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SVETLANKA909090 [29]
3 years ago
12

Some fireplace logs (commercially made) burn with a red and/or green flame. Using the information in this experiment, what eleme

nts could be responsible for these colored flames?
Chemistry
1 answer:
Diano4ka-milaya [45]3 years ago
5 0

Answer:

Because each element has an exactly defined line emission spectrum, scientists are able to identify them by the color of flame they produce. For example, copper produces a blue flame, lithium, and strontium a red flame, calcium an orange flame, sodium a yellow flame, and barium a green flame. When you heat an atom, some of its electrons are "excited* to higher energy levels. When an electron drops from one level to a lower energy level, it emits a quantum of energy. ... The different mix of energy differences for each atom produces different colors. Each metal gives a characteristic flame emission spectrum

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For an exit heroic reaction what effect will increasing the temperature have on the equilibrium?
tankabanditka [31]
We are on fire trust the process
3 0
3 years ago
Are the following statements true or false? (a) Formal charges represent an actual separation of charges. true false (b) ΔH o rx
katrin [286]

Answer:

Explanation:

A) Formal charges represent an actual separation of charges.(FALSE)

(B) ΔHo rxn can be estimated from the bond enthalpies of reactants and products.(TRUE)

C)All second-period elements obey the octet rule in their compounds(FALSE).

(D)The resonance structures of a molecule can be separated from one another in the laboratory.(FALSE)

Bond enthalpy which is also reffered to as bond energy is the amount of energy that is required to break one mole of a bond.

taking the single bond between Oxygen and Hydrogen into considerationthe bond energy between their single bond is 463 kJ/mol.

formal charge is used for the comparison of the number of electrons present around an atom in a particular molecule with the number of electrons present around a neutral

3 0
3 years ago
How to solve for K when given your anode and cathode equations and voltage
Aleks04 [339]

Answer:

See Explanation

Explanation:

In thermodynamics theory the Free Energy (ΔG) of a chemical system is described by the expression ΔG = ΔG° + RTlnQ. When chemical system is at equilibrium ΔG = 0. Substituting into the system expression gives ...

0 = ΔG° + RTlnKc, which rearranges to ΔG° = - RTlnKc.  ΔG° in electrochemical terms gives ΔG° = - nFE°, where n = charge transfer, F = Faraday Constant = 96,500 amp·sec and E° = Standard Reduction Potential of the electrochemical system of interest.

Substituting into the ΔG° expression above gives

-nFE°(cell) = -RTlnKc => E°(cell) = (-RT/-nF)lnKc = (2.303·R·T/n·F)logKc

=> E°(cell) = (0.0592/n)logKc = E°(Reduction) - E°(Oxidation)

Application example:

Calculate the Kc value for a Zinc/Copper electrochemical cell.

Zn° => Zn⁺² + 2e⁻  ;    E°(Zn) = -0.76 volt  

Cu° => Cu⁺² + 2e⁻ ;    E°(Cu) =  0.34 volt

By natural process, charge transfer occurs from the more negative reduction potential to the more positive reduction potential.

That is,

           Zn° => Zn⁺² + 2e⁻ (Oxidation Rxn)

Cu⁺² + 2e⁻ => Cu°             (Reduction Rxn)

E°(Zn/Cu) = (0.0592/n)logKc

= (0.0592/2)logKc = E°(Cu) - E°(Zn) = 0.34v - (-0.76v) = 1.10v

=> logKc = 2(1.10)/0.0592 = 37.2

=> Kc = 10³⁷°² = 1.45 x 10³⁷

3 0
3 years ago
Block A has a mass of 5g and a volume of 15mL. What is the density of block A? Will it float in water (density of water is 1.0 g
iris [78.8K]

Explanation:

The density of a substance can be found by using the formula

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

From the question

mass of block = 5 g

volume = 15 mL

The density of the block is

density =  \frac{5}{15}  =  \frac{1}{3}  \\  = 0.33333333...

We have the final answer as

<h3>0.33 g/mL</h3>

The block will float on water since it's density is less than that of water which is 1 g/mL

Hope this helps you

8 0
4 years ago
There are 1.2 x 1024 formula units
____ [38]

mol CaCl₂ = 1.2 x 10²⁴ : 6.02 x 10²³ = 1.993

mass = 1.993 x 111 g/mol = 221.223 g

4 0
2 years ago
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