Is electrons are shared ionic compounds or molecular compounds

What does a mole of iron weigh

r-ruslan [8.4K]

As the atomic mass of iron is 55.847u

We say that it is the mass of one mole of iron.

By formula it can be find by

No.of mole=mass in g/molar mass

Mass in gram = no.of mole x molar mass

No.of mole =1

Molar mass = 55.847g

Mass in gram = 1x55.847

= 55.847g of Fe

3 0

4 years ago

While hiking in Costa Rica, Bert discovers a new species of arthropod. Which characteristics should he look at to determine in w

alex41 [277]

Answer:

Characteristics outlined below

Explanation:

Anthropods, though they generally possess exoskeleton, segmented bodies and joint appendages, are divided into four (4) groups:

<em>Tracheata</em> (breath through channels in their bodies. E.g. insects)
Chelicerata (More than four legs, one pir of chelicerae i.e. claw or pincer. E.g. Scorpions and spiders)
<em>Crustaceans</em> (Have a large rigid exoskeletal, without a circulatory system. E.g. Shrimps, crabs)
<em>Trilobite</em> (extinct three-headed/lobed marine species)

6 0

3 years ago

Obtain a box of breakfast cereal and read the list of ingredients. What are four chemicals from the list? a. monoglycerides b. c

GREYUIT [131]

Answer:

B. cocamide DEA

C. folic acid

D. iron

G. lauryl glucoside

5 0

3 years ago

C. Use Hess's law and the following equations to calculate ΔH for the reaction 4NH3 (g) + 5O2 (g) 4NO(g) + 6H2 O(g). Show your w

Monica [59]

Considering the Hess's Law, the enthalpy change for the reaction is -906.4 kJ/mol.

Hess's Law indicates that the enthalpy change in a chemical reaction will be the same whether it occurs in a single stage or in several stages. That is, the sum of the ∆H of each stage of the reaction will give us a value equal to the ∆H of the reaction when it occurs in a single stage.

In this case you want to calculate the enthalpy change of:

4 NH₃ + 5 O₂ → 4 NO + 6 H₂O

which occurs in three stages.

You know the following reactions, with their corresponding enthalpies:

Equation 1: 2 N₂ + 6 H₂ → 4 NH₃ ΔH = –183.6 kJ/mol

Equation 2: 2 N₂ + 2 O₂ → 4 NO ΔH = 361.1 kJ/mol

Equation 3: 2 H₂ + O₂→ 2 H₂O ΔH = -483.7 kJ/mol

Because of the way formation reactions are defined, any chemical reaction can be written as a combination of formation reactions, some going forward and some going back.

In this case, first, to obtain the enthalpy of the desired chemical reaction you need 4 moles of NH₃ on reactant side and it is present in first equation on product side. So you need to invert the reaction, and when an equation is inverted, the sign of delta H also changes.

Now, 4 moles of NO must be a product and is present in the second equation, so let's write this as such.

Finally, you need 6 moles of H₂O on the product side, so you need to multiply by 3 the third equation to obtain the amount of water that you need. Since enthalpy is an extensive property, that is, it depends on the amount of matter present, since the equation is multiply by 3, the variation of enthalpy also.

In summary, you know that three equations with their corresponding enthalpies are:

Equation 1: 2 4 NH₃ → N₂ + 6 H₂ ΔH = 183.6 kJ/mol

Equation 2: 2 N₂ + 2 O₂ → 4 NO ΔH = 361.1 kJ/mol

Equation 3: 6 H₂ + 3 O₂→ 6 H₂O ΔH = -1,451.1 kJ/mol

Adding or canceling the reactants and products as appropriate, and adding the enthalpies algebraically, you obtain:

4 NH₃ + 5 O₂ → 4 NO + 6 H₂O ΔH= -906.4 kJ/mol

Finally, the enthalpy change for the reaction is -906.4 kJ/mol.

Learn more about Hess's law:

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