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DerKrebs [107]
3 years ago
10

Which lists the structures, in correct order, through which light passes when it enters the eye?

Chemistry
1 answer:
Mashutka [201]3 years ago
8 0

Answer:

cornea, pupil, lens, vitreous humor

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A 3000 MWt fast reactor has a 42% efficiency. This reactor operates for 23 months followed by a 1 month down time for refueling
n200080 [17]

Answer:

capacity factor = 0.952

Availability factor = 0.958

Explanation:

1) capacity factor

capacity factor = actual power output /  maximum power output

                        = (actual power output)/(efficiency * rated power output)

                       = \frac{1200}{\frac{42}{100}*3000}

= 0.952

2) Availability factor

Availability factor  = Actual operation time period/ total time period

                            = 23/24 = 0.958

8 0
3 years ago
4. Write the electronic configuration of first 20 elements in the periodic table.
DIA [1.3K]

Answer:

Name Atomic Number Electron Configuration Period 1 Hydrogen 1 1s1 Helium 2 1s2 Period 2 Lithium 3 1s2 2s1 Beryllium 4 1s2 2s2 Boron 5 1s2 2s22p1 Carbon 6 1s2 2s22p2 Nitrogen 7 1s2 2s22p3 Oxygen 8 1s2 2s22p4 Fluorine 9 1s2 2s22p5 Neon 10 1s2 2s22p6 Period 3 Sodium 11 1s2 2s22p63s1 Magnesium 12 1s2 2s22p63s2 Aluminum 13 

7 0
3 years ago
Consider the reaction given below.
Drupady [299]

Answer:

  • <u>K =  0.167 s⁻¹</u>

Explanation:

<u>1) Rate law, at a given temperature:</u>

  • Since all the data are obtained at the same temperature, the equilibrium constant is the same.

  • Since only reactants A and B participate in the reaction, you assume that the form of the rate law is:

        r = K [A]ᵃ [B]ᵇ

<u>2) Use the data from the table</u>

  • Since the first and second set of data have the same concentration of the reactant A, you can use them to find the exponent b:

        r₁ = (1.50)ᵃ (1.50)ᵇ = 2.50 × 10⁻¹ M/s

        r₂ = (1.50)ᵃ (2.50)ᵇ = 2.50 × 10⁻¹ M/s

         Divide r₂ by r₁:     [ 2.50 / 1.50] ᵇ = 1 ⇒ b = 0

  • Use the first and second set of data to find the exponent a:

        r₁ = (1.50)ᵃ (1.50)ᵇ = 2.50 × 10⁻¹ M/s

        r₃ = (3.00)ᵃ (1.50)ᵇ = 5.00 × 10⁻¹ M/s

        Divide r₃ by r₂: [3.00 / 1.50]ᵃ = [5.00 / 2.50]

                                  2ᵃ = 2 ⇒ a = 1

         

<u>3) Write the rate law</u>

  • r = K [A]¹ [B]⁰ = K[A]

This means, that the rate is independent of reactant B and is of first order respect reactant A.

<u>4) Use any set of data to find K</u>

With the first set of data

  • r = K (1.50 M) = 2.50 × 10⁻¹ M/s ⇒ K = 0.250 M/s / 1.50 M = 0.167 s⁻¹

Result: the rate constant is K =  0.167 s⁻¹

6 0
4 years ago
How many moles of CO2 are produced from the combustion of 5.25 moles of CH3OH?
iVinArrow [24]

First, we write the reaction for CH3OH combustion

CH3OH+3/2O2--->CO2+2H2O

for 1 mole of methanol, we get 1 mole of CO2, therefore for 5,25 moles of methanol we will get 5,25 moles of CO2

4 0
3 years ago
PLEASE HELP
victus00 [196]

Answer:

A restoration

Explanation:

the action of returning something to a former owner, place, or condition.

7 0
2 years ago
Read 2 more answers
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