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ch4aika [34]
3 years ago
7

The instruction booklet for your pressure cooker indicates that its highest setting is 11.1 psi. You know that standard atmosphe

ric pressure is 14.7 psi, so the booklet must mean 11.1 psi above atmospheric pressure.
At what temperature in degrees Celsius will your food cook in this pressure cooker set on "high"?
Chemistry
1 answer:
Jlenok [28]3 years ago
4 0

<u>Answer:</u> The temperature in the pressure cooker set on high is 206.14°C

<u>Explanation:</u>

STP conditions:

Pressure = 14.7 psi

Temperature = 273 K

To calculate the final temperature of the system, we use the equation given by Gay-Lussac Law. This law states that pressure of the gas is directly proportional to the temperature of the gas at constant pressure.

Mathematically,

\frac{P_1}{T_1}=\frac{P_2}{T_2}

where,

P_1\text{ and }T_1 are the initial pressure and temperature of the gas.

P_2\text{ and }T_2 are the final pressure and temperature of the gas.

We are given:

P_1=14.7psi\\T_1=273K\\P_2=(14.7+11.1)psi=25.8psi\\T_2=?

Putting values in above equation, we get:

\frac{14.7psi}{273K}=\frac{25.8psi}{T_2}\\\\T_2=\frac{25.8\times 273}{14.7}=479.14K

Converting the temperature from kelvins to degree Celsius, by using the conversion factor:

T(K)=T(^oC)+273

479.14=T(^oC)+273\\\\T(^oC)=(479.14-273)=206.14^oC

Hence, the temperature in the pressure cooker set on high is 206.14°C

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What volume (in L) of oxygen will be required to produce 77.4 L of water vapor in the reaction below?
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90.3 L

Explanation:

Given data:

Volume of water produced = 77.4 L

Volume of oxygen required = ?

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Chemical equation:

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It is known that,

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Now we will compare:

                               H₂O           :              O₂    

                               134.5         :              156.9

                                 77.4         :             156.9/134.5×77.4 =90.3 L

So for the production of 77.4 L water 90.3 L oxygen is required.

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If 15.0 mL of phosphoric acid completely neutralizes 38.5 mL of 0.150 mol/L calcium hydroxide, what is the concentration of the
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Let me give it a try.

H3PO4 + Ca(OH)2 = Ca3(PO4)2 + H2O

Balancing this reaction

2H3PO4 + 3Ca(OH)2 == Ca3(PO4)2 + 6H2O.

Moles= Molarity x Volume

Volume = 38.5ml = 0.0385L

Moles of Ca hydroxide = 0.150m/L x 0.0385L

(Notice the units canceling out...leaving moles).

=0.005775moles of Ca(OH)2.

From balanced reaction...

3moles of Ca(OH)2 completely reacts with 2moles of H3PO4

0.005775moles of Ca(OH)2 would completely react with....

= 0.005775 x 2/(3)

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Now we're looking for its Concentration in Mol/L

Molarity=Moles of solute/Volume of solution(in L)

Volume of solution assuming no other additions to the reaction = 15ml + 38.5ml =53.5ml =0.0535L

Molarity = 0.00385/0.0535

=0.072Mol/L.

If this is wrong

then Simply Try The formula for Mixing of solutions

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C2 = 0.11M/L.

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