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dusya [7]
2 years ago
13

A student mixes two solutions together. One is believed to be HI, the other is believed to be HNO3. A precipitate forms. Believi

ng the results may be in error the student repeats the process twice in separate wells. After repeating the trial there was no observed precipitate. What is the most likely reason for the initial precipitate result?
Chemistry
1 answer:
4vir4ik [10]2 years ago
5 0

Answer:

The presence of <em>silver nitrate impurities</em> is the most probable cause for the initial precipitate result.

Explanation:

In test for halide ions, silver nitrate AgNO_{3} is used to test for the presence of soluble halide ions. The presence of AgNO_{3} in the reaction mixture (possibly the HNO3) gave the semblance of test for chlorides, allowing for formation of precipitation in the initial experiment.

This error could have been caused by handling of reagents and used laboratory apparatus during the initial reaction.

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4 0
2 years ago
Bromophenol blue is the indicator used in detecting the endpoint for the antacid analysis in this experiment. what is the expect
Paul [167]
Hello!

For the antacid analysis, the chemical reactions that occur in the titration are the following ones:

First, the antacid (composed of weak bases and carbonates) is completely neutralized by the H⁺ ions in the HCl

2HCl + CaCO₃ → CO₂ + H₂O + 2CaCl₂

HCl + OH⁻ → H₂O + Cl⁻

The titration reaction consists in titrating the excess H⁺ ions that are left in the solution, by the following reaction:

H⁺ + NaOH → H₂O + Na⁺

So, when the equivalence point is reached, the solution will go from acid to basic. As bromophenol blue is yellow in acidic solution and blue in basic solution, you'll expect the indicator to change from yellow to blue.

Have a nice day!
7 0
3 years ago
In Experiment 9, a 1.05 g sample of a mixture of NaCl (MW 58.45) and NaNO2 (MW 69.01) is reacted with excess sulfamic acid. The
Kaylis [27]

Answer:

There is 76.6 mL of nitrogen collected

Explanation:

<u>Step 1: </u>Data given

Mass of the sample = 1. 05 grams

The sample is 40.00% by mass NaNO2

MW of NaCl = 58.45 g/mol

MW of NaNO2 = 69.01 g/mol

Temperature = 22.0 °C

Pressure = 750.0 mmHg = (750/760) atm

The vapor pressure of water at 22.0°C is 19.8 mm Hg = 0.02605 atm

<u>Step 2:</u> Calculate mass of NaNO2

(40/100)*1.05  = 0.42 grams

<u>Step 3:</u> Calculate moles of NaNO2

Moles NaNO2 = 0.42 grams / 69.01 g/mol

Moles NaNO2 = 0.00608 moles

<u>Step 4:</u> Calculate moles of N2

For 2 moles of NaNO2 we'll get 1 mol of N2

For 0.00608 moles of NaNO2 we'll get 0.00608/2 = 0.00304 moles

<u>Step 5:</u> Calculate pressure of N2

P = 750.0 - 19.8 = 730.2 mmHg = (730.2/760)atm = 0.96079 atm = 97352 Pa

<u>Step 6:</u> Calculate volume of N2

PV = nRT

⇒ P = the pressure of N2 =  0.96079

⇒ V = the volume of N2 = TO BE DETERMINED

⇒ n = moles of N2 = 0.00304 moles

⇒ R = the gas constant = 0.08206 L*atm/K*mol

⇒ T = the temperature = 22°C = 295 Kelvin

V = (0.00304*0.08206*295)/0.96079

V = 0.0766 L = 76.6 mL

There is 76.6 mL of nitrogen collected

4 0
3 years ago
Some antacid tables contain aluminum hydroxide. The aluminum hydroxide reacts with stomach acid according to the equation: Al(OH
MatroZZZ [7]

Answer:

26.67 mol HCl

Explanation:

Al(OH)₃ + 3HCl → AlCl₃ + 3H₂O

In order to solve this problem, we need to c<u>onvert Al(OH)₃ moles to HCl moles</u>.

To do so we use the<em> stoichiometric ratios</em> of the balanced reaction:

  • 8.89 mol Al(OH)₃ * \frac{3molHCl}{1molAl(OH)_{3}} = 26.67 mol HCl

Thus 26.67 moles of HCl would react completely with 8.89 moles of Al(OH)₃.

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2 years ago
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Rudiy27
It's botanical name would be Mandragora officianarum, and it has a humanoid shape. Its a murderous plant that grows from blood that grows in Mediterranean.
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3 years ago
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