1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
riadik2000 [5.3K]
3 years ago
12

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

<em>Q1. Mass of Hg </em>

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

<em>Q2. Mass of Pb </em>

<em>Step 1</em>. Calculate the <em>volume of the Pb</em>.

<em>V = lwh</em> = 6.0 m × 3.5 m × 2.0 m = 42.0 m^3

<em>Step 2</em>. Calculate the <em>mass of the Pb</em>.

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

<em>Q3. Volume of displaced water </em>

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

<em>Archimedes</em>: volume of displaced water = volume of Ag = <em>0.106 cm^3</em>

<em>4. Density of metal </em>

<em>Step 1</em>. Convert <em>ounces to grams </em>

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

<em>Step 2</em>. Calculate the <em>volume in cubic inches </em>

<em>V = lwh</em> = 3.0 in × 2.5 in × 0.45 in = 3.38 in^3

<em>Step 3</em>. Convert <em>cubic inches to cubic centimetres</em><em> </em>

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

<em>Step 4</em>. Calculate the <em>density</em>

ρ = <em>m</em>/<em>V</em> = (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 ofBaSO_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

so

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

<span>volume Na2S = 0.0212 L = 21.2 mL</span>

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
Other questions:
  • Consider the reaction 2CO(g) + O2(g)2CO2(g) Using standard thermodynamic data at 298K, calculate the entropy change for the surr
    7·1 answer
  • PLEASE SOMEBODY HELP PLEASE ! I WILL MARK BRAINLY I PROMISE !
    5·1 answer
  • 29. A certain nut crunch cereal contains 11.0 grams of sugar (sucrose, C12H22O11) per serving size of 60.0 grams. How many servi
    12·1 answer
  • 1. 51.2 g of NaClO (pKa of HClO is 7.50) and 25.3 g of KF (pKa of of HF is 3.17) were combined in enough water to make 0.250 L o
    8·1 answer
  • The type of current produced by a maget is
    8·1 answer
  • pls help If we multiply the mass of an object and the acceleration due to gravity, we find the downward force of the object whic
    5·1 answer
  • The pressure of a compressed gas is 1.45 atm. What is this<br> pressure in kPa?<br> kPa
    11·1 answer
  • If one jellybean weighs 0.350g how many jelly beans are in 504g of jellybeans?
    5·1 answer
  • Consider a hypothetical gas which has the following Van der Waals constants:
    11·2 answers
  • How many valence electrons does sr2+ have
    7·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!