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Lelu [443]
3 years ago
6

One plus one hundred and thirty five

Chemistry
1 answer:
Makovka662 [10]3 years ago
5 0

Answer:

Explanation:

136

thxx

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Write a balanced chemical equation for the standard formation reaction of solid lead(II) nitrate PbNO32.
Troyanec [42]

Answer:

Pb(s) + N2(g) + 3O2(g) --> Pb(NO3)2

Explanation:

3 0
3 years ago
HELP FAST ITS MULTIPLE CHOICE WILL MARK BRAINLIEST!!
Alina [70]

Answer: I believe its the last one SPHERES.

Explanation:

5 0
3 years ago
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Which metals may be oxidized by H+ under standard-state conditions? Ag+(aq) + e– → Ag(s) E° = 0.80 V Cu2+(aq) + 2e– → Cu(s) E° =
Debora [2.8K]
Answer is: tin and zinc, because they standard potential as less than zero.
Tin and zinc are oxidized to tin and zinc cations (with +2 charge) and hydrogen anions are reduced to hydrogen molecules with neutral charge.
Zn → Zn²⁺ + 2e⁻; 2H⁺ + 2e⁻ → H₂.
<span>Oxidation is increase of oxidation number  and reduction is decrease of oxidation number.</span>
8 0
3 years ago
How many atoms are in 0.750 moles of zinc?
Ira Lisetskai [31]
1 mole Zn ---------- 6.02 x 10²³ atoms
0.750 moles Zn ----- ? 

atoms = 0.750 * ( 6.02 x 10²³ / 1 )

= 4.515 x 10²³ atoms

hope this helps!


4 0
4 years ago
A gaseous system undergoes a change in temperature and volume. What is the entropy change for a particle in this system if the f
deff fn [24]

Explanation:

Entropy means the amount of randomness present within the molecules of the body of a substance.

Relation between entropy and microstate is as follows.

           S = K_{b} \times ln \Omega

where,      S = entropy

             K_{b} = Boltzmann constant

             \Omega = number of microstates

This equation only holds good when the system is neither losing or gaining energy. And, in the given situation we assume that the system is neither gaining or losing energy.

Also, let us assume that \Omega = 1, and \Omega' = 0.833

Therefore, change in entropy will be calculated as follows.

     \Delta S = K_{b} \times ln \Omega' - K_{b} \times ln \Omega

                 = 1.38 \times 10^{-23} \times ln(0.833) - 1.38 \times 10^{-23} \times \times ln(1)

                 = 1.38 \times 10^{-23} \times (-0.182)

                 = -0.251 \times 10^{-23}

or,             = -2.51 \times 10^{-24}

Thus, we can conclude that the entropy change for a particle in the given system is -2.51 \times 10^{-24} J/K particle.

8 0
3 years ago
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