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3 years ago

7

How much heat does an aluminum block absorb if 10.00 grams are heated from 25.0˚C to 50˚C? The specific heat of aluminum is .900

j/g ˚C
(2 Points)
450
225
-450
-225

1 answer:

OverLord2011 [107]3 years ago

5 0

Answer:

225

Explanation:

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What volume will 1,27 moles of helium gas occupy at 63°C and 75.2 kPa?

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Answer:

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Explanation:

Using the general gas equation as follows:

PV = nRT

Where;

P = pressure (atm)

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R = gas law constant (0.0821 Latm/molK)

T = temperature (K)

According to the information provided in this question,

V = ?

P = 75.2 kPa = 75.2 × 0.00987 = 0.74atm

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Consider the following reaction of the commercial production of SO3 using sulfur dioxide and oxygen: SO2(g) + O2(g) → SO3(g)

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Answer:

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Explanation:

Hello,

STP conditions are P=1 atm and T=273.15 K, thus, the reacting moles are:

$n_{SO_2}=\frac{5.3L*1atm}{0.082\frac{atm*L}{mol*K}*273.15K}=0.2366molSO_2\\n_{O_2}=\frac{4.7L*1atm}{0.082\frac{atm*L}{mol*K}*273.15K}=0.2098molO_2$

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Thus, the exact moles of oxygen that completely react with 0.2366 moles of sulfur dioxide are (limiting reagent identification):

$0.2366molSO_2*\frac{1molO_2}{2molSO_2}=0.1182molO_2$

Since 0.2098 moles of oxygen are available, we stipulate the oxygen is in excess and the sulfur dioxide is the limiting reagent. In such a way, the yielded grams of sulfur trioxide turn out into:

$m_{SO_3}=0.2366molSO_2*\frac{2molSO_3}{2molSO_2}*\frac{80gSO_3}{1molSO_3} \\m_{SO_3}=18.93gSO_3$

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