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3 years ago

Which type of electromagnetic radiation has a lower frequency than infared radiation

Chemistry

1 answer:

eduard3 years ago

8 0

Radio waves have the lowest energies, longest wavelengths, and lowest frequencies of any type of EM radiation.

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I need some information about greenhouse effect, please. Could you help me?

LenKa [72]

Answer: green house affect is the sun on plants it is transfering energy to the plants to help them grow same with every plant, tree, fruit tree. everything needs sun.

Explanation: look up there.

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3 years ago

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Convert 7.72 years into days.

Lelu [443]

Answer:2817.8

Explanation:multiply the value by 365

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3 years ago

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Difference between dipole dipole and hydrogen bonding

riadik2000 [5.3K]

Hydrogen bonds are stronger than the dipole dipole attraction force present in any molecule.

<h3>What is bonding in molecules?</h3>

Bonding is a type of attraction force which is present between the different atoms or elements of any substance.

Dipole dipole attraction force is a weak force as compared to the hydrogen bonding and present between any two oppositely charged atoms.
Hydrogen bond is present between the hydrogen atom and more electronegative atoms like O, S, N and F.

Hence main difference is that hydrogen bond is only present between the hydrogen atom and more electronegative.

To know more about dipole-dipole force, visit the below link:
brainly.com/question/24197168

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8 0

2 years ago

A gas mixture with 4 mol of Ar, x moles of Ne, and y moles

maks197457 [2]

Answer:

a) $\Delta G_{mixing}=\frac{R*T}{12}*[4*ln (1/3) +x*ln (x/12) +(8-x)*ln ((8-x)/12)]$

b) $x=4$

c) $\Delta G_{max}=-2721.9 J/mol$

Explanation:

Gas mixture:

$n_{Ar}= 4 mol$

$n_{Ne}= x mol$

$n_{Xe}= y mol$

$n_{tot}= n_{Ar} + n_{Ne} + n_{Xe}=3*n_{Ar}$

$n_{Ne} + n_{Xe}=2*n_{Ar}$

$x + y=8 mol$

$y=8 mol- x$

Mol fractions:

$x_{Ar}=\frac{4 mol}{12 mol}=1/3$

$x_{Ne}=\frac{x mol}{12 mol}=x/12$

$x_{Xe}=\frac{8 - x mol}{12 mol}=(8-x)/12$

Expression of $\Delta G_{mixing}$

$\Delta G_{mixing}=R*T*\sum_{i]*x_i*ln (x_i)$

$\Delta G_{mixing}=R*T*[1/3*ln (1/3) +x/12*ln (x/12) +(8-x)/12*ln ((8-x)/12)]$

$\Delta G_{mixing}=\frac{R*T}{12}*[4*ln (1/3) +x*ln (x/12) +(8-x)*ln ((8-x)/12)]$

Expression of $\Delta G_{max}$

$\frac{d \Delta G_{mixing}}{dx}=0$

$\frac{d \Delta G_{mixing}}{dx}=\frac{R*T}{12}*[ln (x/12)+12-ln ((8-x)/12)-12]$

$0=\frac{R*T}{12}*[ln (x/12)-ln ((8-x)/12)$

$0=[ln (x/12)-ln ((8-x)/12)$

$ln (x/12)=ln ((8-x)/12)$

$x=(8-x)$

$x=4$

$\Delta G_{max}=\frac{8.314*298}{12}*[4*ln (1/3) +4*ln (4/12) +(8-4)*ln ((8-4)/12)]$

$\Delta G_{max}=\frac{8.314*298}{12}*[4*ln (1/3) +4*ln (1/3) +(4)*ln (1/3)]$

$\Delta G_{max}=\frac{8.314*298}{12}*[12*ln (1/3)]$

$\Delta G_{max}=-2721.9 J/mol$

4 0

3 years ago

10. Copper(i) bromide reacts with magnesium metal: 2 CuBr + Mg → 2 Cu + MgBrz

velikii [3]

Answer:

72.6 grams

Explanation:

I got this answer through stoichiometry. For every 1 mole of Mg, 2 moles of CuBr are consumed. Because of this, multiply the moles of Mg by 2. Then, convert moles to grams.

5 0

3 years ago

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