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12345 [234]
2 years ago
7

How much aluminum nitrate can be found when 21g of aluminum react with lead (II) nitrate? Answer in units of mol.

Chemistry
1 answer:
Bond [772]2 years ago
7 0

0.77 moles of aluminum nitrate can be found when 21g of aluminum react with lead (II) nitrate.

<h3>Equation of reaction</h3>
  • Al + Pb(NO₃)₂  ---> Al(NO₃)₃ + Pb (s)

From the equation of reaction:

1 mole of aluminum reacts with 1 mole of lead nitrate to produce 1 mole of aluminum nitrate

1 mole aluminum has a mass of 27 g

Number of moles of aluminum in 21 g = 21/27 =0.77 moles

0.77 moles of Aluminum produces 0.77 moles of aluminum nitrate

Therefore, 0.77 moles of aluminum nitrate can be found when 21g of aluminum react with lead (II) nitrate.

Learn more moles at: brainly.com/question/13314627

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What process occurs as pieces of dry ice get smaller
fenix001 [56]
Sublimation. It's basically, in simple terms, when a solid changes to a gas without going into liquid form. 
8 0
3 years ago
A sample consisting of 1.00 mol of perfect gas molecules at 27 °C is expanded isothermally from an initial pressure of 3.00 atm
Evgesh-ka [11]

Answer:

a) reversibly

ΔU = 0

q = 2740.16 J

w = -2740.16 J

ΔH = 0

ΔS(total) = 0

ΔS(sys)  =9.13 J/K

ΔS(surr) = -9.13 J/K

b) against a constant external pressure of 1.00 atm

ΔU = 0

w = -1.66 kJ

q = 1.66 kJ

ΔH = 0

ΔS(sys) = 9.13 J/K

ΔS(surr) = -5.543 J/K

ΔS(total) = 3.587 J/K

Explanation:

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

Number of moles = 1.00 mol

Temperature = 27.00 °C = 300 Kelvin

Initial pressure = 3.00 atm

Final pressure = 1.00 atm

The gas constant = 8.31 J/mol*K

<u>(a) reversibly</u>

<u>Step 2:</u> Calculate work done

For ideal gases ΔU depends only on temperature. So as it is an isothermal (T constant).

Since the temperature remains constant:

ΔU = 0

ΔU = q + w

q = -w

w = -nRT ln (Pi/Pf)

⇒ with n = the number of moles of perfect gas = 1.00 mol

⇒ with R = the gas constant = 8.314 J/mol*K

⇒ with T = the temperature = 300 Kelvin

⇒ with Pi = the initial pressure = 3.00 atm

⇒ with Pf = the final pressure = 1.00 atm

w =- 1*8.314 *300 * ln(3)

w = -2740.16 J

q = -w

q = 2740.16 J

<u>Step 3:</u> Calculate change in enthalpy

Since there is no change in energy, ΔH = 0

<u>Step 4:</u> Calculate ΔS

for an isothermal process

ΔS (total) = ΔS(sys) + ΔS(surr)  

ΔS(sys) = -ΔS(surr)

ΔS(sys) = n*R*ln(pi/pf)

ΔS(sys) = 1.00 * 8.314 * ln(3)

ΔS(sys) = 9.13 J/K

ΔS(surr) = -9.13 J/K

ΔS (total) = ΔS(sys) + ΔS(surr) = 0

<u>(b) against a constant external pressure of 1.00 atm</u>

<u>Step 1</u>: Calculate the work done

w = -Pext*ΔV

w = -Pext*(Vf - Vi)

⇒ with Vf = the final volume

⇒ with Vi = the initial volume

We have to calculate the final and initial volume. We do this via the ideal gas law P*V=n*R*T

V = (n*R*T)/P

Initial volume = (n*R*T)/Pi

⇒ Vi = (1*0.08206 *300)/3

   ⇒ Vi = 8.206 L

Final volume = (n*R*T)/Pf

     ⇒ Vf = (1*0.08206 *300)/1

      ⇒ Vf = 24.618 L

The work done w = -Pext*(Vf - Vi)

w = -1.00* ( 24.618 - 8.206)

w = -16.412 atm*L

w = -16 .412 *(101325/1atm*L) *(1kJ/1000J)

w = -1662.9 J = -1.66 kJ

<u>Step 2:</u> Calculate the change in internal energy

ΔU = 0

q = -w

q = 1.66 kJ

ΔH = 0 because there is no change in energy

<u>Step 3: </u>Calculate ΔS

ΔS(sys) = n*R*ln(3)

ΔS(sys) = 1.00 * 8.314 * ln(3)

ΔS(sys) = 9.13 J/K

ΔS(surr) = -q/T

ΔS(surr) = -1662.9J/300K

ΔS(surr) = -5.543 J/K

ΔS(total) = ΔS(surr) +ΔS(sys) = -5.543 J/K + 9.13 J/K = 3.587 J/K

4 0
3 years ago
Calculate the ph of a solution in which [oh–] = 4.5 × 10–9m.
frosja888 [35]

pOH = -LOG([OH])  

pOH = -LOG(4.5*10^-9)  

pOH = 8.34  

pH + pOH = 14  

pH = 14 - 8.34 = 5.65

hope this helps!

(:

5 0
3 years ago
Please, What is the mass of oxygen in 72.0g of water? [ H= 1.0; O = 16.0]
yarga [219]
Convert 72g of water into moles of water using molecular weights.

So water is H2O so add up those molecular weights (H=1 and O=16)

2(1)+(16) = 18 g/mol

Then convert so 72g / (18 g/mol) = 4 mol

Now you can convert mol of water to mol of oxygen. So 4 mol of water is 4 mol of oxygen. Then use oxygen molecular weight to find grams again.

4 mol oxygen * 16 g/mol = 64g of oxygen

If we were doing hydrogen instead of oxygen there would be 8 mol hydrogen in 4 mol of water (2 H’s in every H2O molecule) and since we have 74 grams and oxygen is 64 grams, Hydrogen should be 8 grams. Math to check below

8 mol hydrogen * 1 g/mol = 8g of hydrogen

It all adds to 72 so we are correct.
5 0
3 years ago
Identify the type of reaction shown by this chemical equation:<br>2AI + 6HCI - 2AlCl3 + 3H2​
ArbitrLikvidat [17]

Answer:

Single displacement reaction

Explanation:

2AI + 6HCI —> 2AlCl3 + 3H2

From the above reaction, we see clearly that Al displaces H from HCl. This is clearly a single displacement reaction.

5 0
3 years ago
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