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trasher [3.6K]
4 years ago
7

Iodine + potassium > Symbol equation ___ + ____ > ___ + ____

Chemistry
1 answer:
pentagon [3]4 years ago
5 0

Answer:

Here's what I get.  

Explanation:

                  Word equation: Iodine + potassium yields potassium iodide

              Symbol equation:     I₂     +          K         ⟶                 KI

Balanced symbol equation:     I₂     +        2K         ⟶              2KI

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Each equation is incorrect. find the errors, then rewrite and balance each equation. cl2 + nai → nacl2 + i nh3 → n + h2 na + o2
VikaD [51]
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Chlorine and iodine are diatomic molecules.
2) Balanced chemical reaction: 2NH₃ → N₂ + 3H₂.
Nitrogen and hydrogen are diatomic molecules.
3) Balanced chemical reaction: 4Na + O₂ → 2Na₂O.
Sodium in compounds has oxidation number +1 and oxygen -2.
6 0
4 years ago
What were Lamarck's ideas about evolution and why were those ideas incorrect
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What were Lamarck's ideas about evolution and why were those ideas incorrect
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3 years ago
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If 21.00 mL of a 0.68 M solution of C6H5NH2 required 6.60 mL of the strong acid to completely neutralize the solution, what was
Andru [333]

Answer:

pH = 2.46

Explanation:

Hello there!

In this case, since this neutralization reaction may be assumed to occur in a 1:1 mole ratio between the base and the strong acid, it is possible to write the following moles and volume-concentrations relationship for the equivalence point:

n_{acid}=n_{base}=n_{salt}

Whereas the moles of the salt are computed as shown below:

n_{salt}=0.021L*0.68mol/L=0.01428mol

So we can divide those moles by the total volume (0.021L+0.0066L=0.0276L) to obtain the concentration of the final salt:

[salt]=0.01428mol/0.0276L=0.517M

Now, we need to keep in mind that this is an acidic salt since the base is weak and the acid strong, so the determinant ionization is:

C_6H_5NH_3^++H_2O\rightleftharpoons  C_6H_5NH_2+H_3O^+

Whose equilibrium expression is:

Ka=\frac{[C_6H_5NH_2][H_3O^+]}{C_6H_5NH_3^+}

Now, since the Kb of C6H5NH2 is 4.3 x 10^-10, its Ka is 2.326x10^-5 (Kw/Kb), we can also write:

2.326x10^{-5}=\frac{x^2}{0.517M}

Whereas x is:

x=\sqrt{0.517*2.326x10^{-5}}\\\\x=3.47x10^-3

Which also equals the concentration of hydrogen ions; therefore, the pH at the equivalence point is:

pH=-log(3.47x10^{-3})\\\\pH=2.46

Regards!

6 0
3 years ago
50 POINTS
Trava [24]
My guess is b for the question
3 0
3 years ago
Read 2 more answers
A solution has a [H3O+] of 1x 10-5 M. What is the [OH-] of the solution? (5 points)
White raven [17]

Answer:

1x10^–9 M

Explanation:

From the question given,

Concentration of hydronium ion, [H3O+] = 1x10^-5 M.

Concentration of Hydroxide ion, [OH-] =..?

The concentration of the hydroxide ion, [OH-] can be obtained as follow:

[H3O+] x [OH-] = 1x10^–14

1x10^-5 M x [OH-] = 1x10^–14

Divide both side by 1x10^-5

[OH-] = 1x10^–14 / 1x10^-5

[OH-] = 1x10^–9 M

8 0
3 years ago
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