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3 years ago

Describe and name the parts of the atom.

Chemistry

2 answers:

Elan Coil [88]3 years ago

8 0

Answer:

protons, electrons, and neutrons.

A proton is positively charged and is located in the center or nucleus of the atom.

Electrons are negatively charged and are located in rings or orbits spinning around the nucleus.

The number of protons and electrons is always equal.

Explanation:

Setler79 [48]3 years ago

4 0

Answer:

The parts of an atom are<em><u> protons, electrons, and neutrons.</u></em>

A proton is positively charged and is located in the center or nucleus of the atom.

Electrons are negatively charged and are located in rings or orbits spinning around the nucleus.

The number of protons and electrons is always equal.

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An ethylene glycol solution contains 21.4 g of ethylene glycol (C2H6O2) in 97.6 mL of water. (Assume a density of 1.00 g/mL for

8090 [49]

Answer: The freezing point and boiling point of the solution are $-6.6^0C$ and $101.8^0C$ respectively.

Explanation:

Depression in freezing point:

$T_f^0-T^f=i\times k_f\times \frac{w_2\times 1000}{M_2\times w_1}$

where,

$T_f$ = freezing point of solution = ?

$T^o_f$ = freezing point of water = $0^0C$

$k_f$ = freezing point constant of water = $1.86^0C/m$

i = vant hoff factor = 1 ( for non electrolytes)

m = molality

$w_2$ = mass of solute (ethylene glycol) = 21.4 g

$w_1$ = mass of solvent (water) = $density\times volume=1.00g/ml\times 97.6ml=97.6g$

$M_2$ = molar mass of solute (ethylene glycol) = 62g/mol

Now put all the given values in the above formula, we get:

$(0-T_f)^0C=1\times (1.86^0C/m)\times \frac{(21.4g)\times 1000}{97.6g\times (62g/mol)}$

$T_f=-6.6^0C$

Therefore,the freezing point of the solution is $-6.6^0C$

Elevation in boiling point :

$T_b-T^b^0=i\times k_b\times \frac{w_2\times 1000}{M_2\times w_1}$

where,

$T_b$ = boiling point of solution = ?

$T^o_b$ = boiling point of water = $100^0C$

$k_b$ = boiling point constant of water = $0.52^0C/m$

i = vant hoff factor = 1 ( for non electrolytes)

m = molality

$w_2$ = mass of solute (ethylene glycol) = 21.4 g

$w_1$ = mass of solvent (water) = $density\times volume=1.00g/ml\times 97.6ml=97.6g$

$M_2$ = molar mass of solute (ethylene glycol) = 62g/mol

Now put all the given values in the above formula, we get:

$(T_b-100)^0C=1\times (0.52^0C/m)\times \frac{(21.4g)\times 1000}{97.6g\times (62g/mol)}$

$T_b=101.8^0C$

Thus the boiling point of the solution is $101.8^0C$

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￼ Describe and name the parts of the atom.

Describe and name the parts of the atom.