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2 years ago

How many moles of a gas would occupy 22.4 Liters at 273 K and 1 atm?

Chemistry

1 answer:

Molodets [167]2 years ago

4 0

Answer:

1 mole of a gas would occupy 22.4 Liters at 273 K and 1 atm

Explanation:

An ideal gas is a set of atoms or molecules that move freely without interactions. The pressure exerted by the gas is due to the collisions of the molecules with the walls of the container. The ideal gas behavior is at low pressures, that is, at the limit of zero density. At high pressures the molecules interact and intermolecular forces cause the gas to deviate from ideality.

An ideal gas is characterized by three state variables: absolute pressure (P), volume (V), and absolute temperature (T). The relationship between them constitutes the ideal gas law, an equation that relates the three variables if the amount of substance, number of moles n, remains constant and where R is the molar constant of the gases:

P * V = n * R * T

In this case:

P= 1 atm
V= 22.4 L
n= ?
R= 0.082 $\frac{atm*L}{mol*K}$
T=273 K

Reemplacing:

1 atm* 22.4 L= n* 0.082 $\frac{atm*L}{mol*K}$ *273 K

Solving:

$n=\frac{1 atm* 22.4 L}{0.082 \frac{atm*L}{mol*K} *273 K}$

n= 1 mol

Another way to get the same result is by taking the STP conditions into account.

The STP conditions refer to the standard temperature and pressure. Pressure values at 1 atmosphere and temperature at 0 ° C (or 273 K) are used and are reference values for gases. And in these conditions 1 mole of any gas occupies an approximate volume of 22.4 liters.

1 mole of a gas would occupy 22.4 Liters at 273 K and 1 atm

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The molarity (M) of an aqueous solution containing 22.5 g of sucrose (C12H22O11) in 35.5 mL of solution is ________.

Nonamiya [84]

Answer:

1.86 M

Explanation:

From the question given above, the following data were obtained:

Mass of sucrose (C12H22O11) = 22.5 g

Volume of solution = 35.5 mL

Molarity of solution =?

Next, we shall determine the number of mole in 22.5 g of sucrose (C12H22O11). This can be obtained as follow:

Mass of sucrose (C12H22O11) = 22.5 g

Molar mass of C12H22O11 = (12×12) + (22×1) + (16×11)

= 144 + 22 + 176

= 342 g/mol

Mole of C12H22O11 =?

Mole = mass /Molar mass

Mole of C12H22O11 = 22.5 /342

Mole of sucrose (C12H22O11) = 0.066 mole

Next, we shall convert 35.5 mL to litres (L). This can be obtained as follow:

1000 mL = 1 L

Therefore,

35.5 mL = 35.5 mL × 1 L / 1000 mL

35.5 mL = 0.0355 L

Thus, 35.5 mL is equivalent to 0.0355 L.

Finally, we shall determine the molarity of the solution as follow:

Mole of sucrose (C12H22O11) = 0.066 mole

Volume of solution = 0.0355 L.

Molarity of solution =?

Molarity = mole /Volume

Molarity of solution = 0.066/0.0355

Molarity of solution = 1.86 M

Therefore, the molarity of the solution is 1.86 M.

8 0

3 years ago

The half-life of C-14 is 5470 years. If a particular archaeological sample has one-quarter of its original radioactivity remaini

-Dominant- [34]

10940? Sorry if it isn’t correct

3 0

2 years ago

Read 2 more answers

11. What is the specific heat of a substance with a mass of 25.5 g that requires 412 J

Romashka-Z-Leto [24]

Answer:

297 J

Explanation:

The key to this problem lies with aluminium's specific heat, which as you know tells you how much heat is needed in order to increase the temperature of

1 g

of a given substance by

∘

In your case, aluminium is said to have a specific heat of

0.90

∘

So, what does that tell you?

In order to increase the temperature of

1 g

of aluminium by

∘

, you need to provide it with

0.90 J

of heat.

