All categories

3 years ago

Visible green light has a wavelength around 510 nm. What is its frequency?

Chemistry

1 answer:

madam [21]3 years ago

3 0

Explanation:

freauency = speed of light / wavelen

588235.29=510 ÷ (3×10⁸)

You might be interested in

What is the total number of different elements present in NH4NO3

Nookie1986 [14]

<span>3 elements
Nitrogen
Hydrogen
Oxygen

2 nitrogen 4 hydrogen and 3 oxygens
there are only 3 different elements</span>

4 0

3 years ago

Read 2 more answers

3. How many atoms in 3.91 moles of sodium?

Gala2k [10]

Answer:

Explanation:

The formula for sodium is Na. It does not form a molecule in some way.

1 mol Na = 6.02*10^23 atoms

3.91 mol = x Cross multiply

x = 3.91 * 6.02 * 10^23

x = 23.65 * 10^23

x = 2.365 * 10^24

Scientific notation is always expressed as a number 1 ≤ x < 10

3 0

3 years ago

Wendy made the following diagram shown to represent the solar eclipse.

Pavlova-9 [17]

The moon should be between the sun and Earth

7 0

4 years ago

Read 2 more answers

Examples<br> 1) How many atoms of silver are there in 4.5 moles of silver? [Ag = 107.8682 g/mol]

Anon25 [30]

3.0x10-4..............................

7 0

4 years ago

Read 2 more answers

When 61.6 g of alanine (C3H7NO2) are dissolved in 1150. g of a certain mystery liquid X, the freezing point of the solution is 2

MrRissso [65]

Answer:

Explanation:

From the given information:

TO start with the molarity of the solution:

$= \dfrac{61.6 \ g \times \dfrac{1 \ mol \ C_3H_7 NO_3}{89.1 \ g} }{1150 \ g \times \dfrac{1 \ kg}{1000 \g }}$

= 0.601 mol/kg

= 0.601 m

At the freezing point, the depression of the solution is $\Delta \ T_f = T_{solvent}- T_{solution}$

$\Delta \ T_f = 2.9 ^0 \ C$

Using the depression in freezing point, the molar depression constant of the solvent $K_f = \dfrac{\Delta T_f}{m}$

$K_f = \dfrac{2.9 ^0 \ C}{0.601 \ m}$

$K_f = 4.82 ^0 C / m}$

The freezing point of the solution $\Delta T_f = T_{solvent} - T_{solution}$

$\Delta T_f = 7.3^ 0 \ C$

The molality of the solution is:

$= \dfrac{61.6 \ g \times \dfrac{1 \ mol \ NH_4Cl}{53.5 \ g} }{1150 \ g \times \dfrac{1 \ kg}{1000 \g }}$

Molar depression constant of solvent X, $K_f = 4.82 ^0 \ C/m$

Hence, using the elevation in boiling point;

the Vant'Hoff factor $i = \dfrac{\Delta T_f}{k_f \times m}$

$i = \dfrac{7.3 \ ^0 \ C}{4.82 ^0 \ C/m \times 1.00 \ m}$

$\mathbf {i = 1.51 }$

3 0

3 years ago