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gtnhenbr [62]
3 years ago
13

Chris the chemist was working in the and he wanted to sou what would happen he put copper chloride into a Bunsen flameAs soon as

it the , began 10 amit a green color Which of these BEST describes this outcome
Chemistry
1 answer:
Reptile [31]3 years ago
4 0

The question is incomplete, the complete question is;

Chris the Chemist was working in the lab, and he wanted to see what would happen if he put copper chloride into a Bunsen burner flame. As soon as he put it in the flame, it began to emit a green color. Which of these BEST describes this outcome?

A) The heat split the nucleus and created a new element.

B) As the copper chloride was heated, electrons were turned into neutrons.

C) As the electrons moved to a higher energy level, a photon of light is emitted.

D) When excited electrons return back to the ground state, a photon of light is emitted.

Answer:

D) When excited electrons return back to the ground state, a photon of light is emitted.

Explanation:

According to Bohr's model of the atom, electrons may absorb energy and move up to higher energy levels.

These electrons quickly return to ground state from such higher energy excited states thereby emitting the excess energy absorbed during excitation.

If the higher energy level is E2 and the ground state is E1 then the energy of the photon emitted when the electron descends from energy level E2 to E1 is;

ΔE = E2 - E1

This photon emitted accounts for the green colour of the copper salt observed when it was heated.

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If the mass of a block of wood is 4.6g and the density is 0.98g/mL what is the volume.
irina [24]

Answer:

The answer is

<h2>4.69 mL</h2>

Explanation:

The volume of a substance when given the density and mass can be found by using the formula

volume =  \frac{mass}{density}  \\

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mass of wood = 4.6 g

density = 0.98 g/mL

It's volume is

volume =  \frac{4.6}{0.98}  \\  = 4.69387755...

We have the final answer as

<h3>4.69 mL</h3>

Hope this helps you

3 0
4 years ago
Calculate the fractional saturation for hemoglobin when the partial pressure of oxygen is 40 mm Hg. Assume hemoglobin is 50%% sa
kumpel [21]

Answer:

The fractional saturation for hemoglobin is 0.86

Explanation:

The fractional saturation for hemoglobin can be calculated using the formula

Y_{O_{2} } = \frac{(P_{O_{2} })^{h}  } {(P_{50})^{h}  + (P_{O_{2} })^{h}   }

Where Y_{O_{2} } \\ is the fractional oxygen saturation

{P_{O_{2} } is the partial pressure of oxygen

P_{50} is the partial pressure when 50% hemoglobin is saturated with oxygen

and h is the Hill coefficient

From the question,

{P_{O_{2} } = 40 mm Hg

P_{50} = 22 mm Hg

h = 3

Putting these values into the equation, we get

Y_{O_{2} } = \frac{(P_{O_{2} })^{h}  } {(P_{50})^{h}  + (P_{O_{2} })^{h}   }

Y_{O_{2} } = \frac{40^{3} }{22^{3} + 40^{3}  }

Y_{O_{2} } = \frac{64000 }{10648 + 64000  }

Y_{O_{2} } = \frac{64000 }{74648 }

Y_{O_{2} } = 0.86

Hence, the fractional saturation for hemoglobin is 0.86.

4 0
3 years ago
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