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irinina [24]
3 years ago
15

How many joules (J) are needed to increase the temperature of 10.0 g of lead (Pb) from 30 ˚C to 50 ˚C? (Cp for Pb = 0.128 J/g˚C)

Chemistry
1 answer:
MA_775_DIABLO [31]3 years ago
7 0

Answer:

Q = 25.6 j

Explanation:

Given data:

Energy needed= ?

Mass of lead = 10.0 g

Initial temperature = 30 °C

Final temperature = 50°C

Cp = 0.128 j/g.°C

Solution:

Specific heat capacity:

It is the amount of heat required to raise the temperature of one gram of substance by one degree.

Formula:

Q = m.c. ΔT

Q = amount of heat absorbed or released

m = mass of given substance

c = specific heat capacity of substance

ΔT = change in temperature

ΔT = 50°C - 30°C

ΔT = 20°C

Now we will put the values in formula.

Q = 10 g × 0.128 j/g.°C × 20°C

Q = 25.6 j

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Consider the following system at equilibrium:
Vika [28.1K]

Answer:

1) Rightward shift

2) Rightward shift

3) Leftward shift

4) Leftward shift

5) Leftward shift

6) Rightward shift

7) No shift

8) No shift                                                              

   

Explanation:

To evaluate each case we need to consider Le Chatelier's Principle, which states that the adding of additional reactants or products to a system will shift the equilibrium in the opposite direction, to maintain the equilibrium of the system. On the contrary, if we remove a reactant or a product in the system, the equilibrium will be shifted in the direction of the reactant or product reduced, to produce more of it (and thus maintain balance).        

Taking into account the above, let's see each statement, in the following equation:

A(aq) + B(aq)  ⇄  2C(aq)    (1)

1) Increase A. This will cause a rightward shift in equation 1 in order to consume the reactant added.

2) Increase B. Same as 1), this will cause a rightward in equation 1.

3) Increase C. This will cause a leftward shift in order to consume the excess of product in the system.  

4) Decrease A. This will produce a leftward shift to produce the reactant that is being reduced.

5) Decrease B. Same as 4), a leftward shift.

6) Decrease C. This will produce a rightward shift to produce the product that is being reduced.

7) Double A, half B. The double A will cause a rightward shift and the half B will produce a leftward shift, which results in no shift.

8) Double both B and C. Double B will produce a rightward shift and double C will produce the contrary, a leftward shift, so the final result is no shift.

               

I hope it helps you!

4 0
3 years ago
An ideal gas originally at 0.85 atm and 66°C was allowed to expand until its final volume, pressure and temperature were 94.0mL,
xeze [42]

Answer: The original volume in liters was 0.0707L

Explanation:

Combined gas law is the combination of Boyle's law, Charles's law and Gay-Lussac's law.

The combined gas equation is,

\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}

where,

P_1 = initial pressure of gas = 0.85 atm

P_2 = final pressure of gas = 456 mm Hg = 0.60 atm   (760mmHg=1atm)

V_1 = initial volume of gas = ?

V_2 = final volume of gas = 94.0 ml

T_1 = initial temperature of gas = 66^oC=273+66=339K

T_2 = final temperature of gas = 113^oF=318K  (32^0F=273K)

Now put all the given values in the above equation, we get:

\frac{0.85\times V_1}{339}=\frac{0.60\times 94.0}{318}

V_1=70.7ml=0.0707L   (1L=1000ml)

Thus the original volume in liters was 0.0707L

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