Which of the following measurements is expressed to three significant figures?

Ethanol melts at -114 degree C. The enthalpy of fusion

Brut [27]

Answer: The heat required is 6.88 kJ.

Explanation:

The conversions involved in this process are :

$(1):ethanol(s)(-135^0C)\rightarrow ethanol(s)(-114^0C)\\\\(2):ethanol(s)(-114^0C)\rightarrow ethanol(l)(-114^0C)\\\\(3):ethanol(l)(-114^0C)\rightarrow ethanol(l)(-50^0C)$

Now we have to calculate the enthalpy change.

$\Delta H=[m\times c_{p,s}\times (T_{final}-T_{initial})]+n\times \Delta H_{fusion}+[m\times c_{p,l}\times (T_{final}-T_{initial})]+n\times \Delta H_{vap}+[m\times c_{p,g}\times (T_{final}-T_{initial})]$

where,

$\Delta H$ = enthalpy change = ?

m = mass of ethanol = 25.0 g

$c_{p,s}$ = specific heat of solid ethanol= 0.97 J/gK

$c_{p,l}$ = specific heat of liquid ethanol = 2.31 J/gK

n = number of moles of ethanol = $\frac{\text{Mass of ethanol}}{\text{Molar mass of ethanol}}=\frac{25.0g}{46g/mole}=0.543mole$

$\Delta H_{fusion}$ = enthalpy change for fusion = 5.02 KJ/mole = 5020 J/mole

$T_{final}-T_{initial}=\Delta T$ = change in temperature

The value of change in temperature always same in Kelvin and degree Celsius.

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

$\Delta H=[25.0 g\times 0.97J/gK\times (-114-(-135)K]+0.534mole\times 5020J/mole+[25.0g\times 2.31J/gK\times (-50-(-114))K]$

$\Delta H=6885.93J=6.88kJ$ (1 KJ = 1000 J)

Therefore, the heat required is 6.88 kJ

3 0

3 years ago

If you drop a 50 gram piece of metal with a temperature of 125° Celsius into 1000 grams of water at 20° Celsius, what best descr

Nostrana [21]

Dropping a 50 gram piece of metal with a temperature of 125° Celsius into 1000 grams of water at 20° Celsius would lead to a heat loss by the metal and an heat gain by the water. These loss and gain should be equal since energy cannot be created nor destroyed.

3 0

3 years ago

What does ROCAPV stand for?

Irina-Kira [14]

Answer:Risk Management, Operational Controls, Compliance, and Asset Quality

Explanation:

6 0

3 years ago

Read 2 more answers

In a reaction, 24.9 L of N2 reacts with excess Hy to produce NH3. How many liters of NH3

Amiraneli [1.4K]

Answer:

A. 49.8L of NH3.

B. 33.83g.

Explanation:

We'll begin by writing the balanced equation for the reaction. This is shown below:

N2 + 3H2 —> 2NH3

A. From the balanced equation above,

1L of N2 produced 2L of NH3.

Therefore, 24.9L of N2 will produce = 24.9 x 2 = 49.8L of NH3.

Therefore, 49.8L of NH3 is produced from the reaction.

B. Determination of the mass NH3 produced.

First, we shall determine the number of mole of NH3 produced. This can be obtained as follow:

Volume (V) = 49.8L

Pressure (P) = 97.8 kPa = 97.8/101.325 = 0.97atm

Temperature (T) = 23.7°C = 23.7°C + 273 = 296.7K

Gas constant (R) = 0.082atm.L/Kmol

Number of mole (n) =...?

PV = nRT

n = PV /RT

n = (0.97 x 49.8) / (0.082 x 296.3)

n = 1.99 mole

Next, we shall convert 1.99 mole to NH3 to grams. This is illustrated below:

Number of mole NH3 = 1.99 mole

Molar mass of NH3 = 14 + (3x1) = 17g/mol

Mass of NH3 =..?

Mass = mole x molar Mass

Mass of NH3 = 1.99 x 17

Mass of NH3 = 33.83g

Therefore, 33.83g of NH3 is produced.

6 0

4 years ago

The science of life or living matter in all it’s forms and phenomena? What’s it’s vocab answer?

coldgirl [10]

Answer:

Biology

Explanation:

8 0

3 years ago