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amm1812
3 years ago
5

Write a net ionic equation for the overall reaction that occurs when aqueous solutions of potassium hydroxide and phosphoric aci

d are combined. assume excess base.
Chemistry
1 answer:
tino4ka555 [31]3 years ago
7 0
The complete reaction for potassium hydroxide (KOH) and phosphoric acid (H3PO4) would yield products of water (H2O) and potassium phosphate (K3PO4).

3 KOH + H3PO4 ---> K3PO4 + 3 H2O

The net ionic equation would be obtained when you express the compounds in their ionic forms and eliminating the spectator ions. Spectator ions are the ions that appear in the reactant and product side. They do not actively participate the in the reaction.

3 K+ + 3 OH- + H3PO4  --->  3K+ + PO4^3- + 3 H2O

H3PO4 is a weak acid so it does not dissociate. H2O is not an ionic compound so it also does not dissociate. Therefore, the spectator ion is the potassium ion K+. The net ionic reaction is

3 OH- + 3 H3PO4  ---> PO4^3- + 3 H2O
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Below is the thermochemical equation for the combustion of octane.
Gnom [1K]

5mol produces -10.941KJ

4 mol let produces x KJ

Solve through proportion

  • 5/-10.941=4/x
  • 5x=-43.764
  • x=-43.764/5
  • x=8.756 KJ
8 0
2 years ago
A 20.0-milliliter sample of 0.200 M K2CO3 solution is added to 30.0 milliliters of 0.400 M Ba(NO3)2 solution.
jenyasd209 [6]

Answer:

(B) 0.160 M

Explanation:

Considering:

Molarity=\frac{Moles\ of\ solute}{Volume\ of\ the\ solution}

Or,

Moles =Molarity \times {Volume\ of\ the\ solution}

Given :

For K_2CO_3 :

Molarity = 0.200 M

Volume = 20.0 mL

The conversion of mL to L is shown below:

1 mL = 10⁻³ L

Thus, volume = 20.0×10⁻³ L

Thus, moles of K_2CO_3 :

Moles=0.200 \times {20.0\times 10^{-3}}\ moles

Moles of K_2CO_3 = 0.004 moles

For Ba(NO_3)_2 :

Molarity = 0.400 M

Volume = 30.0 mL

The conversion of mL to L is shown below:

1 mL = 10⁻³ L

Thus, volume =30.0×10⁻³ L

Thus, moles of Ba(NO_3)_2 :

Moles=0.400\times {30.0\times 10^{-3}}\ moles

Moles of Ba(NO_3)_2  = 0.012 moles

According to the given reaction:

K_2CO_3_{(aq)}+Ba(NO_3)_2_{(aq)}\rightarrow BaCO_3_{(s)}+2KNO_3_{(aq)}

1 mole of potassium carbonate react with 1 mole of barium nitrate

0.004 moles potassium carbonate react with 0.004 mole of barium nitrate

Moles of barium nitrate  = 0.004 moles

Available moles of barium nitrate  =  0.012 moles

Limiting reagent is the one which is present in small amount. Thus, potassium carbonate is limiting reagent.

The formation of the product is governed by the limiting reagent. So,

1 mole of potassium carbonate gives 1 mole of barium carbonate

Also,

0.004 mole of potassium carbonate gives 0.004 mole of barium carbonate

Mole of barium carbonate = 0.004 moles

Also, consumed barium nitrate = 0.004 moles  (barium ions precipitate with carbonate ions)

Left over moles = 0.012 - 0.004 moles = 0.008 moles

Total volume = 20.0 + 30.0 mL = 50.0 mL = 0.05 L

So, Concentration = 0.008/0.05 M = 0.160 M

<u>(B) is correct.</u>

4 0
3 years ago
When a planet has the illusion of moving backwards or bouncing up and down it is exhibiting what?
Goryan [66]

Answer:  It is exhibiting retrogade motion.

7 0
3 years ago
__Zn(s)+__CuSO4(aq)---&gt;_______+________
stiks02 [169]

Answer:

Zn + CuSO4 —> ZnSO4 + Cu

Explanation:

Zn is higher than Cu in electrochemical series and so will displaces Cu in solution according to the equation:

Zn + CuSO4 —> ZnSO4 + Cu

5 0
3 years ago
What is the molarity of a 17.0% by mass solution of sodium acetate, NaC2H3O2 (82.0 g/mol), in water? The density of the solution
sattari [20]

Answer:

[NaCH₃COO] = 2.26M

Explanation:

17% by mass is a sort of concentration. Gives the information about grams of solute in 100 g of solution. (In this case, 17 g of NaCH₃COO)

Let's determine the volume of solution, by density

Mass of solution / Volume of solution = Solution density

100 g / Volume of solution = 1.09 g/mL

100 g / 1.09 g/mL = 91.7 mL

17 grams of solute is contained in 91.7 mL

Molarity (M) = Mol of solute /L of solution

91.7 mL / 1000 = 0.0917L

17 g / 82 g/m = 0.207 moles

Molariy = 0.207 moles / 0.0917L → 2.26M

4 0
3 years ago
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