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Oduvanchick [21]
3 years ago
6

The following chemical reaction takes place in aqueous solution: 2FeBr3 (aq) + 3Na2S (aq) → Fe2S3 (s) + 6NaBr (aq) Write the net

ionic equation for this reaction.
Chemistry
1 answer:
lakkis [162]3 years ago
5 0
A net ionic equation simply means to cancel out any ions which appear on both sides of the chemical equation that are not involved in the reaction - they're called spectator ions. 
We'll first write out the full ionic equation, showing all ions and compounds formed, then rewrite and not include spectator ions.

2FeBr3(aq) + 3Na2S(Aq) --> Fe2S3(s) + 6NaBr(aq)   [original eqation]

2Fe3+(aq) + 6Br-(aq) + 3Na+(aq) + 3S2-(aq)--> Fe2S3(s)+6Na+(aq) + 6Br-(aq)
[full ionic equation]

2Fe3+(aq) + 3S2-(aq)--> Fe2S3(s)   [net ionic equation]

notice that Br- and Na+ appear unreacted on both sides of the full ionic equation, so they cancel out and do not appear in the net ionic.

*Please give me a 'brainliest' if you can! Thanks!


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Rhodium crystallizes in a face-centered cubic unit cell. The radius of a rhodium atom is 135 pm. Determine the density of rhodiu
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Answer:

Density of unit cell ( rhodium) = 12.279 g/cm³

Explanation:

Given that:

The radius (r) of a rhodium atom = 135 pm

The atomic mass of rhodium = 102.90 amu

For a face-centered cubic unit cell,

r = \dfrac{a}{2\sqrt{2}}

where;

a = edge length.

Making "a" the subject of the formula:

a = 2 \sqrt{2} \times r

a = 2 \times 1.414 \times 135 \ pm

a = 381.8 pm

to cm, we get:

a = 381.8 × 10⁻¹⁰ cm

However, recall that:

density \ of \ unit \ cell = \dfrac{mass \ of \ unit \ cell}{volume \ of \unit \ cell}

where;

mass of unit cell = mass of atom × numbers of atoms per unit cell

Also;

mass\  of\ atom =\dfrac{ atomic \ mass}{Avogadro  \  number}

mass\  of\ atom =\dfrac{ 102.9}{6.023 \times 10^{23}}

Recall also that number of atoms in a unit cell for a  face-centered cubic = 4

So;

mass \ of \ unit \ cell= \dfrac{102.90}{6.023 \times 10^{23}}\times 4

mass of unit cell = 6.83380375 × 10⁻²² g

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Density of unit cell ( rhodium) = 12.279 g/cm³

3 0
2 years ago
A mass of 100.0 g of NaCl is added to 100.0 mL of water, but not all of it dissolves. A mass of 59.5 grams of NaCl solid remains
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Answer:

The molarity of the dissolved NaCl is 6.93 M

Explanation:

Step 1: Data given

Mass of NaCl = 100.0 grams

Volume of water = 100.0 mL = 0.1 L

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Molar mass NaCl= 58.44 g/mol

Step 2: Calculate the dissolved mass of NaCl

100 - 59. 5 = 40.5 grams

Step 3: Calculate moles

Moles NaCl = 40.5 grams / 58.44 g/mol

Moles NaCl = 0.693 moles

Step 4: Calculate molarity

Molarity = moles / volume

Molarity dissolved NaCl = 0.693 moles / 0.1 L

Molarity dissolved NaCl = 6.93 M

The molarity of the dissolved NaCl is 6.93 M

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