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DENIUS [597]
3 years ago
12

The volume of a single tantalum atom is 1.20×10-23 cm3. What is the volume of a tantalum atom in microliters?

Chemistry
1 answer:
sergiy2304 [10]3 years ago
3 0

Answer:

1.20x10⁻²⁰μL

Explanation:

1cm³ is equal to 1milliliter. As we must know, 1milliliter = 1000 microliters, 1000μL. To convert the 1.20x10⁻²³mL we need to use the conversion factor: 1mL = 1000μL.

The volume of tantalum in μL is:

1.20x10⁻²³mL * (1000μL /1L) = 1.20x10⁻²⁰μL

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yawa3891 [41]

Given:

E = 7.3 × 10–17 Hz                                                                                      

 h= 6.63 × 10–34 J•s

Now <em>E = hf</em>

where E is the energy of the photon                                                          

h is the Planck's constant                                                                          

f is the frequency of the photon

Substituting the values in the equation we get                                        

E= 7.3 × 10^-17 × 6.63 × 10^-34                                                                  

<u>E= 4.8399 × 10^-50  J. </u>                                                                                                      



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4 years ago
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3 years ago
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A gram of gold has the same _____ as a kilogram of gold.
aivan3 [116]
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Freshwater is distributed in both time and space
Neko [114]
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7 0
2 years ago
A certain first-order reaction has a rate constant of 2.15×10−2 s−1 at 20 ∘C. What is the value of k at 55 ∘C if Ea = 72.0 kJ/mo
adell [148]

Answer:

k_2=0.504s^{-1}

Explanation:

Hello there!

In this case, according to the given information, it turns out possible for us to calculate the rate constant at 55 °C by using the temperature-variable version of the Arrhenius equation:

ln(\frac{k_2}{k_1} )=-\frac{Ea}{R}(\frac{1}{T_2} -\frac{1}{T_1} )

Thus, we plug in the temperatures, activation energy and universal constant of gases in consistent units to obtain:

ln(\frac{k_2}{0.0215s^{-1}} )=-\frac{72000\frac{J}{mol}}{8.3145\frac{J}{mol*K}}(\frac{1}{55+273} -\frac{1}{20+273} ) \\\\ln(\frac{k_2}{0.0215s^{-1}} )=3.154\\\\k_2=0.0215s^{-1}exp(3.154)\\\\k_2=0.504s^{-1}

Regards!

3 0
3 years ago
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