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Butoxors [25]
3 years ago
13

What is the net ionic equation for KCl+NaI= KI+NaCl

Chemistry
1 answer:
irinina [24]3 years ago
8 0
K^+ + I^- -> KI (s)
Because kcl, NaCl & naI are aq so they're split apart & KI is a solid so it stays together, then you just cross out the similar ions.
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What is the total number of electrons in As-3?
V125BC [204]

Answer:

Arsenic, As, has atomic number 33, which is the number of protons in the nuclei of its atoms. A neutral As atom would also have 33 electrons. The electron configuration of a neutral arsenic atom is [Ar]3d104s24p3

Explanation:

5 0
3 years ago
MATCH THE TERM WITH THE DEFINITION:
arlik [135]
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these definitions are out of my textbook but in my own words.

4 0
3 years ago
Read 2 more answers
Bronze is an example of which of the following:<br><br> element<br> atom<br> solution<br> solvent
Brut [27]
It is a solution, because it's the alloy of tin and copper 
8 0
4 years ago
What is the formula for the carboxylic acid functional group?
boyakko [2]

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7 0
2 years ago
What is the pH if 1mL of 0.1M HCl is added to 99mL of pure water?
coldgirl [10]

Answer:

pH of buffer after addition of 1 mL of 0,1 M HCl = 7,0

Explanation:

It is possible to use Henderson–Hasselbalch equation to estimate pH in a buffer solution:

pH = pka + log₁₀

Where A⁻ is conjugate base and HA is conjugate acid

The equilibrium of phosphate buffer is:

H₂PO₄⁻ ⇄ HPO4²⁻ + H⁺    Kₐ₂ = 6,20x10⁻⁸; pka=7,2

Thus, Henderson–Hasselbalch equation for 7,00 phosphate buffer is:

7,0 = 7,2 + log₁₀ \frac{[HPO4^{2-}] }{[H2PO4^{-}]}

Ratio obtained is:

0,63 = \frac{[HPO4^{2-}] }{[H2PO4^{-}]}

As the problem said you can assume [H₂PO₄⁻] = 0,1 M and [HPO4²⁻] = 0,063M

As the amount added of HCl is 0,001 M the concentrations in equilibrium are:

H₂PO₄⁻   ⇄   HPO4²⁻ +        H⁺

0,1 M +x      0,063M -x  0,001M -x -<em>because the addition of H⁺ displaces the equilibrium to the left-</em>

Knowing the equation of equilibrium is:

K_{a} = \frac{[HPO_{4}^{2-}][H^{+}]}{[H_{2} PO_{4}^{-}]}

Replacing:

6,20x10⁻⁸ = \frac{[0,063-x][0,001-x]}{[0,1+x]}

You will obtain:

x² -0,064 x + 6,29938x10⁻⁵ = 0

Thus:

x = 0,063 → No physical sense

x = 0,00099990

Thus, [H⁺] in equilibrium is:

0,001 M - 0,00099990 = 1x10⁻⁷

Thus, pH of buffer after addition of 1 mL of 0,1 M HCl =

-log₁₀ [1x10⁻⁷] = 7,0

A buffer is a solution that can resist pH change upon the addition of an acidic or basic components. In this example you can see its effect!

I hope it helps!

5 0
3 years ago
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