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3 years ago

The light emitted from a flame is produced when electrons in an excited state

2 answers:

Studentka2010 [4]3 years ago

8 0

The light emitted from a flame is produced when electrons in an excited state releases energy as they move to lower energy States.

sesenic [268]3 years ago

7 0

Electrons in their ground state when excited absorb energy and move to higher energy states.
Absorption is when energy is absorbed by electrons and they move to a higher energy state
Emission is when electrons release that absorbed energy and move to a lower energy state
This release of energy results in emission of light from a flame
The correct answer is
Release energy as they move to lower energy states

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The actual density of iron is 7.874 g/mL. In a laboratory investigation, Jason finds the density of a piece of iron to be 7.921

mixas84 [53]

The percent error associated with Jason’s measurement is 0.596%.

HOW TO CALCULATE PERCENTAGE ERROR:

The percentage error of a measurement can be calculated by following the following process:

Find the difference between the true value and the measured value of a quantity.
Then, divide by the true value and then multiplied by 100

The true value of the density of iron is 7.874 g/mL
Jason observed value is 7.921 g/mL

Difference = 7.921 g/mL - 7.874 g/mL

Difference = 0.047 g/mL

Percentage error = 0.047/7.874 × 100

Percentage error = 0.596%.

Therefore, the percent error associated with Jason’s measurement is 0.596%.

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5 0

2 years ago

Consider the following information. The lattice energy of CsCl is Δ H lattice = − 657 kJ/mol. The enthalpy of sublimation of Cs

AlladinOne [14]

Answer:

Explanation:

8 0

4 years ago

Read 2 more answers

What is the molarity of ZnCl2 that forms when 25.0 g of zinc completely reacts with CuCl2 according to the following reaction? A

yaroslaw [1]

The ZnCl₂ solution have a molarity of 1.33 M.

Explanation:

We have the following chemical reaction:

Zn (s) + CuCl₂ (aq) → ZnCl₂ (aq) + Cu (s)

number of moles = mass / molar weight

number of moles of Zn = 25 / 65.4 = 0.38 moles

From the chemical equation we see that 1 moles of Zn produces 1 mole of ZnCl₂, so 0.38 moles of Zn will produce 0.38 moles of ZnCl₂.

molarity = number of moles / volume (L)

molarity of the ZnCl₂ solution = 0.38 / 0.285 = 1.33 M

Learn more about:

molarity

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7 0

3 years ago

In glycolysis, if glucose is labeled at the carbon 6 position (see page 1 for numbering of carbons in glucose) A) the carbon wit

Oliga [24]

Answer:

D) the carbon with the low-energy phosphate on it in 1,3 BPG is labeled.

Explanation:

Glycolysis has 2 phase (1) preparatory phase (2) pay-off phase.

(1) Preparatory phase

During preparatory phase glucose is converted into fructose-1,6-bisphosphate. Till this time the carbon numbering remains the same i.e. if we will label carbon at 6th position of glucose, its position will remian the same in fructose-1,6-bisphosphate that means the labeled carbon will still remain at 6th position.

When fructose-1,6-bisphosphate is further catalyzed with the help of enzyme aldolase it is cleaved into two 3 carbon intermediates which are glyceraldehyde 3-phosphate (GAP) and dihyroxyacetone phosphate (DHAP). In this conversion, the first three carbons of fructose-1,6-bisphosphate become carbons of DHAP while the last three carbons of fructose-1,6-bisphosphate will become carbons of GAP. It simply means that GAP will acquire the last carbon of fructose-1,6-bisphosphate which is labeled. Now the last carbon of GAP which has phosphate will be labeled.

(2) Pay-off phase

During this phase, GAP is dehydrogenated into 1,3-bisphosphoglycerate (BPG) with the help of enzyme glyceraldehyde 3-phosphate dehydrogenase. This oxidation is coupled to phosphorylation of C1 of GAP and this is the reason why 1,3-bisphosphoglycerate has phosphates at 2 positions i.e. at position 1 in which phosphate is newly added and position 3rd which already had labeled carbon.

It is pertinent to mention here that BPG has a mixed anhydride and the bond at C1 is a very high energy bond. In the next step, this high energy bond is hydrolyzed into a carboxylic acid with the help of enzyme phosphoglycerate kinase and the final product is 3-phosphoglycerate. Hence, the carbon with low energy phosphate i.e. the carbon at 3rd position remains labeled.

3 0

3 years ago

Ammonia is produced commercially by the Haber reaction: N2(8)+ 3H2(8)

Shalnov [3]

This is an incomplete question, here is a complete question.

Ammonia is produced commercially by the Haber reaction:

$N_2(g)+3H_2(g)\rightarrow 2NH_3(g)+Heat$

The formation of ammonia is favored by

A) an increase in pressure

B) a decrease in pressure

C) removal of N₂(g)

D) removal of H₂(g)

Answer : The correct option is, (A) an increase in pressure.

Explanation :

Any change in the equilibrium is studied on the basis of Le-Chatelier's principle.

Le-Chatelier's principle states that if there is any change in the variables of the reaction, the equilibrium will shift in the direction to minimize the effect.

Increase the pressure :

If the pressure in the equilibrium is increased, the equilibrium will shift in the direction where fewer total molecules are shown in the balanced chemical equation.

In the given reaction, there are 4 molecules present on reactant side and 2 molecules on product side. That means, less molecules present on product side. Thus, the reaction will shift in right direction that is towards the product.

Removal the reactant molecule :

If any of reactant molecule in the equilibrium is removed, then the equilibrium will shift in left direction that is towards the reactant.

Hnece, the formation of ammonia is favored by an increase in pressure.

5 0

4 years ago