But remember, this is how much you need to provide for every gram of aluminium in order to increase its temperature by

∘

. So if you wanted to increase the temperature of

10.0 g

of aluminium by

∘

, you'd have to provide it with

1 gram



0.90 J

1 gram



0.90 J

...

1 gram



0.90 J



10 times

0.90 J

However, you don't want to increase the temperature of the sample by

∘

, you want to increase it by

∘

−

∘

This means that you're going to have to use that much heat for every degree Celsius you want the temperature to change. You can thus say that

∘



0.90 J

∘



0.90 J

...

∘



0.90 J



33 times

0.90 J

Therefore, the total amount of heat needed to increase the temperature of

10.0 g

of aluminium by

∘

will be

10.0

⋅

0.90

∘

⋅

∘

297 J

I'll leave the answer rounded to three sig figs, despite the fact that your values only justify two sig figs.

For future reference, this equation will come in handy

⋅

, where

- the amount of heat added / removed

- the mass of the substance

- the specific heat of the substance

- the change in temperature, defined as the difference between the final temperature and the initial temperature of the sample

6 0

3 years ago

Copper(II) sulfide is formed when copper and sulfur are heated together. In this reaction,

storchak [24]

Answer:

The mass of copper(II) sulfide formed is:

= 81.24 g

Explanation:

The Balanced chemical equation for this reaction is :

$Cu(s) + S\rightarrow CuS$

given mass= 54 g

Molar mass of Cu = 63.55 g/mol

$Moles = \frac{given\ mass}{Molar\ mass}$

$moles=\frac{54}{63.55}$

Moles of Cu = 0.8497 mol

Given mass = 42 g

Molar mass of S = 32.06 g/mol

$Moles = \frac{given\ mass}{Molar\ mass}$

$moles=\frac{42}{32.06}$

Moles of S = 1.31 mol

Limiting Reagent : The reagent which is present in less amount and consumed in a reaction

First find the limiting reagent :

$Cu + S\rightarrow CuS$

1 mol of Cu require = 1 mol of S

0.8497 mol of Cu should require = 1 x 0.8497 mol

= 0.8497 mol of S

S present in the reaction Medium = 1.31 mol

S Required = 0.8497 mol

S is present in excess and Cu is limiting reagent

All Cu is consumed in the reaction

Amount Cu will decide the amount of CuS formed

$Cu + S\rightarrow CuS$

1 mole of Cu gives = 1 mole of Copper sulfide

0.8497 mol of Cu = 1 x 0.8497 mole of Copper sulfide

= 0.8497

Molar mass of CuS = 95.611 g/mol

$Moles = \frac{given\ mass}{Molar\ mass}$

$0.8497 = \frac{given\ mass}{95.611}$

Mass of CuS = 0.8497 x 95.611

= 81.24 g

3 0

3 years ago

When would formation of a derivative be a necessary step in this experiment?

const2013 [10]

The formation of a derivative being a necessary step in the experiment lies in the importance of the derived structure. Often the derived product confers to reaction pathways which uses less reactive starting materials and more easily proceeds to completion. This also allows us to take a small amount of sample. The derived product at times is a general compound allowing its easy analysis. Often we encounter a product but we find it difficult to analyse it in ways we want. Here lies the essence of forming a derivative which often are simpler compounds allowing easier analysis yet having similar functional groups and structural properties. Also sometimes we encounter problems when our desired product is unstable and forms stable degraded products. But if we somehow manage to synthesize a derivative it may be relatively stable and form no degradation products. It would be stable at least for a significant period of time making it easier to study its properties. The derived product also at times are synthesized using general reaction pathways facilitating a way of easier synthesis and helping it to correlate with other similar reaction pathways and products.So the above paragraph accounts for the need of derivatives. When we encounter problems similar to those mentioned above it becomes necessary for a researcher to form rather synthesize a derivative.

6 0

3 years ago