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AleksandrR [38]
3 years ago
5

How do you calculate the number of photons having a wavelength of 10.0 micrometers required to produce 1.0 kilojoules of energy

Chemistry
1 answer:
shtirl [24]3 years ago
4 0

To calculate this,

We know that energy is 1 photon 
E = hc/wavelenth 
wavelength of 10.0 m 

Solution:
h = 6.626 x 10^-34 Jsec 
C = 2.9979 x 10^8 m/sec 
E = 6.626 10^-34 * 2.9979 10^8 / 10 = 1.9864 10^-26J 

Then, the number of photons is computed by:

n = 1000 / 1.9864 10^-26 = 5.04 10^28 photons 

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Aluminum reacts with sulfur gas to produce aluminum sulfide. a) What is the limiting reactant? What is the excess reagent? b) Ho
Sophie [7]

Answer:

a) Limiting: sulfur. Excess: aluminium.

b) 1.56g Al₂S₃.

c) 0.72g Al

Explanation:

Hello,

In this case, the initial mass of both aluminium and sulfur are missing, therefore, one could assume they are 1.00 g for each one. Thus, by considering the undergoing chemical reaction turns out:

2Al(s)+3S_2(g)\rightarrow 2Al_2S_3(s)\\

a) Thus, considering the assumed mass (which could be changed based on the one you are given), the limiting reagent is identified as shown below:

n_S^{available}=1.00gS_2*\frac{1molS_2}{64gS_2} =0.0156molS_2\\n_S^{consumed\ by \ Al}=1.00gAl*\frac{1molAl}{27gAl}*\frac{3molS_2}{2molAl}=0.0556molS_2

Thereby, since there 1.00g of aluminium will consume 0.0554 mol of sulfur but there are just 0.0156 mol available, the limiting reagent is sulfur and the excess reagent is aluminium.

b) By stoichiometry, the produced grams of aluminium sulfide are:

m_{Al_2S_3}=0.0156molS_2*\frac{2molAl_2S_3}{3molS_2} *\frac{150gAl_2S_3}{1molAl_2S_3} =1.56gAl_2S_3

c) The leftover is computed as follows:

m_{Al}^{excess}=(0.0556-0.0156)molS_2*\frac{2molAl}{3molS_2}*\frac{27gAl}{1molAl} =0.72 gAl\\

NOTE: Remember I assumed the quantities, they could change based on those you are given, so the results might be different, but the procedure is quite the same.

Best regards.

7 0
2 years ago
How many moles are in 9.8 grams of calcium?
kondor19780726 [428]
<span>when the number of moles Ca = mass of Ca / molar mass of Ca.

and we can get the molar mass of Ca, it is = 40 g/mol

and we have already the mass of Ca (given) = 9.8 g

so, by substitution: the moles Ca = 9.8 g / 40 g/mol

                                                       = 0.245 moles</span>
4 0
3 years ago
Read 2 more answers
Does vinegar conduct? yes or no
uranmaximum [27]
Im guessing and i think the answer is yes.
8 0
2 years ago
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Which of the following is not a characteristic of metals? A. good conductors B. ductile and malleable C. incompressible D. cryst
Maru [420]
A. good conductors is not a characteristic of metals out of all the other followings.
4 0
3 years ago
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34g aluminum are combined with 39g chlorine gas which is the limiting reactant
vampirchik [111]

The limiting reactant is chlorine (Cl2).

<u>Explanation</u>:

Limiting reactant is the amount of product formed which gets limited by the reagent without continuing it.

        2 Al + 3 Cl2 ==> 2 AlCl3  represents the balanced equation.

Number of moles Al present = 34 g Al x 1 mole Al / 26.98 g

                                                = 1.260 g moles of Al

Number of moles Cl2 present = 39 g Cl2 x 1 mole Cl2 / 35.45 g

                                                  = 1.10 g moles of Cl2

Dividing each reactant by it's coefficient in the balanced equation obtains:

1.260 moles Al / 2 = 0.63 g moles of Al

1.11 moles Cl2 / 3 = 0.36 g moles of Cl2

The reactant which produces a lesser amount of product is called as limiting reactant.

Here the Limiting reactant is Cl2.

6 0
3 years ago
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