All categories

3 years ago

A sample of nitrogen gas expands in volume from 16 L to 5.4 L at constant

Chemistry

1 answer:

xxTIMURxx [149]3 years ago

6 0

a. W = 0 J

b. W = - 308.028 J

<h3>Further explanation</h3>

Given

Nitrogen gas expands in volume from 1.6 L to 5.4 L

Required

The work done

Solution

Isothermal :

W = -P . ΔV

Input the value :

a. At a vacuum, P = 0

So W = 0

b. At pressure = 0.8 atm

W = - 0.8 x ( 5.4 - 1.6)

W = -3.04 L.atm ( 1 L.atm = 101.325 J)

W = - 3.04 x 101.325

W = - 308.028 J

You might be interested in

What are the difficulties encountered in precipitation titration

Mumz [18]

One difficulty encountered in precipitation titration is that it is hard to determine the exact end point of its reaction.

Precipitation titration is a titration in which a reaction occurs from the analyte and titrant to form an insoluble precipitate.

With the use of silver for the titrations, (argentometric) we are able to develop many precipitation reactions.

The precipitation titrimetry methods with the use of argentometry includes

• Mohr’s Method

• Fajan’s Method

• Volhard’s Method

Difficulties encountered in precipitation titration includes

Getting the exact end point is hard.

it is a very slow titration method.

it includes periods of filtration and cooling thereby reducing the reactions available for this type of titration.

See more on Precipitation: brainly.com/question/20628792

8 0

2 years ago

I need to check some chemistry questions. Help with any question is appreciated! :) Please include an explanation with your conc

finlep [7]

1. 4.67 kg; 2. 4.8 ×10^5 kg; 3. 0.106 cm^3; 4. 1.7 g/cm^3

Q1. Mass of Hg

Mass = 345 mL × (13.53 g/1 mL) = 4670 g = 4.67 kg

Q2. Mass of Pb

Step 1. Calculate the volume of the Pb.

V = lwh = 6.0 m × 3.5 m × 2.0 m = 42.0 m^3

Step 2. Calculate the mass of the Pb.

Mass = 42.0 m^3 × (11 340 kg/1 m^3) = 4.8 × 10^5 kg

Q3. Volume of displaced water

Volume of Ag = 0.987 g × (1 cm^3/9.320 g) = 0.106 cm^3

Archimedes: volume of displaced water = volume of Ag = 0.106 cm^3

4. Density of metal

Step 1. Convert ounces to grams

Mass = 3.35 oz × (28.35 g/1 oz) = 94.97 g

Step 2. Calculate the volume in cubic inches

V = lwh = 3.0 in × 2.5 in × 0.45 in = 3.38 in^3

Step 3. Convert cubic inches to cubic centimetres

V = 3.38 in^3 × (2.54 cm/1 in)^3 = 55.3 cm^3

Step 4. Calculate the density

ρ = m/V = (94.97 g/55.3 cm^3) = 1.7 g/cm^3 (magnesium?)

4 0

3 years ago

A 254.5 g sample of a white solid is known to be a mixture of KNO3, BaCl2, and NaCl. When 116.5 g of this mixture is dissolved i

Margaret [11]

Answer:

Mass of KNO3 in the original mix is 146.954 g

Explanation:

mass of $KNO_3$ in original 254.5 mixture.

moles of $BaSO_4 = \frac{mass}{Molecular\ Weight}$

moles of $BaSO_4 = \frac{68.3}{233.38}$

= 0.2926 mol of BaSO4

Therefore,

0.2926 mol of BaCl2,

mass of $BaCl_2 = mol\times molecular weight$

$= 0.2926\times 208.23$

= 60.92 g

the AgCl moles $= \frac{mass}{Molecular\ Weight}$

$= \frac{199.1}{143.32}$

= 1.3891 mol of AgCl

note that, the Cl- derive from both, $BACl_2 and NaCl$

mole of Cl- f NaCl $= (1.3891) - (0.2926\times 2) = 0.8039$ mol of Cl-

mol of NaCl = 0.8039 moles

$mass = mol\times Molecular\ Weight = 0.8039 \times 58 = 46.626\ g \ of \ NaCl$

then

KNO3 mass = 254.5 - 60.92-46.626 = 146.954 g of KNO_3

Mass of KNO3 in the original mix is 146.954 g

8 0

3 years ago

How many milliliters of 0.260 m na2s are needed to react with 35.00 ml of 0.315 m agno3?

allochka39001 [22]

The complete balanced chemical reaction is:

2 AgNO3 + Na2S --> 2 NaNO3 + Ag2S

First let us calculate the number of moles of AgNO3.

moles AgNO3 = 0.315 M * 0.035 L

moles AgNO3 = 0.011025 mol

From the reaction, 1 mole of Na2S is needed for every 2 moles of AgNO3 hence:

moles Na2S required = 0.011025 mol AgNO3 * (1 mol Na2S / 2 mol AgNO3)

moles Na2S required = 5.5125 x 10^-3 mol

Therefore volume required is:

volume Na2S = 5.5125 x 10^-3 mol / 0.260 M

volume Na2S = 0.0212 L = 21.2 mL

6 0

3 years ago

When 100 mL of 1.0 M Na3PO4 is mixed with 100 mL of 1.0 M AgNO3, a yellow precipitate forms and [Ag ] becomes negligibly small.

lana [24]

Answer:

$Na3PO4 + 3AgNO3 -------> Ag3PO4 + 3NaNO3$

In which [Ag+] in negligibly small and the concentration of each reactant is 1.0 M

The answer is A) PO43- < NO3- < Na+

Explanation:

Ag+ is removed from the solution just like PO43-, so there are just 2 possible answers at this point: a or b. Then we can notice that Na3PO4 releases 3 moles of Na+ and just 1 mole of NO3-

We have 100mL of each reactant with the same concentration for both (1.0 M) so:

(0.1)(1)(3)= 0.3 mol Na+

(0.1)(1)= 0.1 mol NO3-

so PO43- < NO3- < Na+

5 0

3 years ago