare 0,2 L of a 0,1M solution of a phosphate buffer to a pH of 7,2 you need to add 1,05 mL of 6M HCl, 2,722 g of K₂HPO₄ and complete 0,2 L with water.

6 0

4 years ago