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1 year ago

How do I draw energy level diagrams?

1 answer:

7 0

Energy levels inside an atom are the specific energies that electrons can have when energy occupies specific orbitals. Electrons can be excited to higher energy levels by absorbing energy from the surroundings, an equivalent light is emitted when an electron returns from a high energy state to a lower one. Representation of this diagrammatic is known as the energy level diagram.

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According to Dalton, if 2 grams of element X combine with 4 grams of element Y to form compound XY, how many grams of element X

adell [148]

The ratio of reactant in grams would be influenced by the reaction equation and the molar mass of the elements. But the ratio of a reactant needed in grams would be same for a reaction.
If 2 grams of element X combine with 4 grams of element Y to form compound XY, it means the ratio of reactant in grams would be 2:4 or 1:2

The number of element X needed for 28grams of element Y would be:
1/2 * 28grams = 14 grams.

8 0

2 years ago

Which element is rolled into a foil and used in your kitchen?

Doss [256]

Aluminum is the correct answer

8 0

2 years ago

Using standard heats of formation, calculate the standard enthalpy change for the following reaction. 2NH3(g) 3N2O(g)4N2(g) 3H2O

trasher [3.6K]

Answer: $\Delta H^{0} = -879.15 kJ/mol$

Explanation: Heats of formation is the amount of heat necessary to create 1 mol of a compound from its molecular constituents. The basic conditions the substance is formed is at standard conditions: 1 atm and 25°C. Each compound has its own heat of formation per mol of compound (kJ/mol), but to an element is assigned a value of zero.

Standard Enthalpy Change is defined as the heat absorbed or released when a reaction takes place. It can be positive or negative, which means reaction is endothermic or exothermic, respectively.

Enthalpy change is calculated as the difference between the sum of heat formation of products and the sum of heat formation of the reactants:

$\Delta H^{0}=\Sigma H^{0}_{f}_{(products)}-\Sigma H^{0}_{f}_{(reactants)}$

For the reaction

2NH₃ + 3N₂O → 4N₂ + 3H₂O

2(-46.2) + 3(82.05) 4(0) + 3(-241.8)

$\Delta H^{0}=3(-241.8)-[ 2(-46.2)+3(82.05)]$

$\Delta H^{0}=-725.4-153.75$

$\Delta H^{0}=-879.15$

The standard enthalpy change for the reaction is $\Delta H^{0}=-879.15$ kJ

8 0

2 years ago

What is the molarity of a solution that has 2.50 moles of NaOH dissolved in 0.500 L of solution?

const2013 [10]

Answer: 5 is the molarity

Explanation:

The molarity formula is moles over liters and that in your case is 2.50 moles divided by .500 L which results in 5 which is your answear hope this helped god bless

5 0

3 years ago

Read 2 more answers

If 20.0 g of NaOH is added to 0.750 L of 1.00 M Cd(NO₃)₂, how many grams of Cd(OH)₂ will be formed in the following precipitatio

bulgar [2K]

Answer:

$m_{Cd(OH)_2}=36.6 gCd(OH)_2$

Explanation:

Hello.

In this case, for the given chemical reaction, in order to compute the grams of cadmium hydroxide that would be yielded, we must first identify the limiting reactant by computing the yielded moles of that same product, by 20.0 grams of NaOH (molar mass = 40 g/mol) and by 0.750 L of the 1.00-M solution of cadmium nitrate as shown below considering the 1:2:1 mole ratios respectively:

$n_{Cd(OH)_2}^{by\ NaOH}=20.0gNaOH*\frac{1molNaOH}{40gNaOH} *\frac{1molCd(OH)_2}{2molNaOH} =0.25molCd(OH)_2\\\\n_{Cd(OH)_2}^{by\ Cd(NO_3)_2}=0.750L*1.00\frac{molCd(NO_3)_2}{L}*\frac{1molCd(OH)_2}{1molCd(NO_3)_2} =0.75molCd(OH)_2$

Thus, since 20.0 grams of NaOH yielded less of moles of cadmium hydroxide, NaOH is the limiting reactant, therefore the mass of cadmium hydroxide (molar mass = 146.4 g/mol) is:

$m_{Cd(OH)_2}=0.25molCd(OH)_2*\frac{146.4gCd(OH)_2}{1molCd(OH)_2} \\\\m_{Cd(OH)_2}=36.6 gCd(OH)_2$

Best regards.

4 0

2 years ago