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3 years ago

What pressure is exerted by 0.625 moles of a gas in a 45.4 L container at -24.0 °C? Find pressure in units of atm.

Chemistry

1 answer:

scoundrel [369]3 years ago

3 0

Answer:

Ok to solve this you will need to use the Ideal Gas Law Formula which is as follows:

PV = nRT

P= pressure

V= volume

n= # of moles

R= Universal Gas Constant (0.0821 L x atm/mol x K)

T= Kelvin temperature

1.Simplify the Ideal Gas Law formula to what you need to solve for:

P = (nRT)/ V

2. List all you components as follows (this makes the process easier):

P = ?

V = 45.4 L

n = 0.625 mol

R = 0.0821 L x atm/ mol x K

T = 249 K

To find the Kelvin temperature K = C + 273

3. Plug in all your components in your set up formula:

P = [(0.625 mol)(0.0821 L x atm/ mol x K)(249 K)] / (45.4 L)

4. Cross out all similar units so the only thing left is atm because you are trying to find pressure.

P = [(0.625)(0.0821atm)(249)] / (45.4)

5. Multiply through and simplify

P = 0.28 atm

B. is the correct answer.

Glad I could help!! If you have any other questions just message me. Hopefully this was helpful.

Explanation:

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3 years ago

Give the effect on the melting point of the presence of a cis double bond in a fatty acid.

vovikov84 [41]

Answer:

The cis double bond present in unsaturated fatty acids acids results in lower melting point when compared to saturated fatty acids of the same chain length.

Explanation:

Melting point of a fatty acids are affected by the length and degree of unsaturation of the hydrocarbon chain.

At room temperature, saturated fatty acids with hydrocarbon chain lengths between 12-24 are waxy solids whereas unsaturated atty acids of the same chain length are liquids. This is due to the nature of the packing of the fatty acid molecules in the saturated and unsaturated compounds.

In the saturated compounds, the molecules are tightly packed side by side with minimal steric hindrance and maximal van der Waals forces of attraction between molecules. However, in unsaturated fatty acids, the cis double bond introduces a bend or kink in the molecules which then interferes with the tight packing of the molecules and reducing interaction between molecules. Therefore, less energy is required to cause a disorder in the arrangement of unsaturated fatty acids, leading to a lowering of melting point.

8 0

3 years ago

When nonpolar solutes are placed in water, the hydrogen bonding network of water is disrupted, and there are fewer ways for wate

satela [25.4K]

Answer: increase in entropy

Explanation:

Entropy is the measure of randomness or disorder of a system. If a system moves from an ordered arrangement to a disordered arrangement, the entropy is said to decrease and vice versa.

$\Delta S$ is positive when randomness increases and $\Delta S$ is negative when randomness decreases.

When the non polar solutes are placed in water, the hydrogen bonding network of water is disrupted, and there are fewer ways for water to hydrogen-bond with itself. That means the water molecules are more randomly arranged and thus have more entropy and thus $\Delta S$ is positive.

5 0

3 years ago

Because copper is a metal, it is what?

oksano4ka [1.4K]

Copper is a mineral and an element essential to our everyday lives. It is a major industrial metal because of its high ductility, malleability, thermal and electrical conductivity and resistance to corrosion. It is an essential nutrient in our daily diet. And, its antimicrobial property is becoming increasingly important to the prevention of infection. It ranks third after iron and aluminum in terms of quantities consumed in the USA.

6 0

3 years ago

Lt takes 4 hr 39 min for a 2.00-mg sample of radium-230 to decay to 0.25 mg. what is the half-life of radium-230?

Rufina [12.5K]

Radioactive decay => C = Co { e ^ (- kt) |

Data:

Co = 2.00 mg
C = 0.25 mg
t = 4 hr 39 min

Time conversion: 4 hr 39 min = 4.65 hr

1) Replace the data in the equation to find k

C = Co { e ^ (-kt) } => C / Co = e ^ (-kt) => -kt = ln { C / Co} => kt = ln {Co / C}

=> k = ln {Co / C} / t = ln {2.00mg / 0.25mg} / 4.65 hr = 0.44719

2) Use C / Co = 1/2 to find the hallf-life

C / Co = e ^ (-kt) => -kt = ln (C / Co)

=> -kt = ln (1/2) => kt = ln(2) => t = ln (2) / k

t = ln(2) / 0.44719 = 1.55 hr.

Answer: 1.55 hr

6 0

4 years ago