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Artemon [7]
3 years ago
5

The range in size of most atomic radii is approximately _____.

Chemistry
2 answers:
Mashutka [201]3 years ago
8 0
The range in size of most atomic radii is approximately <span>5 × 10−21 m to 2 × 10−20 m. The rest of the choices do not answer the question above.</span>
Novay_Z [31]3 years ago
3 0

The atomic radii is approximately between 5 × 10−11 m to 2 × 10−10 m



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An aqueous solution contains 32.7% kcl (weight/weight %). how many grams of water are contained in 100 g of this solution
igor_vitrenko [27]
Weight/weight percentage is the percentage of mass of solute from the mass of the whole solution
w/w % = 32.7 %
this means that in a solution of 100 g - mass of KCl is 32.7 g 
since solution is made of water and KCl
then the mass of water is - 100 - 32.7 = 67.3 g
therefore in 100 g of solution - 67.3 g of the solution is water
5 0
3 years ago
Petrol does not catch fire on its own at room temperature. Give Reason
tino4ka555 [31]
Because room temperature is below the ignition temperature of petrol. At this temperature energy in the form of heat, flame, or electric spark is needed to start the reaction.
3 0
3 years ago
The image shows sedimentary rock layers with index fossils and a fault.
GrogVix [38]

Answer:

Layer 2 and layer 9 is the same relative age since it is the same type of rock and has the same fossils.

Hope this helped!

4 0
3 years ago
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

4 0
3 years ago
Using the thermodynamic information in the ALEKS Data tab, calculate the standard reaction entropy of the following chemical rea
Aleksandr [31]

Answer:

ΔS° = - 47.2 J/mol.K

Explanation:

  • P4O10 + 6H2O → 4H3PO4

ΔS°= 4(S°mH3PO4) - 6(S°mH2O) - S°mP4O10

∴ S°mH2O(l) = 69.9 J/mol.K

∴ S°mP4O10 = 231 J/mol.K

∴ S°mH3PO4 = 150.8 J/mol.K

⇒ ΔS° = 4*(150.8) - 6*(69.9) - 231

⇒ ΔS° = - 47.2 J/mol.K

5 0
3 years ago
Read 2 more answers
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