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3 years ago

You wish to set up an electrochemical cell that has a copper anode and an

Chemistry

1 answer:

professor190 [17]3 years ago

4 0

I think cathode should be silver or gold
you can use silver nitrate solution or mercury nitrate solution

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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:

Step 1: 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

Step 2: Calculate mass of NaNO2

(40/100)*1.05 = 0.42 grams

Step 3: Calculate moles of NaNO2

Moles NaNO2 = 0.42 grams / 69.01 g/mol

Moles NaNO2 = 0.00608 moles

Step 4: 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

Step 5: Calculate pressure of N2

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

Step 6: 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

I recently started a science experiment and need help coming up with a catchy title. Can you help me? Here is my problem stateme

kirill115 [55]

I think Salty Boiling would be a good one lol :)

7 0

3 years ago

Read 2 more answers

During a combustion reaction, 9.00 grams of oxygen reacted with 3.00 grams of CH4.

Monica [59]

Answer:

0.74 grams of methane

Explanation:

The balanced equation of the combustion reaction of methane with oxygen is:

CH₄ + 2 O₂ → CO₂ + 2 H₂O

it is clear that 1 mol of CH₄ reacts with 2 mol of O₂.

firstly, we need to calculate the number of moles of both

for CH₄:

number of moles = mass / molar mass = (3.00 g) / (16.00 g/mol) = 0.1875 mol.

for O₂:

number of moles = mass / molar mass = (9.00 g) / (32.00 g/mol) = 0.2812 mol.

it is clear that O₂ is the limiting reactant and methane will leftover.

using cross multiplication

1 mol of CH₄ needs → 2 mol of O₂

??? mol of CH₄ needs → 0.2812 mol of O₂

∴ the number of mol of CH₄ needed = (0.2812 * 1) / 2 = 0.1406 mol

so 0.14 mol will react and the remaining CH₄

mol of CH₄ left over = 0.1875 -0.1406 = 0.0469 mol

now we convert moles into grams

mass of CH₄ left over = no. of mol of CH₄ left over * molar mass

= 0.0469 mol * 16 g/mol = 0.7504 g

So, the right choice is 0.74 grams of methane

3 0

3 years ago

What 3 sub types of covergent plate boundaries?

Virty [35]

(1) Ocean to Continent
(2)Continent to Continent
(3)Ocean to Ocean
are the three sub types of convergent plate boundaries.

5 0

3 years ago

Which of the following objects does not decompose?

Anettt [7]

B) Plastic Water Bottle

6 0

3 years ago

Read 2 more answers