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iogann1982 [59]
3 years ago
10

Write the balanced chemical equations, including states, for the neutralization reactions between: HI (aq) and NaOH (aq) H2CO3 (

aq) and Sr(OH)2 (s) *note the strontium compound will be a solid Ca(OH)2 (s) and H2SO4 (aq)
Chemistry
1 answer:
rewona [7]3 years ago
8 0

Answer:

* HI (aq) +NaOH (aq) \rightarrow NaI(aq)+H_2O(l)

* H_2CO_3 (aq) + Sr(OH)_2\rightarrow SrCO_3(s)+2H_2O(l)

* Ca(OH)_2 (s) +H_2SO_4 (aq)\rightarrow CaSO_4(aq)+2H_2O(l)

Explanation:

Hello!

In this case, since neutralization chemical reactions are carried out when an acid and a base react to produce a salt and water, for the first reaction, HI (aq) and NaOH (aq), we can write:

HI (aq) +NaOH (aq) \rightarrow NaI(aq)+H_2O(l)

Whereas the salt, sodium iodide remains aqueous and water liquid. Next, when carbonic acid and strontium hydroxide react, strontium carbonate as the salt precipitates out and water is also produced:

H_2CO_3 (aq) + Sr(OH)_2\rightarrow SrCO_3(s)+2H_2O(l)

Finally, when calcium hydroxide and sulfuric acid react, the compound calcium sulfate as the salt and water are produced as shown below:

Ca(OH)_2 (s) +H_2SO_4 (aq)\rightarrow CaSO_4(aq)+2H_2O(l)

Best regards!

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The molar mass of a solid carboxylic acid is determined by titrating a known mass of the acid with a standardized solution of Na
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In this example we can take acetic acid as carboxylic acid; basic salt sodium acetate CH₃COONa is formed from the reaction between weak acid (in this example acetic acid CH₃COOH) and strong base (in this example sodium acetate NaOH).  

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2 years ago
By pipet, 11.00 mL of a 0.823 MM stock solution of potassium permanganate (KMnO4) was transferred to a 50.00-mL volumetric flask
tangare [24]

Answer:

1) 0.18106 M is the molarity of the resulting solution.

2) 0.823 Molar is the molarity of the solution.

Explanation:

1) Volume of stock solution = V_1=11.00 mL

Concentration of stock solution = M_1=0.823 M

Volume of stock solution after dilution = V_2=50.00 mL

Concentration of stock solution after dilution = M_2=?

M_1V_1=M_2V_2 ( dilution )

M_2=\frac{0.823 M\times 11.00 mL}{50 ,00 mL}=0.18106 M

0.18106 M is the molarity of the resulting solution.

2)

Molarity of the solution is the moles of compound in 1 Liter solutions.

Molarity=\frac{\text{Mass of compound}}{\text{Molar mas of compound}\times Volume (L)}

Mass of potassium permanganate = 13.0 g

Molar mass of potassium permangante = 158 g/mol

Volume of the solution = 100.00 mL = 0.100  L ( 1 mL=0.001 L)

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0.823 Molar is the molarity of the solution.

6 0
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