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3 years ago

11

An electron in the ground state absorbs a single photon of light and then relaxes back to the ground state by emitting an infrar

ed photon (1200 nm) followed by an orange photon (600 nm). What is the wavelength of the absorbed photon?
A. 400 nm
B. 600 nm
C. 1800 nm

1 answer:

vladimir2022 [97]3 years ago

7 0

B. 600 nm would be the wavelength

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Four springs are stretched to the same distance from the equilibrium position. The spring constants are listed in the table. A 2

svp [43]

Answer:

The correct option is;

X, W, Y, Z

Explanation:

The parameters given are;

Spring (S), Spring Constant (N/m)

W, 24

X, 35

Y, 22

Z, 15

The equation for elastic potential energy, $E_e$ , is $E_e = 0.5 \times k \times x^2$

The above equation can also be written as $E_e =\dfrac{1}{2} \times k \times x^2$

Where:

k = The spring constant in (N/m)

x = The spring extension

Therefore, since the elastic potential energy, $E_e$ , of the spring is directly proportional to the spring constant, k, we have the springs with higher spring constant will have higher elastic potential energy, $E_e$ , therefore the correct order is as follows;

X > W > Y > Z

9 0

3 years ago

Photosynthesis by land plants leads to the fixation each year of about 1 kg of carbon on the average for each square meter of an

Rama09 [41]

Answer:

a) mass of carbon directly above 1 ( each) square meter of the earth is 1.65kg

b) all CO₂ will definitely be used up from the atmosphere directly above a forest in 1.65 years

Explanation:

first we calculate the moles of carbon

moles = mass/molar mass

= 1kg/12gmol⁻¹

= 1000g/12gmol⁻¹

= 83.33 mol

now using the ideal gas equation

we find the volume of co₂required based on 83.33 moles

PVco₂ = nRT

Vco₂ = nRT/P

Vco₂ = (83.22mol × 0.0821L atm k⁻¹ mol⁻¹ 298 K) / 1 atm

Vco₂ = 2083.73 L

so since CO₂ in air is 0.0390% by volume in the atmosphere, we find the the total amount of air required to obtain 1kg carbon

therefore

Vair × 0.0390/100 = 2038.73L

Vair = (2038.73L × 100) / 0.0390

Vair = 5.23 × 10⁶L

therefore 5.23 × 10⁶ L of air will be required to obtain 1kg carbon

a)

Here we calculate the mass of air over 1 square meter of surface.

Remember that atmospheric pressure is the consequence of the force exerted by all the air above the surface; 1 bar is equivalent to 1.020×10⁴kgm⁻²

NOW

mass of air = 1.020×10⁴kgm⁻² × 1m²

= 1.020×10⁴kg

= 1.020×10⁷g [1kg = 10³g]

we now find the moles of air associated with it

moles = mass/molar mass

= 1.020 × 10⁷g / ( 20%×Mo₂ + 80%×Mn₂)

= 1.020 × 10⁷g / ( 20%×32gmol⁻¹ + 80%×28gmol⁻¹)

= 1.020 × 10⁷g / 28.8 gmol⁻¹

= 354166.67mol

so based on the question, for each mole (air), there is 0.0390% of CO₂

now to calculate the moles of CO₂ we say;

MolesCo₂ = 0.0390/100 × 354166.67mol

= 138.125 moles

Now we calculate mass of CO₂ from the above findings

Moles = mass/molar mass

mass = moles × molar mass

= 138.125 moles × 12gmol⁻¹

= 1657.5g

we covert to KG

= 1657.5g / 1000

mass = 1.65kg

therfore mass of carbon directly above 1 ( each) square meter of the earth is 1.65kg

b)

to find the number years required to use up all the CO₂, WE SAY

Number of years = total carbon per m² of the forest / carbon used up per m² from the forest per year

Number of years = 1.65kgm⁻² / 1kg²year⁻¹

Number of years = 1.65 years

Therefore all CO₂ will definitely be used up from the atmosphere directly above a forest in 1.65 years

6 0

4 years ago

an aqueous solution contains 1.2mM of total ions. if the solution is NaCl (aq), what is the concentration of chloride ions?

Genrish500 [490]

Explanation:

As the total concentration is given as 1.2 mM. And, it is also given that salt present in the solution is NaCl.

As sodium chloride is an ionic compound so, when it is added to water then it will dissociate into sodium and chlorine ions as follows.

$NaCl \rightarrow Na^{+} + Cl^{-}$

So, it means in total there will be formation of 2 ions when one molecules of NaCl dissociates.

Therefore, concentration of chlorine ions will be calculated as follows.

Concentration of $Cl^{-}$ ions = $\frac{1.2mM}{2}$

= 0.6 mM

Thus, we can conclude that the concentration of chloride ions is 0.6 mM.

5 0

3 years ago

A stock solution of sodium sulfate NaSO4 has a concentrate of 1.00 m. The volume of this solution is 50 ml. What volume of 0.25

mylen [45]

<h3>Answer:</h3>

200 mL

<h3>Explanation:</h3>

Concept tested: Dilution formula

We are given;

Concentration of stock solution as 1.00 M
Volume of the stock solution as 50 mL
Molarity of the dilute solution as 0.25 M

We are required to calculate the volume of diluted solution;

The stock solution is the original solution before dilution while diluted solution is the solution after dilution.
Using the dilution formula we can determine the volume of diluted solution;

M1V1 = M2V2

Rearranging the formula;

V2 = M1V1 ÷ M2

= (1.00 M × 50 mL) ÷ 0.25 M

= 200 mL

Therefore, a volume of 200mL of 0.25 M solution could be made from the stock solution.

5 0

3 years ago

3. At 10.0°C and 2.00 atm, the volume of the gas is 75.0 mL. If the temperature of the gas increases to

alex41 [277]

Explanation:

c.

$from \\ \frac{P _{1} V _{1} }{T _{1}} = \frac{P _{2} V _{2}}{T _{2}} \\ \frac{2 \times 75}{(10+273)} = \frac{V _{2} \times 1}{(35+273)} \\V _{2} = \frac{2 \times 75 \times (35+273)}{(10+273) \times 1} \\ V _{2} = 163 \: ml$

6 0

3 years ago