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slamgirl [31]
4 years ago
6

How many joules of heat are needed to raise the temperature of 47.5 g of aluminum from 21°C to 94°C , if the specific heat of al

uminum is 0.90 J/g°C ?
STEP BY STEP EXPLANATION PLEASE i actually want to understand it.
Chemistry
1 answer:
Montano1993 [528]4 years ago
5 0

Answer:

3120.75J

Explanation:

So, we have the formula q = mc\Delta t. For this example, q is the heat energy in Joules, m is the mass in grams, c is the specific heat capacity in \frac{J}{{g\°C}}, and \Deltat is the change in temperature. In this case, m = 47.5g, c = 0.9  \frac{J}{{g\°C}}, and \Deltat  = 94-21 = 73°C. Plugging in the values, we get the joules of heat required to raise 47.5g of Al from 21°C to 94°C which is stated above. You can double check my answer but that should be it. An important thing to be aware of are the units. Sometimes, the heat capacity may not be \frac{J}{{g\°C}}. I may be in Kelvin or something. Anyways, hope this helps.

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Answer:

A.  a new substance is being produced.

Explanation:

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In a chemical change, new substances are usually produced. They are accompanied by the evolution or absorption of energy.

The reaction of Zinc with a strong acid to produce bubbles on the surface of the metal indicates a chemical change and the formation of a new kind of substance.

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Since Zn is higher than Hydrogen in the activity series, it will displace it from HCl and liberate hydrogen gas as a product. This will cause the bubbles observed in the reaction.

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B and D are wrong because they are both physical changes.

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8 0
3 years ago
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Ascorbic acid (H2C6H6O6; H2Asc for this problem), known as vitamin C, is a diprotic acid (Ka1= 1.0x10–5 and Ka2= 5x10–12) found
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Answer:

The concentrations are :

[HAsc^-]=0.000702 M

[Asc^{2-}]=5.92\times 10^{-8} M

The pH of the solution is 3.15.

Explanation:

H_2Asc\rightleftharpoons HAs^-+H^+         K_{a1}=1.0\times 10^{-5}

Initial

c                0              0

Equilibrium

c-x                x          x

K_{a1}=\frac{[HAs^-][H^+]}{[H_2Asc]}

1.0\times 10^{-5}=\frac{x\times x}{(c-x)}

1.0\times 10^{-5}=\frac{x^2}{(0.050-x)}

Solving for x:

x = 0.000702 M

[HAsc^-]=0.000702 M

HAsc^-\rightleftharpoons As^{2-}+H^+        K_{a2}=5\times 10^{-12}

Initially

x                0          0

At equilibrium ;

(x - y)            y         y

K_{a2}=\frac{[As^{2-}][H^+]}{[HAsc^-]}

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5\times 10^{-12}=\frac{y^2}{(0.000702 -y)}

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pH=-\log[H^+]

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Answer:

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