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3 years ago

What mass of O2 is needed to completely react with 500 g of Fe

Chemistry

1 answer:

Zina [86]3 years ago

8 0

Answer:

solution given:

Explanation:

4Fe +3O2------------>2Fe2O3

4mole. 3mole. 2 mole

224g. 96g. 320g

we have

224 g of fe needed to react completely with 96 g of O2

500g of fe needed to react completely with

96 ×500/224=214.3g of O2

214.3g of O2 is a required answer.

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Describe how the properties of elements in the same group differ to the properties of elements in the same period

Fofino [41]

Answer:

(a) (i) All the elements of a group have similar chemical properties because they have same no. of valence electrons in their outermost shell. (ii) All the elements of a period have different chemical properties because they have different no. of valence electrons in their atoms.

5 0

3 years ago

Identify the state of matter for A, B, and C.

drek231 [11]

Explanation:

A) particles are close together in random positions with about equal kinetic energy and intermolecular forces.

These points are about liquid state.

B) particles are close together in fixed positions with low kinetic energy

These points satisfy the qualities of Solid state

C)particles are far apart with greater kinetic energy and low intermolecular forces.

The above qualities are for Gaseous state of matter

A) Liquid

B)Solid

C)Gas

7 0

3 years ago

"How much NH4Cl, when present in 2.00 liters of 0.200 M ammonia, will give a solution with pH = 8.20? For NH3, Kb = 1.8 x 10-5"

Andru [333]

Answer:

245.66g of NH₄Cl is the mass we need to add to obtain the desire pH

Explanation:

The mixture of NH3/NH4Cl produce a buffer. We can find the pH of a buffer using H-H equation:

pH = pKa + log [A⁻] / [HA]

Where [A⁻] is the molar concentration of the base, NH₃, and [HA] molar concentration of the acid, NH₄⁺. This molar concentration can be taken as the moles of each chemical

First, we need to find pKa of NH₃ using Kb. Then, the moles of NH₃ and finally replace these values in H-H equation to solve moles of NH₄Cl we need to obtain the desire pH.

pKa NH₃/NH₄⁺

pKb = - log Kb

pKb = -log 1.8x10⁻⁵ = 4.74

pKa = 14 - pKb

pKa = 14 - 4.74

pKa = 9.26

Moles NH₃

2.00L ₓ (0.200mol NH₃ / L) = 0.400 moles NH₃

H-H equation:

pH = pKa + log [NH₃] / [NH₄Cl]

8.20 = 9.26 + log [0.400 moles] / [NH₄Cl]

-1.06 = log [0.400 moles] / [NH₄Cl]

0.0087 = [0.400 moles] / [NH₄Cl]

[NH₄Cl] = 0.400 moles / 0.0087

[NH₄Cl] = 4.59 moles of NH₄Cl we need to add to original solution to obtain a pH of 8.20. In grams (Using molar mass NH₄Cl=53.491g/mol):

4.59 moles NH₄Cl ₓ (53.491g / mol) =

<h3>245.66g of NH₄Cl is the mass we need to add to obtain the desire pH</h3>

3 0

4 years ago

Free-energy change, ΔG∘, is related to cell potential, E∘, by the equationΔG∘=−nFE∘where n is the number of moles of electrons t

mart [117]

Answer:

a) $\Delta G=372490 J$

b) $\Delta G=-568614 J$

Explanation:

a) The reaction:

$Mg(s) +Fe^{2+}(aq) \longrightarrow Mg^{2+}(aq) + Fe(s)$

The free-energy expression:

$\Delta G=-n*F*E$

$E=E_{red}-E_{ox]$

The element wich is reduced is the Fe and the one that oxidates is the Mg:

$-0.44V=E_{red}-(-2.37V)=1.93V$

The electrons transfered (n) in this reaction are 2, so:

$\Delta G=-2mol*96500 C/mol * 1.93 V$

$\Delta G=-372490 J$

b) If you have values of enthalpy and enthropy you can calculate the free-energy by:

$\Delta G=\Delta H - T* \Delta S$

with T in Kelvin

$\Delta G=-675 kJ*\frac{1000J}{kJ} - 298K*-357 J/K$

$\Delta G=-568614 J$

7 0

3 years ago

An egg is placed in a graduated cylinder to determine its volume. The following data are obtained. What is the volume of the egg

Natalka [10]

If egg is dipped in cylinder then the volume of egg will be difference in the volumes before dipping egg (initial volume) and volume after dipping egg (final volume)

Volume of egg= 58.5-50.0 = 8.5 mL

7 0

3 years ago