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3 years ago

Which statements best describe wavelengths of sunlight that are longer than 700 nanometers

Chemistry

2 answers:

fredd [130]3 years ago

8 0

Answer:

These wavelengths are longer than wavelength of visible light.

These wavelengths form the infra-red part of the spectrum.

Explanation:

Solar spectrum has lies in three parts of the electromagnetic spectrum - Ultraviolet, visible and infra-red.

Visible light lies from 400 nm to 700 nm.

Ultraviolet region precedes visible spectrum and has shorter wavelength than it.

Beyond visible spectrum, infrared part of the spectrum starts and has longer wavelength than it.

The wavelength of sunlight that are greater than 700 nm wavelength have longer wavelength than visible light and lie in the infra-red part of the spectrum.

Oxana [17]3 years ago

5 0

Answer:

1. These wavelengths are longer than the wavelengths of visible light

2. These wavelengths form the infared part of the spectrum.

Explanation:

Just took the test uwu

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Food is the only limiting factor that keeps populations from growing too large true or false?

topjm [15]

FALSE

There are other limiting factors like lack of space, diseases, and com petition.

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3 years ago

Read 2 more answers

Calculate the standard enthalpy change for the reaction at 25 ∘ 25 ∘ C. Standard enthalpy of formation values can be found in th

WINSTONCH [101]

Answer: The standard enthalpy change of the reaction is coming out to be -16.3 kJ

Explanation:

Enthalpy change is defined as the difference in enthalpies of all the product and the reactants each multiplied with their respective number of moles. It is represented as $\Delta H$

The equation used to calculate enthalpy change is of a reaction is:

$\Delta H_{rxn}=\sum [n\times \Delta H_f(product)]-\sum [n\times \Delta H_f(reactant)]$

For the given chemical reaction:

$Mg(OH)_2(s)+2HCl(g)\rightarrow MgCl_2(s)+2H_2O(g)$

The equation for the enthalpy change of the above reaction is:

$\Delta H_{rxn}=[(1\times \Delta H_f_{(MgCl_2(s))})+(2\times \Delta H_f_{(H_2O(g))})]-[(1\times \Delta H_f_{(Mg(OH)_2(s))})+(2\times \Delta H_f_{(HCl(g))})]$

We are given:

$\Delta H_f_{(Mg(OH)_2(s))}=-924.5kJ/mol\\\Delta H_f_{(HCl(g))}=-92.30kJ/mol\\\Delta H_f_{(MgCl_2(s))}=-641.8kJ/mol\\\Delta H_f_{(H_2O(g))}=-241.8kJ/mol$

Putting values in above equation, we get:

$\Delta H_{rxn}=[(1\times (-641.8))+(2\times (-241.8))]-[(1\times (-924.5))+(2\times (-92.30))]\\\\\Delta H_{rxn}=-16.3kJ$

Hence, the standard enthalpy change of the reaction is coming out to be -16.3 kJ

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3 years ago

The standard molar heat of fusion of ice is 6020 j/mol. calculate q, w, and ∆e for melting 1.00 mol of ice at 0◦c and 1.00 atm p

zysi [14]

Answer : q = 6020 J, w = -6020 J, Δe = 0

Solution : Given,

Molar heat of fusion of ice = 6020 J/mole

Number of moles = 1 mole

Pressure = 1 atm

Molar heat of fusion : It is defined as the amount of energy required to melt 1 mole of a substance at its melting point. There is no temperature change.

The relation between heat and molar heat of fusion is,

$q=\Delta H_{fusion}(\frac{Mass}{\text{ Molar mass}})$ (in terms of mass)