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ioda
3 years ago
5

A 0.300 g piece of copper is heated and fashioned into a bracelet. The amount of energy transferred by heat to the copper is 66,

300 J. If the specific heat of copper is 390 J/g 0C, what Is the change of the copper's temperature?
Chemistry
1 answer:
Alja [10]3 years ago
3 0

Answer:

X=600 degrees celsius

Explanation:

We can evaluate by using the equation ==> Q=m(c)(delta T)

-We know the mass is 0.3 g

-Q= 66,300 J

-C= 390 j/g 0C

-Therefore, we need to solve for delta T (which will be represented by x)

 Q= mct

66,300 J= 0.3 g (390 J/g 0C) (t)

The unit of measurement (g-grams) can be crossed out on the left side of our expression.

66,300 J=0.3(390 J 0C) (t)

-If we convert into a proportional relationship:

66,300 J                117 J/0C x

------------       =        -----------

117 J/0C                 117 J/0C

-Joules on both sides can be crossed off

*117 J/0C came from the product in terms of 0.3(390 J 0C)

*If we find the quotient of 66,300 J and 117 J/0C, we get 566.6 repeating

-You may keep the answer as it is, although for simplicity reasons, we can round to the nearest hundred.

-Therefore, x=600.

-----Remember, x represented the delta t in our equation. (I made notice of that when evaluating).

Hope this helps! :)

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In a aqueous solution of 4-chlorobutanoic acid , what is the percentage of -chlorobutanoic acid that is dissociated
lord [1]

Answer:

Explanation:

Let assume that the missing aqueous solution of 4-chlorobutanoic acid = 0.76 M

Then, the dissociation of 4-chlorobutanoic acid can be expressed as:

\mathsf{C_3H_6ClCO_2H }          ⇄      \mathsf{C_3H_6ClCO_2^-}      +      \mathsf{H^+}

The ICE table can be computed as:

                   \mathsf{C_3H_6ClCO_2H }          ⇄      \mathsf{C_3H_6ClCO_2^-}      +      \mathsf{H^+}

Initial              0.76                                 -                           -

Change            -x                                  +x                         +x

Equilibrium   0.76 - x                              x                          x  

K_a = \dfrac{[\mathsf{C_3H_6ClCO_2^-}] [\mathsf{H^+}]}{\mathsf{[C_3H_6ClCO_2H ]}}

K_a = \dfrac{[x] [x]}{ [0.76-x]}

where:

K_a = 3.02*10^{-5}

3.02*10^{-5} = \dfrac{x^2}{ [0.76-x]}

however, the value of x is so negligible:

0.76 -x = 0.76

Then:

3.02*10^{-5}*0.76 = x^2

x=\sqrt{3.02*10^{-5}*0.76 }

x = 0.00479 M

∴

x = \mathsf{[C_3H_6ClCO_2^-] = [H^+]=} 0.00479 M

\mathsf{C_3H_6ClCO_2H }  = (0.76 - 0.00479) M

= 0.75521 M

Finally, the percentage of the acid dissociated is;

= ( 0.00479 / 0.76) × 100

= 0.630 M

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Answer:

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Explanation:

The p sublevel has 3 orbitals, so can contain 6 electrons max. The d sublevel has 5 orbitals, so can contain 10 electrons max. And the 4 sublevel has 7 orbitals, so can contain 14 electrons max.

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