Identify indicators of a chemical reaction. Check all of the boxes that apply.

Where else can you observe colorful light emisiions? are these emission applications related?

aksik [14]

Explanation:

Some examples of colorful light emissions are street lights, neon signs and, of course, fireworks. Neon signs emit light when an electric current passes through the neon gas.

4 0

3 years ago

What is the outcome of the experoment?

Vikki [24]

The outcome of a experiment is the result

3 0

3 years ago

Based on your energy level calculations, predict the photon energies that would be emitted by transitions between the hydrogen e

mote1985 [20]

Answer:

The formula defining the photon energies that would be emitted by transitions between the hydrogen energy levels are given by the equation:

ΔE = -13.6Z²[1/n₁² - 1/n₂²]eV, where (1 eV = 1.602×10-19 Joules) and n₁ = initial energy level and n₂ = final energy level.

Z = Atomic number of the atom

8 0

3 years ago

Potassium thiocyanate, KSCN, is often used to detect the presence of Fe3+ ions in solution through the formation of the red Fe(H

Natalija [7]

Answer:

Ferric ions left in the solution at equilibrium is $8.0\times 10^{-6} M$ .

Explanation:

Moles of ferric nitrate in 0.700 L = n

Volume of the solution = V = 0.700 L

Molarity of ferric nitrate = M = 0.00150 M

$n=M\times V=0.00150 M\times 0.700 L=0.00105 mol$

Moles of potassium thiocyanate in 0.700 L = n'

Volume of the potassium thiocyanate solution = V' = 0.700 L

Molarity of potassium thiocyanate = M' = 0.200 M

$n'=M'\times V'=0.200 M\times 0.700 L=0.140 mol$

Molarity of ferric ions after mixing :

1 mol of ferric nitrate gives 1 mol of ferric ions.Then 0.00105 mol ferric nitrate will :

Moles of ferric ions = 0.00105 mol

$M_1=\frac{0.00105 mol}{0.700 L+0.700 L}=0.00075 M$

Molarity of thiocyanate ions after mixing :

1 mol of potassium thiocyanate gives 1 mol of thiocyanate ions.Then 0.140 mol potassium thiocyanate will give:

Moles of thiocyanate ions = 0.140 mol

$M_2=\frac{0.140 mol}{0.700 L+0.700 L}=0.1 M$

Complex equation:

$Fe^{3+}+SCN^-\rightleftharpoons [Fe(SCN)]^{2+}$

0.00075 M 0.1 M 0

At equilibrium:

(0.00075 M -x) (0.1 M-x) x

The formation constant of the given complex = $K_f=8.9\times 10^2$

$K_f=\frac{[[Fe(SCN)]^{2+}]}{[[Fe^{3+}]][SCN^{-}]}$

$8.9\times 10^2=\frac{x}{(0.00075 M -x)\times (0.1 M-x)}$

Solving for x:

x = 0.000742 M

Ferric ions left in the solution at equilibrium :

= (0.00075 M -x) = (0.00075 M - 0.000742 M)= $8.0\times 10^{-6} M$

5 0

3 years ago

PLZ HELP ASAP WILL

amid [387]

Answer:

A) Mass = 32 g of KCl

Explanation:

Given data:

Mass of potassium chloride produced = ?

Mass of potassium chlorate = 52 g

Solution:

Chemical equation:

2KClO₃ → 2KCl + 3O₂

Number of moles of KClO₃:

Number of moles = mass/molar mass

Number of moles = 52 g/ 122.55 g/mol

Number of moles = 0.424 mol

Now we will compare the moles of KClO₃ and KCl

KClO₃ : KCl

2 : 2

0.424 : 0.424

Mass of KCl:

Mass = number of moles × molar mass

Mass = 0.424 mol × 74.55 g/mol

Mass = 32 g

4 0

3 years ago