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Oksi-84 [34.3K]
3 years ago
12

Two litres of an ideal gas at a pressure of 10 atm expands isothermally into a vacuum until its total volume is 10 litres. How m

uch heat is absorbed and how much work is done in the expansion ?
Chemistry
1 answer:
nlexa [21]3 years ago
4 0

Answer:

The work done and heat absorbed are both -8,1 kJ

Explanation:

The work done in an isobaric process is defined as:

W = -P (Vf - Vi)

Where P is pressure ( 10 atm)

Vf = 10 L

Vi = 2 L

Thus, <em>W = -80 atm×L ≡ -8,1 kJ</em>

This is the work done in expansion of the gas.  As the gas remains at the same temperature, there is no change in internal energy doing that all work was absorbed as heat.

I hope it helps!

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How many milliliters of a 1.25 molar HCL solution would be needed to react completely with 60 g of calcium metal
Svet_ta [14]
Ca + 2HCl = CaCl₂ + H₂

m(Ca)=60 g
c=1.25 mol/l
M(Ca)=40g/mol
v-?

m(Ca)/M(Ca)=m(HCl)/[2M(HCl)]=n(HCl)/2

n(HCl)=2m(Ca)/M(Ca)

n(HCl)=cv

cv=2m(Ca)/M(Ca)

v=2m(Ca)/{cM(Ca)}

v=2·60g/mol/{1.25mol/l·40g/mol}=2.4 l = 2400 ml

2400 milliliters of a 1.25 molar HCl solution would be needed

3 0
3 years ago
What is the molarity of a 250.0 milliliter aqueous solution of sodium hydroxide that contains 15.5 grams of solute
tensa zangetsu [6.8K]

There are a number of ways to express concentration of a solution. This includes molarity. Molarity is expressed as the number of moles of solute per volume of the solution. The concentration of the solution is calculated as follows:

 <span> </span><span>Molarity = 15.5 g NaOH (1 mol NaOH / 40 g NaOH)  / .250 L solution</span>

<span>Molarity = 1.55 M</span>

5 0
3 years ago
The density of a 3.37M MgCl2 (FW = 95.21) is 1.25 g/mL. Calulate the molality, mass/mass percent, and mass/volume percent. So fa
Dafna1 [17]

Answer : The molality, mass/mass percent, and mass/volume percent are, 0.0381 mole/Kg, 25.67 % and 32.086 % respectively.

Solution : Given,

Density of solution = 1.25 g/ml

Molar mass of MgCl_2 (solute) = 95.21 g/mole

3.37 M magnesium chloride means that 3.37 gram of magnesium chloride is present in 1 liter of solution.

The volume of solution = 1 L = 1000 ml

Mass of MgCl_2 (solute) = 3.37 g

First we have to calculate the mass of solute.

\text{Mass of }MgCl_2=\text{Moles of }MgCl_2\times \text{Molar mass of }MgCl_2

\text{Mass of }MgCl_2=3.37mole\times 95.21g/mole=320.86g

Now we have to calculate the mass of solution.

\text{Mass of solution}=\text{Density of solution}\times \text{Volume of solution}=1.25g/ml\times 1000ml=1250g

Mass of solvent = Mass of solution - Mass of solute = 1250 - 320.86 = 929.14 g

Now we have to calculate the molality of the solution.

Molality=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Mass of solvent}}=\frac{3.37g\times 1000}{95.21g/mole\times 929.14g}=0.0381mole/Kg

The molality of the solution is, 0.0381 mole/Kg.

Now we have to calculate the mass/mass percent.

\text{Mass by mass percent}=\frac{\text{Mass of solute}}{\text{Mass of solution}}\times 100=\frac{320.86}{1250}\times 100=25.67\%

The mass/mass percent is, 25.67 %

Now we have to calculate the mass/volume percent.

\text{Mass by volume percent}=\frac{\text{Mass of solute}}{\text{Volume of solution}}\times 100=\frac{320.86}{1000}\times 100=32.086\%

The mass/volume percent is, 32.086 %

Therefore, the molality, mass/mass percent, and mass/volume percent are, 0.0381 mole/Kg, 25.67 % and 32.086 % respectively.

8 0
3 years ago
What is the mass of 2.23 × 10^23 atoms of aluminum?​
rjkz [21]

Answer:

10

Explanation:

In image

Take atomic mass or molar mass

of Al =27

7 0
3 years ago
Robert has pure samples of both D-ribose and D-arabinose, but he forgot to label them. He only has some nitric acid, the reagent
Sauron [17]

Answer:

Following are the responses to the given question:

Explanation:

Since HN03 is an oxidation substance D-ribose u.ith oxidized to form in rubric acid Ribose is chiral, but rubric acid is achiral because of its symmetry mirror level, Hence no infrared roster in the sample holder is observed.

Please find the attached file.

D-Arabinose, on either hand, gives optical aldaric acid with such a net optical rotation observed inside the polarimeter for diagnosis with HN03.

4 0
3 years ago
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