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jek_recluse [69]
3 years ago
7

When a solution of NaI reacts with a solution of AgNO3, what is the net ionic equation?

Chemistry
1 answer:
Leni [432]3 years ago
3 0

Answer:

Ag⁺(aq) + I⁻(aq) → AgI(s)

Explanation:

Net ionic equation is a way to write a chemical equation in which you are listing only the species that are participating in the reaction.

In the reaction:

AgNO₃(aq) + NaI(aq) → AgI(s) + NaNO₃(aq).

The ionic equation is:

Ag⁺(aq) + NO₃⁻(aq) + Na⁺(aq) + I⁻(aq) → AgI(s) + Na⁺(aq) + NO₃⁻(aq).

Now, listing only the species that are participating in the reaction:

<h3>Ag⁺(aq) + I⁻(aq) → AgI(s)</h3>

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4 0
3 years ago
Explain why the symbol for the element sulfur and the formula for a molecule of sulfur differ.
Alexeev081 [22]

Answer:

Symbol : S

molecule : S₈

Explanation:

The symbol of an element is just the representation with few alphabets (one or two) with no digit as we don't represent it for an atom or molecule. Just a symbol.

So for example

Symbol for Helium is He

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5 0
4 years ago
Read 2 more answers
Please answer this question
tankabanditka [31]
A.0.103M
B. 0.0168mol
C.24.6ml
Work shown in picture
3 0
3 years ago
sphingomyelin contains a fatty acid linked by an ________ bond and phosphocholine linked via the c-3 hydroxyl group of sphingosi
irga5000 [103]
The answer is amide



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6 0
3 years ago
If anyone could please help me i'd greatly appreciate it! I have tried over and over again and can not figure this out.
Papessa [141]

Answer:

%Ionization = 1.63%

Explanation:

Hydrazine in aqueous media theoretically forms a difunctional hydroxyl system. However, for this problem assume only monofunctional ionization occurs. A second hydroxyl ionization would not likely occur as the formal cationic charge formed in the 1st ionization would inhibit a second ionization.

H₂NNH₂ + 2H₂O => HONHNHOH => HONHNH⁺ + OH⁻; Kb = 1.3 x 10⁻⁶

So, assuming all OH⁻ and HONHNH⁺ are delivered in the 1st ionization then a good estimate of the %ionization can be calculated.

               HONHNHOH => HONHNH⁺  +  OH⁻

C(i) =>              0.490M                  0M           0M

ΔC =>                  -x                         +x             +x

C(eq) =>         0.490 - x                    x               x

                      ≅0.490M* => *x is dropped as Conc H₂NNH₂/Kb > 100

Kb = [HONHNH⁺][OH⁻]/[HONHNHOH]

1.3 x 10⁻⁶ = x²/0.490

=> x = [OH⁻] = [HONHNH⁺] = √[(1.3 x 10⁻⁶)(0.490)] = 8 x 10⁻⁴

=> %Ionization = (x/0.490)100% = (8 x 10⁻⁴/0.490)100% = 1.63%

7 0
3 years ago
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