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2 years ago

You are given 2 samples of different elements. One is a perfect cube and the other is an

Chemistry

1 answer:

Julli [10]2 years ago

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Write the equation of the chemical reaction of burning glucose (C6H12O6)

kotegsom [21]

Answer:

C6H12O6 + 6O2 → 6CO2 + 6H2O.

Explanation:

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2 years ago

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The methylene chloride solution is filtered through sodium sulfate to:

Vlad [161]

Answer:

C. Dry the methylene chloride by removing water

Explanation:

Anhydrous sodium sulfate is known for its high capacity to absorb water, for this reason it is widely used in laboratories as a drying agent.

Sodium sulfate is a neutral molecule so it cannot be used to neutralize and is very stable, so it is difficult to precipitate organic molecules.

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2 years ago

For the procedural error, indicate if the error will affect the actual yield of copper(II) saccharinate product and if it does,

jolli1 [7]

Answer:

Lowers the actual yield

Explanation:

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2 years ago

This is about rocks and/or minerals

vampirchik [111]

I need the same answer

5 0

2 years ago

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The energy of a photon needed to cause ejection of an electron from a photoemissive metal is expressed as the sum of the binding

slega [8]

Answer:

$Binding\ energy=43.43\times 10^{-20}\ J$

Explanation:

Using the expression for the photoelectric effect as:

$E=h\nu_0+\frac {1}{2}\times m\times v^2$

Also, $E=\frac {h\times c}{\lambda}$

$\nu_0=\frac {c}{\lambda_0}$

Applying the equation as:

$\frac {h\times c}{\lambda}=\frac {hc}{\lambda_0}+\frac {1}{2}\times m\times v^2$

Where,

h is Plank's constant having value $6.626\times 10^{-34}\ Js$

c is the speed of light having value $3\times 10^8\ m/s$

$\lambda$ is the wavelength of the light being bombarded

Given, $\lambda=4.00\times 10^{-7}\ m$

$\frac {hc}{\lambda_0}$ is the binding energy or threshold energy

$\frac {1}{2}\times m\times v^2$ is the kinetic energy of the electron emitted. = $6.26\times 10^{-20}\ J$

Thus, applying values as:

$\frac{h\times c}{\lambda}=Binding\ Energy+Kinetic\ Energy$

$\frac{6.626\times 10^{-34}\times 3\times 10^8}{4.00\times 10^{-7}}\ J=Binding\ Energy+6.26\times 10^{-20}\ J$

$\frac{19.878}{10^{19}\times \:4}\ J=Binding\ Energy+6.26\times 10^{-20}\ J$

$49.69\times 10^{-20}\ J=Binding\ Energy+6.26\times 10^{-20}\ J$

$Binding\ energy=43.43\times 10^{-20}\ J$

5 0

2 years ago