Any part of an experiment that cannot be changed is considered to be a(n)

Ionization, also known as ionization, would be the phenomenon by which an atom or molecule gains or loses electrons to acquire a negative and positive charge, frequently in conjunction with those other chemical changes. Ions are the name for the electrically charged molecular or atom that results.

These chemicals are known as ionic compounds, with table salt serving as an example. Ionization would be the process of removing electrons from such an element as well as a molecule as well as the dissociation of an ionic material, such as salt, itself into the individual ions in a solution like water.

The energy of hydrogen atom will be calculated by using the formula:

$E_{n} = -13.6 ev / n^{2}$

where, $E_{n}$ = energy of nth level, n is the state.

It is given that, n = 3 .

Now put the value of n in above equation.

$E_{4} = -13.6 ev / 4^{2}\\E_{4} = -13.6 ev / 16\\E_{4} = -0.85 ev$

Therefore, the energy of the hydrogen atom will be -0.85 ev.

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4 0

1 year ago

1. Which kingdom is made up of only autotrophs?

Reika [66]

Answer: I believe it's C

Hope this helped<3

Can you please make my answer brainly

3 0

3 years ago

50.00 mL of 0.10 M HNO 2 (nitrous acid, K a = 4.5 × 10 −4) is titrated with a 0.10 M KOH solution. After 25.00 mL of the KOH sol

boyakko [2]

Answer:

b. 3.35

Explanation:

To calculate the pH of a solution containing both acid and its salt (produced as a result of titration) we need to use Henderson’s equation i.e.

pH = pKa + log ([salt]/[acid]) (Eq. 01)

Where

pKa = -log(Ka) (Eq. 02)

[salt] = Molar concentration of salt produced as a result of titration

[acid] = Molar concentration of acid left in the solution after titration

Let’s now calculate the molar concentration of HNO2 and KOH considering following chemical reaction:

HNO2 + KOH ⇆ H2O + KNO2 (Eq. 03)

This shows that 01 mole of HNO2 and 01 mole of KOH are required to produce 01 mole of KNO2 (salt). And if any one of them (HNO2 and KOH) is present in lower amount then that will be considered the limiting reactant and amount of salt produced will be in accordance to that reactant.

Moles of HNO2 in 50 mL of 0.01 M HNO2 solution = 50/1000x0.01 = 0.005 Moles

Moles of KOH in 25 mL of 0.01 M KOH solution = 25/1000x0.01 = 0.0025 Moles

As it can be seen that we have 0.0025 Moles of KOH therefore considering Eq. 03 we can see that 0.0025 Moles of KOH will react with only 0.0025 Moles of HNO2 and will produce 0.0025 Moles of KNO2.

Therefore

Amount of salt produced i.e [salt] = 0.0025 moles (Eq. 04)

Amount of acid left in the solution [acid] = 0.005 - 0.0025 = 0.0025 moles (Eq.05)

Putting the values in (Eq. 01) from (Eq.02), (Eq. 04) and (Eq. 05) we will get the following expression:

pH= -log(4.5x10 -4) + log (0.0025/0.0025)

Solving above we get

pH = 3.35

5 0

3 years ago