For C2H5OH (liquid)So = 161 J/K.mol; for C2H5OH (gas)So = 282.6 J/K.mol. What is ΔS if 92 g of C2H5OH is evaporated?

Electronic and electrical devices differ in the way they use _____.

Katena32 [7]

Election current because voltage is a measurement, information doesn't apply to all electrical devices and the wires within are usually copper bc it conducts and hardly ever will the wires be anything different because copper is cheap

3 0

2 years ago

PLEASE HELP

lisabon 2012 [21]

Answer:

The answer is B.

Explanation:

Trust me i took the test already its b.

6 0

3 years ago

A 10,0-L cylinder of gas is stored at room temperature (20.0°C) and a pressure of 1800 psi. If the gas is

makvit [3.9K]

Considering the Charles' law, the gas would have a temperature of -109.2 C.

<h3>Charles' law</h3>

Finally, Charles' law establishes the relationship between the volume and temperature of a gas sample at constant pressure. This law says that the volume is directly proportional to the temperature of the gas. That is, if the temperature increases, the volume of the gas increases, while if the temperature of the gas decreases, the volume decreases.

Charles' law is expressed mathematically as:

$\frac{V}{T} =k$

If you want to study two different states, an initial state 1 and a final state 2, the following is true:

$\frac{V1}{T1} =\frac{V2}{T2}$

<h3>Temperature of the gas in this case</h3>

In this case, you know:

P1= 1800 psi
V1= 10 L
T1= 20 C= 293 K (being 0 C= 273 K)
P2= 1800 psi
V2= 6 L
T2= ?

You can see that the pressure remains constant, so you can apply Charles's law.

Replacing in the Charles's law:

$\frac{10 L}{293 K} =\frac{6 L}{T2}$

Solving:

$\frac{10 L}{293 K} T2=6 L$

$T2=\frac{6 L}{\frac{10 L}{293 K} }$

The gas would have a temperature of -109.2 C.

Learn more about Charles's law:

brainly.com/question/4147359?referrer=searchResults

7 0

2 years ago

Which is the electron configuration for lithium

creativ13 [48]

Answer:

The electron configuration is [HE] 2s1

7 0

3 years ago

Sulfur dioxide, SO 2 ( g ) , can react with oxygen to produce sulfur trioxide, SO 3 ( g ) , by the reaction 2 SO 2 ( g ) + O 2 (

Alex

Answer:

ΔHorxn = - 11.79 KJ

Explanation:

2 SO 2 ( g ) + O 2 ( g ) ⟶ 2 SO 3 ( g )

The standard enthalpies of formation for SO 2 ( g ) and SO 3 ( g ) are Δ H ∘ f [ SO 2 ( g ) ] = − 296.8 kJ / mol Δ H ∘ f [ SO 3 ( g ) ] = − 395.7 kJ / mol

From the reaction above, 2 mol of SO2 reacts to produce 2 mol of SO3. Assuming ideal gas behaviour,

1 mol = 22.4l

x mol = 2.67l

Upon cross multiplication and solving for x;

x = 2.67 / 22.4 = 0.1192 mol

0.1192 mol of SO2 would react to produce 0.1192 mol of SO3.

Amount of heat is given as;

ΔHorxn = ∑mΔHof(products) − ∑nΔHof(reactants)

Because O2(g) is a pure element in its standard state, ΔHοf [O2(g)] = 0 kJ/mol.

ΔHorxn = 0.1192 mol * (− 395.7 kJ / mol) - 0.1192 mol * ( − 296.8 kJ / mol)

ΔHorxn = - 47.17kj + 35.38kj

ΔHorxn = - 11.79 KJ

5 0

3 years ago