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statuscvo [17]
3 years ago
5

Punnet squares, help how do I do this

Chemistry
1 answer:
vladimir2022 [97]3 years ago
6 0

Explanation:

Im not sure what the other trait is so I couldn't do phenotype. I hope this still helps though

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Over the course of many years, a rural farm town becomes more urban. Which of the following will most likely happen in the town
MrRa [10]

Answer:

Explanation:

Rivers may be contaminated by sewage.

I do hope I helped you! :)

6 0
3 years ago
Convert from 1.56×1030 particles of sodium chloride (NaCl) to grams of sodium chloride.
Anika [276]

Answer:

15.14×10⁷ g

Explanation:

Given data:

Number of particles of NaCl = 1.56×10³⁰ particles

Mass of sodium chloride = ?

Solution:

The given problem will solve by using Avogadro number.

It is the number of atoms , ions and molecules in one gram atom of element, one gram molecules of compound and one gram ions of a substance.  The number 6.022 × 10²³ is called Avogadro number.

1 mole = 6.022 × 10²³ particles

1.56×10³⁰ particles × 1 mol / 6.022 × 10²³ particles

0.259 ×10⁷ mol

Mass in gram:

Mass = number of moles × molar mass

Mass = 0.259 ×10⁷ mol × 58.44 g/mol

Mass = 15.14×10⁷ g

8 0
3 years ago
What compoponents of the atom are found outside of nucleus
liberstina [14]

Electrons are the smallest of the three particles that make up atoms. Electrons are found in shells or orbitals that surround the nucleus of an atom. Protons and neutrons are found in the nucleus. They group together in the center of the atom.

5 0
3 years ago
Calculate ΔH and ΔStot when two copper blocks, each of mass 10.0 kg, one at 100°C and the other at 0°C, are placed in contact in
ira [324]

Explanation:

The given data is as follows.

        m = 10.0 kg = 10,000 g    (as 1 kg = 1000 g)

      Initial temp. of block 1, T_{1} = 100^{o}C = (100 + 273) K = 373 K  

      Initial temp. of block 2, T_{2} = 0^{o}C = (0 + 273) K = 273 K

So, heat released by block 1 = heat gained by block 2

            mC \Delta T = mC \times \Delta T

  10000 g \times 0.385 \times (T_{f} - 100)^{o}C = 10000 g \times 0.385 \times (0 - T_{f})^{o}C

                  T_{f} - 100^{o}C = 0^{o}C - T_{f}    

                   2T_{f} = 100^{o}C

                          T_{f} = 50^{o}C

Convert temperature into kelvin as (50 + 273) K = 323 K.              

Also, we know that the relation between enthalpy and temperature change is as follows.

             \Delta H = mC \Delta T

                         = 10000 g \times 0.385 J/K g \times 323 K

                         = 1243550 J

or,                     = 1243.5 kJ

Now, calculate entropy change for block 1 as follows.

     \Delta S_{1} = mC ln \frac{T_{f}}{T_{i}}

            = 10000 g \times 0.385 J/K g \times ln \frac{323}{373}

            = 10000 g \times 0.385 J/K g \times -0.143

            = -554.12 J/K

Now, entropy change for block 2 is as follows.

   \Delta S_{2} = mC ln \frac{T_{f}}{T_{i}}

           = 10000 g \times 0.385 J/K g \times ln \frac{323}{273}

           = 10000 g \times 0.385 J/K g \times 0.168

           = 647.49 J/K

Hence, total entropy will be sum of entropy change of both the blocks.

            \Delta S_{total} = \Delta S_{1} + \Delta S_{2}

                       = -554.12 J/K + 647.49 J/K

                       = 93.37 J/K

Thus, we can conclude that for the given reaction \Delta H is 1243.5 kJ and \Delta S_{total} is 93.37 J/K.

6 0
3 years ago
When water changes from one state to another each process has a name what are the names of the processes that occur in each stat
Nastasia [14]
Ice, freezing, or melting then liquid evaporation or condensation, the vapor, deposition the ice again
4 0
3 years ago
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