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3 years ago

What would be the final temperature of a 73.174g sample

Chemistry

1 answer:

madreJ [45]3 years ago

3 0

The final temperature is -138 °C.

Explanation:

Using the equation of specific heat

$Q = m*c* del T$

We can easily find the final temperature of a 73.174 g of copper sample. As we know that specific heat is the amount of energy required to raise the temperature of the object to 1°C.

The specific heat of copper is known as 0.387 J/g°C and the initial temperature is said as 102 °C . The mass is given as 73.174 g. The heat released is 6800 J.

Since the heat is released the Q value will be negative.

$-6800=0.387*73.174*(T_{final}-102)$

$T_{final}-102=\frac{-6800}{0.387*73.174} =-240$

$T_{final} = -240+102=-138$

Thus, the final temperature is -138 °C.

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When 1.98g of a hydrocarbon is burned in a bomb calorimeter, the temperature increases by 2.06∘C. If the heat capacity of the ca

schepotkina [342]

Answer:

8.3 kJ

Explanation:

In this problem we have to consider that both water and the calorimeter absorb the heat of combustion, so we will calculate them:

q for water:

q H₂O = m x c x ΔT where m: mass of water = 944 mL x 1 g/mL = 944 g

c: specific heat of water = 4.186 J/gºC

ΔT : change in temperature = 2.06 ºC

so solving for q :

q H₂O = 944 g x 4.186 J/gºC x 2.06 ºC = 8,140 J

For calorimeter

q calorimeter = C x ΔT where C: heat capacity of calorimeter = 69.6 ºC

ΔT : change in temperature = 2.06 ºC

q calorimeter = 69.60J x 2.06 ºC = 143.4 J

Total heat released = 8,140 J + 143.4 J = 8,2836 J

Converting into kilojoules by dividing by 1000 we will have answered the question:

8,2836 J x 1 kJ/J = 8.3 kJ

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2 years ago

A 32.2 g iron rod, initially at 21.9 C, is submerged into an unknown mass of water at 63.5 C. in an insulated container. The fin

lorasvet [3.4K]

Answer : The mass of the water in two significant figures is, $3.0\times 10^1g$

Explanation :

In this case the heat given by the hot body is equal to the heat taken by the cold body.

$q_1=-q_2$

$m_1\times c_1\times (T_f-T_1)=-m_2\times c_2\times (T_f-T_2)$

where,

$c_1$ = specific heat of iron metal = $0.45J/g^oC$

$c_2$ = specific heat of water = $4.18J/g^oC$

$m_1$ = mass of iron metal = 32.3 g

$m_2$ = mass of water = ?

$T_f$ = final temperature of mixture = $59.2^oC$

$T_1$ = initial temperature of iron metal = $21.9^oC$

$T_2$ = initial temperature of water = $63.5^oC$

Now put all the given values in the above formula, we get

$32.3g\times 0.45J/g^oC\times (59.2-21.9)^oC=-m_2\times 4.18J/g^oC\times (59.2-63.5)^oC$

$m_2=30.16g\approx 3.0\times 10^1g$

Therefore, the mass of the water in two significant figures is, $3.0\times 10^1g$

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2 years ago

A sample of a compound contains 0.300 mol of carbon and 1.20 mol of hydrogen. What is the empirical formula of the comp

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Answer:

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The number of moles of carbon and hydrogen has been given as follows:

C = 0.300 mol

H = 1.20 mol

Next, we divide each mole value by the smallest (0.300)

C = 0.300 ÷ 0.300 = 1

H = 1.20 ÷ 0.300 = 4

The empirical ratio of Carbon and Hydrogen is 1:4, hence, the empirical formula is CH4

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3 years ago

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Explanation:

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2 years ago

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Answer:

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