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3 years ago

When baking, the rods in an oven heat up and turn red. This is an example of

Chemistry

1 answer:

Misha Larkins [42]3 years ago

7 0

Answer:

The most correct option is A

Explanation:

The correct option is A because all objects emit thermal radiation when there temperature is higher than absolute zero - this is based on the Stefan-Boltzmann law. Hence, when the rods in an oven heat up above absolute zero, they begin to start emitting radiation. It should noted at this point that objects do not "emit" conduction and thus all options that involved "emitting conduction" are subsequently wrong.

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Explain how the following experiment shows the direction that gravity pulls an object, specifically showing that it does not pul

balu736 [363]

Answer:that cant happen it get smaller ever bounce it docent get higher the first bounce it the highest.

Explanation:

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3 years ago

A grating has 470 lines/mm. how many orders of the visible wavelength 538 nm can it produce in addition to the m = 0 order?

maria [59]

Three complete orders on each side of the m=0 order can be produced in addition to the m = 0 order.

The ruling separation is

d=1 / (470mm −1) = 2.1×10⁻³ mm

Diffraction lines occur at angles θ such that dsinθ=mλ, where λ is the wavelength and m is an integer.

Notice that for a given order, the line associated with a long wavelength is produced at a greater angle than the line associated with a shorter wavelength.

We take λ to be the longest wavelength in the visible spectrum (538nm) and find the greatest integer value of m such that θ is less than 90°.

That is, find the greatest integer value of m for which mλ<d.

since d / λ = 538×10⁻⁹m / 2.1×10 −6 m ≈ 3

that value is m=3.

There are three complete orders on each side of the m=0 order.

The second and third orders overlap.

Learn more about diffraction here : brainly.com/question/16749356

#SPJ4

7 0

2 years ago

Calculate the number of sodium ions, perchlorate ions, Cl atoms and O atoms in 17.8 g of sodium perchlorate. Enter your answers

Degger [83]

17.8 g of sodium perchlorate contains 8.73 × 10²² Na⁺ ions, 8.73 × 10²² ClO₄⁻ ions, 8.73 × 10²² Cl atoms and 3.49 × 10²³ O atoms.

First, we will convert 17.8 g of NaClO₄ to moles using its molar mass (122.44 g/mol).

$17.8 g \times \frac{1mol}{122.44g} = 0.145 mol$

Next, we will convert 0.145 moles to molecules of NaClO₄ using Avogadro's number; there are 6.02 × 10²³ molecules in 1 mole of molecules.

$0.145 mol \times \frac{6.02 \times 10^{23}molecules }{mol} = 8.73 \times 10^{22}molecules$

NaClO₄ is a strong electrolyte that dissociates according to the following equation.

NaClO₄ ⇒ Na⁺ + ClO₄⁻

The molar ratio of NaClO₄ to Na⁺ is 1:1. The number of Na⁺ in 8.73 × 10²² molecules of NaClO₄ is:

$8.73 \times 10^{22}moleculeNaClO_4 \times \frac{1 ion Na^{+} }{1moleculeNaClO_4} = 8.73 \times 10^{22}ion Na^{+}$

The molar ratio of NaClO₄ to ClO₄⁻ is 1:1. The number of ClO₄⁻ in 8.73 × 10²² molecules of NaClO₄ is:

$8.73 \times 10^{22}moleculeNaClO_4 \times \frac{1 ion ClO_4^{-} }{1moleculeNaClO_4} = 8.73 \times 10^{22}ion ClO_4^{-}$

The molar ratio of ClO₄⁻ to Cl is 1:1. The number of Cl in 8.73 × 10²² ions of ClO₄⁻ is:

$8.73 \times 10^{22}ion ClO_4^{-} \times \frac{1 atomCl }{1ion ClO_4^{-}} = 8.73 \times 10^{22}atom Cl$

The molar ratio of ClO₄⁻ to O is 1:1. The number of O in 8.73 × 10²² ions of ClO₄⁻ is:

$8.73 \times 10^{22}ion ClO_4^{-} \times \frac{4 atomO }{1ion ClO_4^{-}} = 3.49 \times 10^{23}atom O$

17.8 g of sodium perchlorate contains 8.73 × 10²² Na⁺ ions, 8.73 × 10²² ClO₄⁻ ions, 8.73 × 10²² Cl atoms and 3.49 × 10²³ O atoms.

You can learn more Avogadro's number here: brainly.com/question/13302703

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2 years ago

(r)-2-butanol reacts with potassium dichromate (k2cro4) in aqueous sulfuric acid to give a (c4h8o). treatment of a with sodium b

weqwewe [10]

Potassium dichromate reacts with sulfuric acid to form chromic acid, H₂CrO₄ which is a very strong oxidizing agent. The secondary alcohol, (R)-2-butanol will be oxidized in the presence of chromic acid, but it can only be as oxidized as far as the ketone, which is the product shown, 2-butanone.

Sodium borohydride is a reducing agent that will reduce a ketone or aldehyde to an alcohol. When sodium borohydride reacts with 2-butanone, it reduces it to 2-butanol. However, the alcohol is no longer chiral as it was in the beginning since the sodium borohydride can add a hydride to either face of the carbonyl, which results in a racemic mixture of alcohols. This explains why the product has the same refractive index and boiling point as (R)-2-butanol, however, the product formed would no longer be optically active.