All categories

3 years ago

What is the total number of orbitals found in the second energy level?

Chemistry

1 answer:

erik [133]3 years ago

6 0

Principal Energy Light Type of sub level Maximum number of electrons
P 8
2
S 18
3
P
Hope It really Helps

You might be interested in

Will mark as Brainliest.

LUCKY_DIMON [66]

Gas i took the test plz mark brainliest!

8 0

4 years ago

Read 2 more answers

I need help badly what is the individual element that make up HCI

maxonik [38]

Answer:

It is two elements, Hydrogen and Chlorine

Explanation:

Hydrochloric Acid is made with 1 hydrogen atom and 1 chlorine atom bonded with a single covalent bond

7 0

3 years ago

Read 2 more answers

What is a rise in seawater levels pushed to shore during a hurricane called?

zalisa [80]

Answer:

A storm surge

Explanation:

A rise in sea level that happens during tropical cyclones, and storms like typhoons or hurricanes is called a Storm Surge.

5 0

3 years ago

What is the total energy change for the following reaction:CO+H2O-CO2+H2

Alekssandra [29.7K]

Answer:

$\large \boxed{\text{-41.2 kJ/mol}}$

Explanation:

Balanced equation: CO(g) + H₂O(g) ⟶ CO₂(g) + H₂(g)

We can calculate the enthalpy change of a reaction by using the enthalpies of formation of reactants and products

$\Delta_{\text{rxn}}H^{\circ} = \sum \left( \Delta_{\text{f}} H^{\circ} \text{products}\right) - \sum \left (\Delta_{\text{f}}H^{\circ} \text{reactants} \right)$

(a) Enthalpies of formation of reactants and products

$\begin{array}{cc}\textbf{Substance} & \textbf{$\Delta_{\text{f}}$H/(kJ/mol}) \\\text{CO(g)} & -110.5 \\\text{H$_{2}$O} & -241.8\\\text{CO$_{2}$(g)} & -393.5 \\\text{H$_{2}$(g)} & 0 \\\end{array}$

(b) Total enthalpies of reactants and products

$\begin{array}{ccr}\textbf{Substance} & \textbf{Contribution)/(kJ/mol})&\textbf{Sum} \\\text{CO(g)} & -110.5& -110.5 \\\text{H$_{2}$O(g)} &-241.8& -241.8\\\textbf{Total}&\textbf{for reactants} &\mathbf{ -352.3}\\&&\\\text{CO}_{2}(g) & -393.5&-393.5 \\\text{H}_{2} & 0 & 0\\\textbf{Total}&\textbf{for products} & \mathbf{-393.5}\end{array}$

(c) Enthalpy of reaction $\Delta_{\text{rxn}}H^{\circ} = \sum \left( \Delta_{\text{f}} H^{\circ} \text{products}\right) - \sum \left (\Delta_{\text{f}}H^{\circ} \text{reactants} \right)= \text{-393.5 kJ/mol - (-352.3 kJ/mol}\\= \text{-393.5 kJ/mol + 352.3 kJ/mol} = \textbf{-41.2 kJ/mol}\\ \text{The total enthalpy change is $\large \boxed{\textbf{-41.2 kJ/mol}}$}$

4 0

3 years ago

Calculate the quotient [co32–]/[hco3–] at ph 10.75.

KATRIN_1 [288]

The chemical equation is:

HCO3^ -> H^+ + CO3^2-

We know that the formula for Ka is:
Ka = [H^+][CO3^2-]/[HCO3^-]

log Ka = log[H^+] + log[CO3^2-]/[HCO3^-]
pKa = pH - log[CO3^2-]/[HCO3^-]
log[CO3^2-]/[HCO3^-] = pH - pKa = 10.75 - 10.329 = 0.421
[CO3^2-]/[HCO3^-] = Antilog (0.421) = 2.636

Answer:

2.636

3 0

3 years ago