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3 years ago

50 POINTS! ANSWER FAST!

2 answers:

7 0

Hello there! The calculations above are correct, however, the answer is: C. The electron moved up to an energy level and has an energy of 10.92*10^-19 J

Korvikt [17]3 years ago

6 0

Given:

The initial energy of the electron Einitial = 16.32 * 10⁻¹⁹ J

The energy released i.e the change in energy ΔE = 5.4 * 10⁻¹⁹ J

To determine:

The final energy state Efinal of the electron

Explanation:

Since energy is being released, this suggests that Efinal < Einitial

i.e. ΔE = Einitial - Efinal

Efinal = Einitial - ΔE = (16.32 - 5.4)*10⁻¹⁹ = 10.92 * 10⁻¹⁹ J

Ans: A)

The electron moved down to an energy level and has an energy of 10.92 * 10⁻¹⁹ J

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What gas does photosynthesis produce

denis23 [38]

The answer would be Oxygen. Hope this helps!

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3 years ago

Read 2 more answers

A student needed to make a 3 g/L NaCl solution. The student weighed 3 g of NaCl in a beaker and measured 1 L of water in a 1L vo

hoa [83]

Answer:

The answer is in the explanation.

Explanation:

A solution is defined as the homogeneous mixture of a solute (In this case, NaCl) and the solvent (water).

To prepare 1L of the solution, the student can weigh the 3g of NaCl in the volumetric flask but need to add slowly water to dissolve the NaCl (That is very soluble in water). When all NaCl is dissolved the student must transfer the solution to the 1L volumetric flask. Then, you must add more water to the beaker until "Clean" all the solute of the beaker to transfer it completely to the volumetric flask.

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2 years ago

Photosynthesis is a cell process used by various organisms. Which of the following best

Dmitry [639]

Answer:

Carbon dioxide and water are turned into sugar and oxygen

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2 years ago

How many moles of aluminum oxide (Al2O3) can be produced from 12.8 moles of oxygen gas (02)

zhannawk [14.2K]

Answer:

Theoretical Yield

Percent yield

Example stoichiometry problem

How much oxygen can be prepared from 12.25 g KClO3 . (Use molar mass KClO3 = 122.5 g.)

Most stoichiometry problems can be solved using the following steps.

Step 1.

Write and balance the equation for the decomposition of KClO3 with heat (∆). 2KClO3 + ∆ → 2KCl + 3O2

Step 2.

Convert what you have (in this case g KClO3) to moles.

# moles = grams/molar mass = 12.25 g /122.5 = 0.100 mole KClO3.

Step 3.

Using the coefficients in the balanced equation, convert moles of what you have (moles KClO3) to moles of what you want (in this case moles oxygen).

0.100 mol KClO3 x (3 moles O2/2 moles KClO3) = 0.100 x (3/2) = 0.150 mole O2.

Step 4.

Convert moles from step 3 to grams.

moles x molar mass = grams

0.150 mole O2 x (32.0 g O2/mole O2) = 4.80 g O2 produced from 12.25 g KClO3. This is the theoretical yield. If the ACTUAL yield is 4.20 grams, calculate percent yield. Percent yield = (actual yield/theoretical yield) x 100 = (4.20/4.80) x 100 = 87.5% yield

NOTE: In step 1, moles can be obtained other ways; in step 4 moles can be converted to other units.

a. For solutions, M x L = moles (or mL x M = millimoles).

b. For gases, L/22.4 = moles

4 0

3 years ago

Suppose a student started with 142.0 mg of trans-cinnamic acid, 412 mg of pyridinium tribromide, and 2.30 mL of glacial acetic a

nirvana33 [79]

Answer: Theoretical Yield = 0.2952 g

Percentage Yield = 75.3%

Explanation:

Calculation of limiting reactant:

n-trans-cinnamic acid moles = (142mg/1000) / 148.16 = 9.584*10⁻⁴ mol

pyridium tribromide moles = (412mg/1000) / 319.82= 1.288*10⁻³ mol

n-trans-cinnamic acid is the limiting reactant

The molar ratio according to the equation mentioned is equals to 1:1

The brominated product moles is also = 9.584*10⁻⁴ mol

Theoretical yield = (9.584*10⁻⁴ mol) * (Mr of brominated product)

= (9.584*10⁻⁴ mol) * (307.97) = 0.2952 g

Percentage Yield is : Actual Yield / Theoretical Yield = 0.2223/0.2952

= 75.3%

4 0

3 years ago