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ipn [44]
2 years ago
11

Which element is the most reactive? sodium nickel carbon oxygen

Chemistry
1 answer:
leva [86]2 years ago
4 0

Sodium .

  • Atomic no-11
  • Electronic configuration=\sf 1s^22s^23s^1
  • Short name=Na
  • Latin name=Natrum
  • Period=3rd
  • Group=1st
  • Block=s-block
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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

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The other options are <em>incorrect </em>because they are <em>weak force</em>s. They would cause <em>low melting points</em>.

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