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2 years ago

What two forces act in the nucleus to create a 'nuclear tug-of-war'?

Chemistry

1 answer:

Alex787 [66]2 years ago

5 0

Answer:

Electrostatic repulsion, strong nuclear force

Explanation:

The nucleus consists of protons and neutrons. protons are positively charged while neutrons possess no charge.

Since protons are positively charged, they repel each other strongly (like charges repel). This strong repulsion of like charges makes the nucleus somewhat unstable leading to spontaneous fission of heavy nuclei.

However, an opposing force called nuclear attractive force tends to hold the nucleons together. This attraction occurs when two nucleons are bonded by a particle called a π meson.

Hence, the two forces that act in the nucleus to create a 'nuclear tug-of-war' are electrostatic repulsion and a strong nuclear force.

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When the energy is removed what will happen to a gas?

creativ13 [48]

Answer:

Hey mate 
I hope it helps

Explanation:

<h3>Removing Energy: Removing energy will cause the particles in a liquid to begin locking into place. A. Boiling and Evaporation: Evaporation is the change of a substance from a liquid to a gas. Boiling is the change of a liquid to a vapor, or gas, throughout the liquid.</h3>

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2 years ago

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Determine the percentage composition of Ptl2

Hitman42 [59]

Answer:

Percent Composition

1. Find the molar mass of all the elements in the compound in grams per mole.

2. Find the molecular mass of the entire compound.

3. Divide the component's molar mass by the entire molecular mass.

4. You will now have a number between 0 and 1. Multiply it by 100% to get percent composition.

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1 year ago

Draw the Lewis structure for the Se and 2 H atoms.

fomenos

When drawing lewis dot diagram structure we consider the number of valence electrons of the atom.Se has six valence electrons since it is in group six thus it Lewis dot diagram is as follows
..
: Se:
Hydrogen is in group one hence has one valence electron.The lewis dot diagram for 2H is therefore
H:H

3 0

2 years ago

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As the pressure on the surface of a liquid decreases, the temperature at which the liquid will boil

ankoles [38]

Answer: option (1) decreases.

Explanation:

May be you have experienced that: when you go to the beach, where the atmposhpere pressure is greater than the atmosphere pressure in places that are at higher altitudes, the water takes longer to boil. That is because the boiling temperature is greater, and you need more total heat (more time) to permit the liquid to reach that temperature.

The reason why that happens is because substances boil when the vapor pressure (the pressure of the particles of vapor over the liquid) equals the atmosphere pressure. So, when the atmposhere pressure increases, the temperature at which the vapor pressure reaches the atmosphere pressure also increases, and when the atmosphere pressure decreases, the temperature at which the vapor pressure reaches the atmosphere pressure decreases.

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2 years ago

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Calculate the freezing point and boiling point of a solution containing 8.15 g of ethylene glycol (C2H6O2) in 96.3 mL of ethanol

pishuonlain [190]

Answer: The freezing point of solution is -117.54°C and the boiling point of solution is 80.48°C

Explanation:

To calculate the mass of ethanol, we use the equation:

$\text{Density of substance}=\frac{\text{Mass of substance}}{\text{Volume of substance}}$

Density of ethanol = 0.789 g/mL

Volume of ethanol = 96.3 mL

Putting values in above equation, we get:

$0.789g/mL=\frac{\text{Mass of ethanol}}{96.3mL}\\\\\text{Mass of ethanol}=(0.789g/mL\times 96.3mL)=75.98g$

Calculating the freezing point:

Depression in freezing point is defined as the difference in the freezing point of pure solution and freezing point of solution.

The equation used to calculate depression in freezing point follows:

$\Delta T_f=\text{Freezing point of pure solution}-\text{Freezing point of solution}$

To calculate the depression in freezing point, we use the equation:

$\Delta T_f=iK_fm$

Or,

$\text{Freezing point of pure solution}-\text{Freezing point of solution}=i\times K_f\times \frac{m_{solute}\times 1000}{M_{solute}\times W_{solvent}\text{ (in grams)}}$

where,

Freezing point of pure solution = -114.1 °C

i = Vant hoff factor = 1 (For non-electrolytes)

$K_f$ = molal freezing point elevation constant = 1.99°C/m

$m_{solute}$ = Given mass of solute (ethylene glycol) = 8.15 g

$M_{solute}$ = Molar mass of solute (ethylene glycol) = 62 g/mol

$W_{solvent}$ = Mass of solvent (ethanol) = 75.98 g

Putting values in above equation, we get:

$-114.1-\text{Freezing point of solution}=1\times 1.99^oC/m\times \frac{8.15\times 1000}{62g/mol\times 75.98}\\\\\text{Freezing point of solution}=-117.54^oC$

Hence, the freezing point of solution is -117.54°C

Calculating the boiling point:

Elevation in boiling point is defined as the difference in the boiling point of solution and freezing point of pure solution.

The equation used to calculate elevation in boiling point follows:

$\Delta T_b=\text{Boiling point of solution}-\text{Boiling point of pure solution}$

To calculate the elevation in boiling point, we use the equation:

$\Delta T_b=iK_bm$

Or,

$\text{Boiling point of solution}-\text{Boiling point of pure solution}=i\times K_b\times \frac{m_{solute}\times 1000}{M_{solute}\times W_{solvent}\text{ in grams}}$