If it is going to be published it must first go to a

Consider the following precipitation reaction (balanced). precipitation reaction: 2NH4Br(aq)+Pb(C2H3O2)2(aq)⟶2NH4C2H3O2(aq)+PbBr

sweet [91]

Answer:

Pb2+ (aq) & 2Br- (aq) --> PbBr2 (s).

Explanation:

Equation of the reaction:

Pb(C2H32O2)2 (aq) + 2 NH4Br (aq) --> 2NH4C2H3O2 (aq) + PbBr2 (s)

Ionic equation:

Pb+2(aq) + 2(C2H3O2)-1 (aq) + 2(NH4+) (aq) + 2Br-1 (aq) --> 2(NH4+) (aq) + 2(C2H3O2-) (aq) + PbBr2 (s)

2(NH4)+1(aq) & 2(C2H3O2)-1 (aq) cancel out from both sides, you are left with the net ionic equation :

Pb2+ (aq) & 2Br- (aq) --> PbBr2 (s).

6 0

3 years ago

Which atom is most likely to accept electrons to form an ionic bond?

Alinara [238K]

Halogens (atoms with 7 valence electrons) and Hydrogen

or generally, atoms with their shells almost full

5 0

3 years ago

Which of the following buffer systems would you use if you wanted to prepare a solution having a pH of approximately 9.5?Which o

zalisa [80]

Answer:

b. 0,08M NH₄⁺ / 0,12M NH₃

Explanation:

The buffers are:

a. 0,08M H₂PO₄⁻ / 0,12M HPO₄²⁻

b. 0,08M NH₄⁺ / 0,12M NH₃

It is possible to find out the pH of a buffer using Henderson-Hasselbalch formula:

pH = pka + log₁₀ [A⁻] / [HA] (1)

Where A⁻ is the conjugate base of the weak acid, HA.

a. For the system H₂PO₄⁻ / HPO₄²⁻ pka is 7,198. Replacing in (1)

pH = 7,198 + log₁₀ [0,12] / [0,08]

pH = 7,37

That means this buffer system don't have a pH of approximately 9,5

b. For the system NH₄⁺ / NH₃ pka is 9,3. Replacing in (1)

pH = 9,3 + log₁₀ [0,12] / [0,08]

pH = 9,50

That means this buffer system is the buffer you need to use.

I hope it helps!

7 0

3 years ago

6. If 0.500 mol NaN3 react, what mass in grams of nitrogen will result?

lions [1.4K]

Answer:

21 g of N₂ are produced by the decomposition

Explanation:

The reaction is: 2 NaN3 → 2 Na + 3 N2

2 moles of sodium nitride decompose in order to produce 2 moles of Na and 3 moles of nitrogen gas.

According to stoichiometry, ratio is 2:3. Therefore we say,

2 moles of nitride can produce 3 moles of N₂

Then, 0.5 moles of NaN₃ will produce (0.5 . 3) / 2 = 0.75 moles of N₂

We convert the moles to mass, to find the answer

0.75 mol . 28 g / 1 mol = 21 g

7 0

3 years ago

Problem page write a balanced half-reaction describing the oxidation of aqueous chromium(ii) cations to aqueous chromium(iv) cat

shutvik [7]

The balanced half-reaction: Cr²⁺⁺(aq) → Cr⁴⁺(aq) + 2e⁻.
Chromium(II) cations two electrons and became chromium(IV) cations.
Oxidation reaction is increasing of oxidation number of element, because element or ion lost electrons in chemical reaction.
Reduction is lowering oxidation number because element or ions gain electrons.

8 0

3 years ago

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