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Goshia [24]
3 years ago
7

What is the frequency of a particular type of electromagnetic

Chemistry
1 answer:
gizmo_the_mogwai [7]3 years ago
4 0

Answer:

f = 0.6× 10¹⁶ Hz

Explanation:

Given data:

Wavelength of radiation = 5.00 × 10⁻⁸ m

Frequency of radiation = ?

Solution:

Formula:

Speed of light = wavelength × frequency

speed of light = 3 × 10⁸m/s

Now we will put he values in formula.

3 × 10⁸m/s = 5.00 × 10⁻⁸ m ×  f

f = 3 × 10⁸m/s / 5.00 × 10⁻⁸ m

f = 0.6× 10¹⁶ s⁻¹

s⁻¹ = Hz

f = 0.6× 10¹⁶ Hz

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What is the mass of a 5.00cm3 piece of copper having a density of 8.96g/cm3
r-ruslan [8.4K]

Answer:

44.8 grams

Explanation:

The density exists as a ratio comparing the mass of a piece of copper per 1 cm³. Since you have been given a new volume, you can set up a proportion to find the new mass.

\frac{8.96 grams}{1.00 cm^3} =\frac{?grams}{5.00cm^3}                                         <----- Proportion

44.8 = (1.00cm^3)(?grams)                              <----- Cross-multiply

44.8= ?grams                                                 <----- Divide both sides by 1.00

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2 years ago
What is the total number of electrons that can occupy the d sublevel?
VikaD [51]
10 electrons Are in d
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4 years ago
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Gold forms a substitutional solid solution with silver. Compute the number of gold atoms per cubic centimeter for a silver-gold
Mama L [17]

<u>Answer:</u> The number of gold atoms per cubic centimeters in the given alloy is 1.83\times 10^{22}

<u>Explanation:</u>

To calculate the number of gold atoms per cubic centimeters for te given silver-gold alloy, we use the equation:

N_{Au}=\frac{N_AC_{Au}}{(\frac{C_{Au}M_{Au}}{\rho_{Au}})+(\frac{M_{Au}(100-C_{Au})}{\rho_{Ag}})}

where,

N_{Au} = number of gold atoms per cubic centimeters

N_A = Avogadro's number = 6.022\times 10^{23}atoms/mol

C_{Au} = Mass percent of gold in the alloy = 42 %

\rho_{Au} = Density of pure gold = 19.32g/cm^3

\rho_{Ag} = Density of pure silver = 10.49g/cm^3

M_{Au = molar mass of gold = 196.97 g/mol

Putting values in above equation, we get:

N_{Au}=\frac{(6.022\times 10^{23}atoms/mol)\times 48\%}{(\frac{48\%\times 196.97g/mol}{19.32g/cm^3})+(\frac{196.97g/mol\times 58\%}{10.49g/cm^3})}\\\\N_{Au}=1.83\times 10^{22}atoms/cm^3

Hence, the number of gold atoms per cubic centimeters in the given alloy is 1.83\times 10^{22}

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3 years ago
What does it mean when a system or a reaction is endothermic?
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It means the system absorbs energy from surrounding objects; usually, but not always, involving heat
4 0
3 years ago
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HELP Which of the following fractions can be used in the conversion of 32 m3 to the unit mm3?
poizon [28]
We know that each millimeter contains 10⁻³ meters. Writing this as a ratio:
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We require a conversion from m³ to mm³, so we must take the cube of the ratio we have made:
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Therefore, the conversion used will be:
(1 mm / 10⁻³ m)³

When we multiply by this conversion, we will get:
32 m³ = 32 x 10⁹ mm³
7 0
3 years ago
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