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xxMikexx [17]
2 years ago
9

How many varmints does it take to ruin a chemist's lawn

Chemistry
1 answer:
I am Lyosha [343]2 years ago
3 0
This was a riddle that has been included in a chemistry test.

The answer was a combination of letters that served as answers to previous questions.

How many varmints does it take to ruin a chemist's lawn? MOLES
You might be interested in
Ammonia gas is compressed from 21°C and 200 kPa to 1000 kPa in an adiabatic compressor with an efficiency of 0.82. Estimate the
Evgen [1.6K]

Explanation:

It is known that efficiency is denoted by \eta.

The given data is as follows.

     \eta = 0.82,       T_{1} = (21 + 273) K = 294 K

     P_{1} = 200 kPa,     P_{2} = 1000 kPa

Therefore, calculate the final temperature as follows.

         \eta = \frac{T_{2} - T_{1}}{T_{2}}    

         0.82 = \frac{T_{2} - 294 K}{T_{2}}    

          T_{2} = 1633 K

Final temperature in degree celsius = (1633 - 273)^{o}C

                                                            = 1360^{o}C

Now, we will calculate the entropy as follows.

       \Delta S = nC_{v} ln \frac{T_{2}}{T_{1}} + nR ln \frac{P_{1}}{P_{2}}

For 1 mole,  \Delta S = C_{v} ln \frac{T_{2}}{T_{1}} + R ln \frac{P_{1}}{P_{2}}

It is known that for NH_{3} the value of C_{v} = 0.028 kJ/mol.

Therefore, putting the given values into the above formula as follows.

     \Delta S = C_{v} ln \frac{T_{2}}{T_{1}} + R ln \frac{P_{1}}{P_{2}}

                = 0.028 kJ/mol \times ln \frac{1633}{294} + 8.314 \times 10^{-3} kJ \times ln \frac{200}{1000}

                = 0.0346 kJ/mol

or,             = 34.6 J/mol             (as 1 kJ = 1000 J)

Therefore, entropy change of ammonia is 34.6 J/mol.

3 0
1 year ago
Determine the PH at the point in the titration of 40.0ml of 0.200M HC4H7o2 with 0.100 M Sr(OH)2 after 100ml of the strong base h
Dmitriy789 [7]

Answer:

Check the explanation

Explanation:

Mols HC4H7O2 = (volume in L)*(molarity) = (40.0 mL)*(0.200 M)

= (40.0 mL)*(1 L)/(1000 mL)*(0.200 M)

= 8.00*10-3 mol.

Mols Sr(OH)2 corresponding to 10.0 mL of 0.100 M solution =

(volume in L)*(molarity)

= (10.0 mL)*(0.100 M)

= (10.0 mL)*(1 L)/(1000 mL)*(0.100 M)

= 1.00*10-3 mol.

Consider the ionization of Sr(OH)2 as below.

Sr(OH)2 (aq) ----------> Sr2+ (aq) + 2 OH- (aq)

As per the stoichiometric equation,

1 mol Sr(OH)2 = 2 mols OH-.

Therefore,

0.0010 mol Sr(OH)2 = [0.0010 mol Sr(OH)2]*(2 mols OH-)/[1 mole Sr(OH)2]

= 0.0020 mol

= 2.00*10-3 mol

Set up the ICE charts as below.

HC4H7O2 (aq) + OH- (aq) ------------> H2O (l) + C4H7O2- (aq)

Before (mol)        8.00*10-3         2.00*10-3                           -                -

Change (mol)      -2.00*10-3       -2.00*10-3                           -        +2.00*10-3

After (mol)           6.00*10-3                0                                  -          2.00*10-3

The change in a pure substance, e.g., H2O is not considered in an acid-base reaction.

Volume of the solution = (40.0 + 10.0) mL = 50.0 mL = (50.0 mL)*(1 L)/(1000 mL) = 0.05 L.

The initial concentrations are obtained by dividing the numbers of moles by the volume, 0.05 L.

Set up the ICE charts as below.

HC4H7O2 (aq) + OH- (aq) ------------> H2O (l) + C4H7O2- (aq)

Initial (M)             0.160               0.0400                             -                -

Change (M)        -0.0400            -0.0400                            -           +0.0400

Equilibrium (M) 0.120                     0                                  -            0.0400

The acid-ionization constant is written as

Ka = [H3O+][C4H7O2-]/[HC4H7O2] = 1.5*10-5

Plug in the known values and get

Ka = [H3O+]*(0.0400)/(0.120) = 1.5*10-5

======> [H3O+] = (1.5*10-5)*(0.120)/(0.0400) (ignore units)

======> [H3O+] = 4.5*10-5

The proton concentration of the solution is 4.5*10-5 M.

pH = -log (4.5*10-5 M)

= 4.346

≈ 4.35 (ans).

8 0
1 year ago
Which of the following is not a nanoscale object?
KiRa [710]
If I am correct I would believe that it would be a muscle cell.
6 0
1 year ago
Read 2 more answers
Temperatures in Earth's interior increase with increasing depth. Seismic waves help to provide information on temperatures in Ea
Valentin [98]

Answer:

Temperature affects Seismic Wave speed.

Explanation:

Both temperature and pressure affect the speed of Seismic waves. The Speed of Seismic waves increases uniformly as pressure increases, meaning that as depth increases, pressure also increases which causes Seismic Wave speeds to increase as well. This can be calculated and the data can be gathered. Temperature on the other hand decreases the speed of Seismic Waves, therefore we can calculate the difference of speed between what the Seismic Wave should be at a certain pressure with the actual speed gathered. This difference in speed will allow us to determine the actual temperature at that level.

3 0
2 years ago
A lab director asks two students, Lydia and Damien, to each select a bottle of a concentrated weak base. When they reach the
Hunter-Best [27]

Answer:

First one is 5.0 M ammonia and the Second one ?

Explanation:

6 0
2 years ago
Read 2 more answers
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