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Ivan
3 years ago
13

Balance this equation. Pb(NO3)2(aq)+NaCl(aq) -> NaNO3(aq)+PbCl2(s)

Chemistry
1 answer:
const2013 [10]3 years ago
5 0

Answer:

Pb(NO3)2(aq) + 2NaCl(aq) -> 2NaNO3(aq)+PbCl2(s)

Explanation:

Pb(NO3)2(aq)+NaCl(aq) -> NaNO3(aq)+PbCl2(s)

This is how it starts out.

Left:

  • 2 NO3s
  • 1 Pb
  • 1 Na
  • 1 Cl

Right

  • 1 Na
  • 1 NO3
  • 1 Pb
  • 2 Cl

So the place to start with this equation is to bring the Cls up to 2

Pb(NO3)2(aq)+2NaCl(aq) -> NaNO3(aq)+PbCl2(s)

But the Nas are now out of kilter.

Pb(NO3)2(aq)+ 2NaCl(aq) -> NaNO3(aq)+PbCl2(s)

Now the right has a problem. There's only 1 Na

Pb(NO3)2(aq) + 2 NaCl(aq) -> 2NaNO3(aq)+PbCl2(s)

Check it out. It looks like we are done.

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Describe how the particles of two substances interact when mixed. Be sure to discuss the activation energy and how the particles
Charra [1.4K]
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It is common knowledge that chemical reactions occur more rapidly at higher temperatures. Everyone knows that milk turns sour much more rapidly if stored at room temperature rather than in a refrigerator, butter goes rancid more quickly in the summer than in the winter, and eggs hard-boil more quickly at sea level than in the mountains. For the same reason, cold-blooded animals such as reptiles and insects tend to be noticeably more lethargic on cold days.

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8 0
3 years ago
The hydrogen chloride (HCl) molecule has an internuclear separation of 127 pm (picometers). Assume the atomic isotopes that make
natta225 [31]

Answer:

the energy of the third excited rotational state \mathbf{E_3 = 16.041 \ meV}

Explanation:

Given that :

hydrogen chloride (HCl) molecule has an intermolecular separation of 127 pm

Assume the atomic isotopes that make up the molecule are hydrogen-1 (protium) and chlorine-35.

Thus; the reduced mass μ = \dfrac{m_1 \times m_2}{m_1 + m_2}

μ = \dfrac{1 \times 35}{1 + 35}

μ = \dfrac{35}{36}

∵ 1 μ = 1.66 × 10⁻²⁷ kg

μ  = \\ \\ \dfrac{35}{36} \times 1.66 \times 10^{-27} \ \  kg

μ  = 1.6139 × 10⁻²⁷ kg

r_o = 127 \ pm = 127*10^{-12} \ m

The rotational level Energy can be expressed by the equation:

E_J = \dfrac{h^2}{8 \pi^2 I } \times J ( J +1)

where ;

J = 3 ( i.e third excited state)  &

I = \mu r^2_o

E_J= \dfrac{h^2}{8  \pi  \mu r^ 2 \mur_o } \times J ( J +1)

E_3 = \dfrac{(6.63 \times 10^{-34})^2}{8  \times  \pi ^2  \times 1.6139 \times 10^{-27} \times( 127 \times 10^{-12}) ^ 2  } \times 3 ( 3 +1)

E_3= 2.5665 \times 10^{-21} \ J

We know that :

1 J = \dfrac{1}{1.6 \times 10^{-19}}eV

E_3= \dfrac{2.5665 \times 10^{-21} }{1.6 \times 10^{-19}}eV

E_3 = 16.041  \times 10 ^{-3} \ eV

\mathbf{E_3 = 16.041 \ meV}

8 0
3 years ago
When the pH value of a solution is changed from 2
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increases my factor of 10

6 0
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Answer:

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3 0
3 years ago
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Calculate the mass of khc8h4o4 that reacts with 15 ml of the naoh solution
dlinn [17]
The balanced chemical reaction would be:

 KHC8H4O4<span> (aq) + </span>NaOH<span> (aq) → NaKC8H4O4 (aq) + H2O.

The concentration of the NaOH is equal 0.1 M. We use this and the volume given above to determine the mass of KH</span>C8H4O4. We do as follows:

0.1 mol / L NaOH (.015 L) ( 1 mol KHC8H4O4 / 1 mol NaOH) (204 g / 1 mol) = 0.306 g KHC8H4O4


6 0
3 years ago
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