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erik [133]
3 years ago
10

Alguien sabe cómo se recicla el sodio?

Chemistry
1 answer:
Liono4ka [1.6K]3 years ago
8 0

Answer:

No se lo siento

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Which of the following is true of atoms with very low electronegativity
Katarina [22]

Answer:

well there is no chose here is my best option i would chose

1 Electronegativity is the ability of an atom to attract electrons.

Explanation:

5 0
3 years ago
A voltaic cell is constructed with two silver-silver chloride electrodes, where the half-reaction is
NemiM [27]
The silver chloride electrode usually functions as a redox electrode where the equilibrium is achieved between silver and its salt (silver chloride).
The half reaction for this electrode is as follow:
<span>AgCl(s)+e−→Ag(s)+Cl−(aq) where:
</span>(s) refers to solid state
(aq) refers to the aqueous state and 
e- is the electron
5 0
3 years ago
1. Calculate the attractive bonding force for NaF and for MgO when the ions just touch. Based on the attractive bonding force, w
Andreas93 [3]

Answer:

The attractive force is negative and MgO has a higher melting point

Explanation:

From Couloumb's law:

Energy of interaction, E = k \frac{q1q2}{r}

where q1 and q2 are the charges of the ions, k is Coulomb's constant and r is the distance between both ions, i.e the atomic radii of the ions.

If you look at Coulomb's law, you note that in the force is negative (because q1 is negative while q2 is positive).

In addition to that, the compounds MgO and NaF have similar combined ionic radii, then we can determine the melting point trend from the amount of energy gotten

The melting point of ionic compounds is determined by 1. charge on the ions 2. size of ions. while NaF has smaller charges (+1 and -1), MgO (+2 and -2) has larger charges and greater combined atomic radii. This implies that the compound with greater force would have a higher melting point.

Hence the compound MgO would have a higher melting point than NaF.

5 0
3 years ago
What's the empirical and molecular weight of C14H22N4O8? What's the molecular formula of C14H22N4O8?
Inessa05 [86]

374u

187u

C₁₄H₂₂N₄O₈

Explanation:

To find the molecular weight of the compound C₁₄H₂₂N₄O₈ we simply sum that atomic masses of the given elements in the compound.

 The empirical weight is determined by using the simplest ratio of the elements involved in the compound;

Molecular weight of C₁₄H₂₂N₄O₈;

atomic mass of C = 12g/mol

                           H = 1g/mol

                            N = 14g/mol

                            O = 16g/mol

 Molecular weight = 14(12) + 22(1) + 4(14) + 8(16)

                                = 168 + 22 + 56 + 128

                                 = 374u

Empirical weight:

  Empirical formula:

                        C₁₄    H₂₂      N₄     O₈

                         14  :    22 :    4  :     8

   divide by 2:

                          7   :    11    :    2  :    4

     empirical formula  C₇H₁₁N₂O₄

     empirical weight = \frac{molecular weight}{2} = \frac{374}{2} = 187u

The molecular formula is the actual combination of atoms in a compound. so the molecular formula of the compound is C₁₄H₂₂N₄O₈

learn more:

Molecular mass brainly.com/question/5546238

#learnwithBrainly

8 0
3 years ago
Calculate the standard entropy change for the following reactions at 25°C.
Art [367]

Answer:

(a) ΔSº = 216.10 J/K

(b) ΔSº = - 56.4 J/K

(c) ΔSº = 273.8 J/K

Explanation:

We know the standard entropy change for a given reaction is given by the sum of the entropies of the products minus the entropies of reactants.

First we need to find in an appropiate reference table the standard  molar entropies entropies, and then do the calculations.

(a)        C2H5OH(l)          +        3 O2(g)         ⇒        2 CO2(g)     +    3 H2O(g)

Sº            159.9                          205.2                         213.8                  188.8

(J/Kmol)

ΔSº = [ 2(213.8) + 3(188.8) ]   - [ 159.9  + 3(205.) ]  J/K

ΔSº = 216.10 J/K

(b)         CS2(l)               +         3 O2(g)               ⇒      CO2(g)      +      2 SO2(g)

Sº          151.0                              205.2                         213.8                 248.2

(J/Kmol)

ΔSº  = [ 213.8 + 2(248.2) ] - [ 151.0 + 3(205.2) ] J/K = - 56.4 J/K

(c)        2 C6H6(l)           +        15 O2(g)                     12 CO2(g)     +     6 H2O(g)

Sº           173.3                           205.2                           213.8                    188.8

(J/Kmol)  

ΔSº  = [ 12(213.8) + 6(188.8) ] - [ 2(173.3) + 15( 205.2) ] = 273.8 J/K

Whenever possible we should always verify if our answer makes sense. Note that the signs for the entropy change agree with the change in mol gas. For example in reaction (b) we are going from 4  total mol gas reactants to 3, so the entropy change will be negative.

Note we need to multiply the entropies of each substance by  its coefficient in the balanced chemical equation.

5 0
3 years ago
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