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Klio2033 [76]
2 years ago
15

Calculate the wavelength (in nm) of the blue light emitted by a mercury lamp with a frequency of 6.32 × 1014 Hz.

Chemistry
1 answer:
tester [92]2 years ago
7 0

Answer:

474 nm or 4.74 x 10^2 nm

Explanation:

c = λv

c (speed of light) = 2.998 x 10^8 m/s

λ = ?

v = 6.32 × 1014 Hz = 6.32 × 1014 1/s

2.998 x 10^8 m/s = (λ)(6.32 × 10^14 1/s)

λ = (2.998 x 10^8 m/s) / (6.32 × 10^14 1/s)

λ = 4.74 x 10^-7 m

λ = 4.74 x 10^-7 m x (1 x 10^9 nm/1 m) = 474 nm

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So to find mass in gram multiply the no.mole by Molar mass
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Explain how and the type of a bond that forms between sodium and chlorine in sodium chloride (NaCl)
Alexandra [31]

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7 0
2 years ago
Which would exert more pressure on the floor, a 100kg man wearing snow shoes or a 60kg woman wearing high heels? PLEASE EXPLAIN
iris [78.8K]

Pressure can be defined as the force acting on a perpendicular surface per unit area.

Force exerted by a man of mass 100 kg wearing snow shoes = m.a

Where m = mass of the man = 100 kg

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Force exerted by the man of mass 100 kg = 100 kg(9.8 m/s^{2}) = 980 N

Force exerted by woman of mass 60 kg = 60 kg(9.8 m/s^{2}) = 588 N

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5 0
2 years ago
Find percent yield:
saveliy_v [14]

<u>Answer:</u> The percent yield of the reaction is 91.8 %

<u>Explanation:</u>

To calculate the number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}     .....(1)

  • <u>For B_5H_9 :</u>

Given mass of B_5H_9 = 4.0 g

Molar mass of B_5H_9 = 63.12 g/mol

Putting values in equation 1, we get:

\text{Moles of }B_5H_9=\frac{4g}{63.12g/mol}=0.0634mol

  • <u>For oxygen gas:</u>

Given mass of oxygen gas = 10.0 g

Molar mass of oxygen gas = 32 g/mol

Putting values in equation 1, we get:

\text{Moles of oxygen gas}=\frac{10g}{32g/mol}=0.3125mol

The chemical equation for the reaction of B_5H_9 and oxygen gas follows:

2B_5H_9+12O_2\rightarrow 5B_2O_3+9H_2O

By Stoichiometry of the reaction:

12 moles of oxygen gas reacts with 2 moles of B_2H_5

So, 0.3125 moles of oxygen gas will react with = \frac{2}{12}\times 0.3125=0.052mol of B_2H_5

As, given amount of B_2H_5 is more than the required amount. So, it is considered as an excess reagent.

Thus, oxygen gas is considered as a limiting reagent because it limits the formation of product.

By Stoichiometry of the reaction:

12 moles of oxygen gas produces 5 moles of B_2O_3

So, 0.3125 moles of oxygen gas will produce = \frac{5}{12}\times 0.3125=0.130moles of water

Now, calculating the mass of B_2O_3 from equation 1, we get:

Molar mass of B_2O_3 = 69.93 g/mol

Moles of B_2O_3 = 0.130 moles

Putting values in equation 1, we get:

0.130mol=\frac{\text{Mass of }B_2O_3}{69.63g/mol}\\\\\text{Mass of }B_2O_3=(0.130mol\times 69.63g/mol)=9.052g

To calculate the percentage yield of B_2O_3, we use the equation:

\%\text{ yield}=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100

Experimental yield of B_2O_3 = 8.32 g

Theoretical yield of B_2O_3 = 9.052 g

Putting values in above equation, we get:

\%\text{ yield of }B_2O_3=\frac{8.32g}{9.052g}\times 100\\\\\% \text{yield of }B_2O_3=91.8\%

Hence, the percent yield of the reaction is 91.8 %

6 0
3 years ago
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