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3 years ago

Compare the wavelength and energy of ultraviolet rays to visible light

Chemistry

1 answer:

Nadusha1986 [10]3 years ago

6 0

Answer:

Visible wavelengths range from 0.0007 milimeters for red light, through orange, yellow, green, and blue, to 0.0004 milimeters for violet light.

Ultraviolet is shorter wavelengths than violet.

Hope This Helps.

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What is the pH of a solution after the addition of 30.0 mL of 0.100 M NaOH to 50.0 mL of 0.10 M HBr?

Pani-rosa [81]

Answer:

pH = 1.6

Explanation:

HBr + NaOH ⇒ NaBr + H2O

0 mL NaOH:

HBr ↔ H3O+ + Br-

⇒ [ H3O+] = M HBr = 0.1 M

⇒pH = -log [H3O+] = 1

30 mL NaOH:

⇒ mol NaOH = 0.1 mol / L * 0.03 L = 3 E-3 mol

⇒ mol HBr = 0.05 L * 0.1 mol/L = 5 E-3 mol

⇒ M HBr = ( 5 E-3 mol - 3 E-3 mol) / 0.08 L = 0.025 M

⇒ pH = - log (0.025) = 1.6

7 0

2 years ago

What are the elements in group 3-12 called

Wittaler [7]

Answer:

Transition Metals

Explanation:

The elements in groups 3-12 are called Transition Metals. These groups contain metals that usually form multiple cations. All other groups on the table (1, 2, 13-18) are called Main Group Elements.

5 0

2 years ago

What is the main cause of an earthquake

Natali5045456 [20]

Answer:

I thinks it's stress caused by plate movement hope this helps

4 0

3 years ago

Read 2 more answers

Does anybody know this ?

geniusboy [140]

Explanation:

density =mass/volume

density = 603/281

density = 2.1459 g/cm³

3 0

3 years ago

If 3 moles of a compound use 12 J of energy in a reaction, what is the Hreaction in kJ/mol

igomit [66]

Answer:

$\Delta _RH=4x10^{-3}\frac{kJ}{mol}$

Explanation:

Hello,

In this case, the molar enthalpy of reaction is obtained by dividing the involved energy by the reacting moles:

$\Delta _RH=\frac{12J}{3mol} =4\frac{J}{mol}$

Thus, it is important to notice that the compound "uses" the energy, it means that it absorbs the energy, for that reason the sign is positive. Moreover, computing the result in kJ/mol we finally obtain:

$\Delta _RH=4\frac{J}{mol}*\frac{1kJ}{1000J} =4x10^{-3}\frac{kJ}{mol}$

Best regards.

5 0

2 years ago