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Sonja [21]
3 years ago
12

A substance releases 2,192 J of heat, changing its temperature from 50 degrees C to 5 degrees C. The mass of the substance is 73

.9 grams. What is the specific heat of the substance? Round your answer to TWO digits after the decimal point.
Chemistry
1 answer:
Gre4nikov [31]3 years ago
3 0

The specific heat capacity of the given substance is -0.66 J/g°C.

<u>Explanation:</u>

The heat absorbed by any substance is the product of its mass, specific heat capacity and change in temperature.

q = m × c × ΔT

m is the mass in grams

q = amount of heat released or absorbed in J

ΔT = change in temperature in °C = 5 -50 = -45°C

c = specific heat capacity in J/g°C

c = $\frac{q}{m \times dT}

Plugin the values, we will get,

c = $\frac{2192 }{73.9 \times-45}

= -0.66 J/g°C

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Water is an amphoteric substance, meaning it can serve as both an acid and a base,
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Answer:

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3 years ago
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How many grams of ammonia are produced when 1.0 mole of nitrogen reacts
Varvara68 [4.7K]

Answer: N2(g) + 3H2-> 2NH3(g)  This is the balanced equation

Note the mole ratio between N2, H2 and NH3.  It is 1 : 3 : 2  This will be important.

 

moles N2 present = 28.0 g N2 x 1 mole N2/28 g = 1 mole N2 present

moles H2 present = 25.0 g H2 x 1 mole H2/2 g = 12.5 moles H2 present

Based on mole ratio, N2 is limiting in this situation because there is more than enough H2 but not enough N2.

 

moles NH3 that can be produced = 1 mole N2 x 2 moles NH3/mole N2 = 2 moles NH3 can be produced  

grams of NH3 that can be produced = 2 moles NH3 x 17 g/mole = 34 grams of NH3 can be produced

 

NOTE:  The key to this problem is recognizing that N2 is limiting, and therefore limits how much NH3 can be produced.

Explanation: here you go!! good luck! hope this helped

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3 years ago
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3 years ago
Using the following reaction (depicted using molecular models), large quantities of ammonia are burned in the presence of a plat
Mila [183]

Answer:

17.65 grams of O2 are needed for a complete reaction.

Explanation:

You know the reaction:

4 NH₃ + 5 O₂ --------> 4 NO + 6 H₂O

First you must know the mass that reacts by stoichiometry of the reaction (that is, the relationship between the amount of reagents and products in a chemical reaction). For that you must first know the reacting mass of each compound. You know the values ​​of the atomic mass of each element that form the compounds:

  • N: 14 g/mol
  • H: 1 g/mol
  • O: 16 g/mol

So, the molar mass of the compounds in the reaction is:

  • NH₃: 14 g/mol + 3*1 g/mol= 17 g/mol
  • O₂: 2*16 g/mol= 32 g/mol
  • NO: 14 g/mol + 16 g/mol= 30 g/mol
  • H₂O: 2*1 g/mol + 16 g/mol= 18 g/mol

By stoichiometry, they react and occur in moles:

  • NH₃: 4 moles
  • O₂: 5 moles
  • NO: 4 moles
  • H₂O: 6 moles

Then in mass, by stoichiomatry they react and occur:

  • NH₃: 4 moles*17 g/mol= 68 g
  • O₂: 5 moles*32 g/mol= 160 g
  • NO: 4 moles*30 g/mol= 120 g
  • H₂O: 6 moles*18 g/mol= 108 g

Now to calculate the necessary mass of O₂ for a complete reaction, the rule of three is applied as follows: if by stoichiometry 68 g of NH₃ react with 160 g of O₂, 7.5 g of NH₃ with how many grams of O₂ will it react?

mass of O_{2} =\frac{7.5 g of NH_{3} * 160 g of O_{2} }{68 g of NH_{3} }

mass of O₂≅17.65 g

<u><em>17.65 grams of O2 are needed for a complete reaction.</em></u>

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3 years ago
A quantity of 0.0250 mol of a gas initially at 0.050 L and 19.0°C undergoes a constant-temperature expansion against a constant
KiRa [710]

Answer:

V_2=2.995L\\\\W=248.5J

Explanation:

Hello,

In this case, for us to compute the final volume we apply the Boyle's law that analyzes the pressure-volume temperature as an inversely proportional relationship:

P_1V_1=P_2V_2

So we solve for V_2 by firstly computing the initial pressure:

P_1=\frac{nRT}{V_1}=\frac{0.025mol*0.082\frac{atm*L}{mol*K}*(19+273.15)K}{0.050L}  =11.98atm

V_2=\frac{P_1V_1}{P_2}=\frac{11.98atm*0.050L}{0.200atm}\\ \\V_2=2.995L

Finally, we can compute the work by using the following formula:

W=nRTln(\frac{V_2}{V_1} )=0.025mol*8.314\frac{J}{mol*K}*(19.0+273.15)K*ln(\frac{2.995L}{0.050L}) \\\\W=248.5J

Best regards.

4 0
3 years ago
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