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igomit [66]
3 years ago
9

Suppose a snack bar is burned in a calorimeter and heats 2,000 g water by 20 °C. How much heat energy was released? (Hint: Use t

he specific heat equation.) Give your answer in both joules and calorie
Chemistry
2 answers:
GenaCL600 [577]3 years ago
7 0

Answer:

Explanation:

Remember that the key is the water so

using the formula q=m*s*deltT

We have the mass 2000g

we have the specific heat of water which is 1.00 cal/ g C

and the final temperature 20 C

Remember that delt T is the change in temperature meaning (Tfinal - Tinitial) Assuming that the water is at room  temperature (25 C)  before starting the reaction then

q= (2000 g)(1.00 cal/g C)(20-25)

q= -10, 000 cal  

if you want you can convert to joules by multiplying by 4.314 J

In more simpler terms - If all the heat released was absorbed by the water, and it takes 1 calorie of heat to raise

the temperature of 1 gram of water by 1°C that is a total of 20 times 2000 or 40,000 cal.

or 40 kilocalories. If you want it in joules, there are 4.184 joules to one calorie so it is

4.184 times 40 or 167.36 kilojoules.

Hope that helped

Salsk061 [2.6K]3 years ago
6 0

The answer is: 167360

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Answer:

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Explanation:

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The molecular formula : As₄S₆

<h3>Further explanation</h3>

Given

Rate of effusion of arsenic(III) sulfide = 0.28 times the rate of effusion of Ar atoms

Required

The molecular formula

Solution

Graham's law: the rate of effusion of a gas is inversely proportional to the square root of its molar masses or  

the effusion rates of two gases = the square root of the inverse of their molar masses:  

\rm \dfrac{r_1}{r_2}=\sqrt{\dfrac{M_2}{M_1} }

or  

\rm M_1\times r_1^2=M_2\times r_2^2

Input the value :

1 = Arsenic(III) sulfide

2 = Ar

MM Ar = 40 g/mol

0.28 = √(40/M₁)

M₁=40 : 0.28²

M₁=510 g/mol

The empirical formula of arsenic(III) sulfide =  As₂S₃

(Empirical formula)n = molecular formula

( As₂S₃)n = 510 g/mol

(246.02 g/mol)n = 510 g/mol

n = 2

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