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4 years ago

The pictures to the right, show two different models

Chemistry

2 answers:

OverLord2011 [107]4 years ago

8 0

Answer:model B

Explanation:

Edge 2020

Svetllana [295]4 years ago

6 0

Answer:

The pictures to the right, show two different models of the atom. Type in the letter of your answer.

Which model best represents Dalton’s atomic theory?

Which model best represents the modifications to the theory that Thomson’s results made necessary?

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Hello how are u guys today?

Lyrx [107]

Answer:

hello

Explanation:

I am good and you , hope your doing great lol

8 0

3 years ago

Find the cell potential for a system whose ∆G = +55 kJ and 3 moles of electrons are exchanged.

rosijanka [135]

Answer:- cell potential = -0.19 volts

Solution:- The equation that shows the connection between $\Delta G$ and cell potential, E is written as:

$\Delta G=-nFE$

in this equation, n stands for moles of electrons, E stands for cell potential and F stands for faraday constant and it's value is $\frac{96485C}{mol}$ .

It asks to calculate the value of E, so let's rearrange the equation:

$E=-\frac{\Delta G}{nF}$

Let's plug in the values in it:

$E=-\frac{55kJ}{3mol*96485C.mol^-}$

$E=-\frac{0.00019kJ}{C}$

since, $\frac{1kJ}{C}=1000V$

Where C stands for coulombs and V stands for volts.

So, $E=-\frac{0.00019kJ}{C}(\frac{1000V}{\frac{1kJ}{C}})$

E = -0.19 V

So, the cell potential is -0.19 volts.

4 0

3 years ago

Need this ASAP!

AleksandrR [38]

Answer:

B. Lower than 100 °C because hydrogen sulfide has dipole-dipole interactions instead of hydrogen bonding.

Explanation:

Boiling point is a physical property which is usually a product of breaking intermolecular bonds.

Both dipole-dipole attractions are intermolecular bonds and they have serious effect on boiling point of a substance.

Hydrogen bonds are very strong intermolecular bonds compared to dipole-dipole attractions. In hydrogen bonding hydrogen atom is directly joined to a highly electronegative atom.

Dipole-Dipole attraction exists between molcules that are polar. Such molecules line up such that the positive pole of one molecule attracts the negative pole of another.

Hydrogen bonds in water are much stronger than the dipole-dipole attraction of hydrogen sulfide.

5 0

3 years ago

Read 2 more answers

Would an earthquake support the principle of uniformitarianism or the principle of catastrophism? Explain.

liberstina [14]

Question: Would an earthquake support the principle of uniformitarianism or the principle of catastrophism? Explain.

Answer: Catastrophism was the theory that the Earth had largely been shaped by sudden, short-lived, violent events, possibly worldwide in scope. This was in contrast to uniformitarianism (sometimes described as gradualism), in which slow incremental changes, such as erosion, created all the Earth's geological features.

Hope this helps, have a good day. c;

3 0

3 years ago

PLSSSSS HELP I DONT GET THIS PROBLEMMMM

Aleks [24]

Answer:

C. 7370 joules.

Explanation:

There is a mistake in the statement. Correct form is described below:

<em>Using the above data table and graph, calculate the total energy in Joules required to raise the temperature of 15 grams of ice at -5.00 °C to water at 35 °C. </em>

The total energy needed to raise the temperature is the combination of latent and sensible heats, all measured in joules, and represented by the following model:

$Q = m\cdot [c_{i} \cdot (T_{2}-T_{1})+L_{f} + c_{w}\cdot (T_{3}-T_{2})]$ (1)

Where:

$m$ - Mass of the sample, in grams.

$c_{i}$ - Specific heat of ice, in joules per gram-degree Celsius.

$c_{w}$ - Specific heat of water, in joules per gram-degree Celsius.

$L_{f}$ - Latent heat of fusion, in joules per gram.

$T_{1}$ - Initial temperature of the sample, in degrees Celsius.

$T_{2}$ - Melting point of water, in degrees Celsius.

$T_{3}$ - Final temperature of water, in degrees Celsius.

$Q$ - Total energy, in joules.

If we know that $m = 15\,g$ , $c_{i} = 2.06\,\frac{J}{g\cdot ^{\circ}C}$ , $c_{w} = 4.184\,\frac{J}{g\cdot ^{\circ}C}$ , $L_{f} = 334.72\,\frac{J}{g}$ , $T_{1} = -5\,^{\circ}C$ , $T_{2} = 0\,^{\circ}C$ and $T_{3} = 35\,^{\circ}C$ , then the final energy to raise the temperature of the sample is:

$Q = (15\,g)\cdot \left[\left(2.06\,\frac{J}{g\cdot ^{\circ}C} \right)\cdot (5\,^{\circ}C)+ 334.72\,\frac{J}{g} + \left(4.184\,\frac{J}{g\cdot ^{\circ}C}\right)\cdot (35\,^{\circ}C) \right]$

$Q = 7371.9\,J$

Hence, the correct answer is C.

8 0

3 years ago