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Lady bird [3.3K]
3 years ago
15

Which net ionic equation would match this description of a chemical reaction: When a

Chemistry
1 answer:
krok68 [10]3 years ago
6 0

Answer:

3Mg^{2+}(aq.)+2PO_{4}^{3-}(aq.)\rightarrow Mg_{3}(PO_{4})_{2}(s)

Explanation:

Mg(NO_{3})_{2}, Na_{3}PO_{4} are strong electrolytes. Hence they are fully ionized in aqueous solution.

Mg_{3}(PO_{4})_{2} is a sparingly soluble salt. Hence it remains undissociated in aqueous solution.

So, total ionic equation:

3Mg^{2+}(aq.)+6NO_{3}^{-}(aq.)+6Na^{+}(aq.)+2PO_{4}^{3-}(aq.)\rightarrow Mg_{3}(PO_{4})_{2}(s)+6Na^{+}(aq.)+6NO_{3}^{-}(aq.)

Net ionic equation is written by omitting spectator ions from total ionic equation.

Here, Na^{+} and NO_{3}^{-} ions are spectator ions as they remain present on both side of total ionic equation.

So, net ionic equation:

3Mg^{2+}(aq.)+2PO_{4}^{3-}(aq.)\rightarrow Mg_{3}(PO_{4})_{2}(s)

So, option (d) is correct.

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What mass of carbon dioxide could be made from 100 tonnes of calcium carbonate?
MA_775_DIABLO [31]
Total mass of CaCO3 = 40 amu of Ca + 12amu of C + 16×3 amu of oxygen = 100amu of CaCO3




i.e 100 tonnes of CaCO3 .


mass of CO2 = 12amu of C + 2× 16amu of O = 44 amu of CO2




mass % of CO2 in CaCO3 = (44/100)×100 =44%


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therefore, 44% of CO2 is present in CaCO3.








3 0
3 years ago
Which substance is the limiting reactant when 2.0 g of sulfur reacts with 3.0 g of oxygen and 4.0 g of sodium hydroxide accordin
Rudik [331]

Answer:

The limiting reactant is NaOH (option B)

Explanation:

2S(s)  +  3O₂(g)  +  4NaOH(aq)   →   2Na₂SO₄(aq)  +  2H₂O(l)

The reaction is ballanced. OK

We need to know how many moles do we have from each compound.

Mass / Molar weight = Mol

Molar weight S = 32 g/m

Molar weight O₂ = 32 g/m

Molar weight NaOH = 40 g/m

Mol S: 2g/ 32g/m = 0.0625 mol

Mol O₂: 3g / 32 g/m = 0.09375 mol

Mol NaOH: 4g/ 40g/m = 0.1 mol

Now, we can play with the reactants. The base is: 2 moles of S, react with 3 mol of O₂ and 4 moles of hydroxide to make 2 moles of sulfate and 2 moles of water. Pay attention to the rules of three.

2 moles of S __ react with __ 3 moles of O₂ __ and __ 4 moles of NaOH

0.0625 moles S __________ 0.09375 moles O₂ ___ 0.125 moles NaOH

The limiting reactant is the NaOH. I need to use 0.125 moles and I only have 0.1 moles.

Let's do the same with O₂

3 moles of O₂ __ react with __ 2 moles of S __ and __ 4 moles of NaOH

0.09375 moles of O₂ _______ 0.0625 mol of S _____ 0.125 moles NaOH

5 0
3 years ago
A buffered solution containing dissolved aniline, C6H5NH2, and aniline hydrochloride, C6H5NH3Cl, has a pH of 5.57 . A. Determine
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Answer:

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Explanation:

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pOH = pKb + log (salt/base)

As we have value of pH, we need to determine the pOH

14 - pH = pOH

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Now we replace data:

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-0.7 = log (  C₆H₅NH₃⁺ / 0.2 )

10⁻⁰'⁷ = C₆H₅NH₃⁺ / 0.2

0.19952 = C₆H₅NH₃⁺ / 0.2

C₆H₅NH₃⁺ = 0.19952 . 0.2  = 0.0399 M

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