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kvasek [131]
3 years ago
8

Which process releases the energy that eventually produces lighting in a thunderstorm

Chemistry
1 answer:
Elena L [17]3 years ago
5 0
The process that releases energy which causes lightning is known as Glaciation
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An element's atomic number is 87. how many protons would an atom of this element have
andreev551 [17]
atomic number is equal to proton number
so the proton number will be 87
5 0
4 years ago
In the equilibrium constant expression for the reaction below what is the correct exponent for N2O4?
irga5000 [103]
As we have the balanced reaction equation is:

N2O4 (g) ↔ 2NO2(g)

from this balanced equation, we can get the equilibrium constant expression

KC = [NO2]^2[N2O4]^1

from this expression, we can see that [NO2 ] is with 2 exponent of  the stoichiometric and we can see that from the balanced equation as NO2
is 2NO2 in the balanced equation.

and [N2O4] is with 1 exponent of the stoichiometric and we can see that from the balanced equation as N2O4 is 1 N2O4 in the balanced equation. 

∴ the correct exponent for N2O4 in the equilibrium constant expression is 1 
7 0
4 years ago
Which scientist was the first to devise the rules of chemical reactions that are explained by modern atomic theory?
valina [46]

Answer:

John Dalton is the correct answer.

Explanation:

3 0
3 years ago
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6 0
3 years ago
A 200.0 mL solution of 0.40 M ammonium chloride was titrated with 0.80 M sodium hydroxide. What was the pH of the solution after
choli [55]

Answer:

9.25

Explanation:

Let first find the moles of NH_4Cl and NaOH

number of moles of NH_4Cl = 0.40  mol/L × 200 ×  10⁻³L

= 0.08 mole

number of moles of NaOH = 0.80  mol/L × 50 ×  10⁻³L

= 0.04 mole

The equation for the reaction is expressed as:

NH^+_{4(aq)} \ + OH^-_{(aq)} ------> NH_{3(g)} \ + H_2O_{(l)}

The ICE Table is shown below as follows:

                            NH^+_{4(aq)} \ + OH^-_{(aq)} ------> NH_{3(g)} \ + H_2O_{(l)}

Initial (M)              0.08            0.04                            0

Change (M)         - 0.04          -0.04                          + 0.04

Equilibrium (M)      0.04             0                                0.04

K_a*K_b = 10^{-14} \ at \ 25^0C

K_a = \frac{10^{-14}}{K_b}

K_a = \frac{10^{-14}}{1.76*10^{-5}}

K_a= 5.68*10^{10}

pK_a = - log \ (K_a)

pK_a = - log \ (5.68*10^{-10})

pK_a = 9.25

pH = pKa + \ log (\frac{HB}{HA} )   for buffer solutions

pH = pKa + \ log (\frac{moles \ of \ base }{ moles\ of \ acid} ) since they are in the same solution

pH = 9.25 + \ log (\frac{0.04 }{ 0.04} )

pH = 9.25

8 0
3 years ago
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