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Studentka2010 [4]
4 years ago
9

If the exact outer limit of an isolated atom cannot be measured, what criterion can we use to determine atomic radii? What is th

e difference between a covalent radius and a metallic radius?
Chemistry
1 answer:
Alik [6]4 years ago
7 0

Answer:

Calculate the atomic radii of two touching or overlapping atoms.

Explanation:

No doubt, we can't find the atomic boundary of a single atom, but when atoms are in the form of pairs it becomes very easy to measure the atomic radii of two and then dividing it by 2 to get an estimate of atomic radius of a single atom.

It is also called as covalent radius which is half of the total inter-nuclear distance between two same bonded atoms (Homo-nuclear).

If two adjacent mettalic ions are joined by such pairing then the same half of the distance between the nucleus is termed as metallic radii.

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Consider the exothermic reaction
alisha [4.7K]

Answer:

The answer to your question is -2855 J

Explanation:

Reaction

                     2C₂H₆  +  7O₂   ⇒   4CO₂  +  6H₂O

Formula

Heat of reaction = ΔHrxn = ΣΔHrxn products - ΣΔHrxn reactants

Substitution

ΔHrxn = { 4(-393.5) + 6(-241.8)} - {2(-84.7) + 7(0)}

ΔHrxn = {-1574 -1450.8} - {-169.4}

ΔHrxn = -3024.8 + 169.4

ΔHrxn = -2855.4 J

4 0
3 years ago
Convert 45 joules to heat calories
skad [1K]

Answer:

45 joules to calories= 10 (10.755258) calories

3 0
4 years ago
Calculate the activation energy (ea in kj/mol for a reaction if the rate constant (k is 2.7 × 10-4 (m−1sec−1 at 600 k and 3.5 ×
djyliett [7]
Below are the choices:

a. −166 kJ/mol 
<span>b. 166 kJ/mol </span>
<span>c. 1.64 kJ/mol </span>
<span>d. 1.66 × 10^5 kJ/mol
</span>
To calculate the activation energy of a reaction, we use the Arrhenius equation.  You may want to look it up to see how and why it works.  In the problem you posted, there are two temperatures and two rate constants.  After some rearranging and substitution of the Arrhenius equation, we have Ea = R T1 T2/(T1-T2) ln(k1/k2) = 8.314 J/mol K (600 K)(650 K)/(600 K-650 K) ln(2.7×10^-4 M^−1sec^−1/3.5×10^−3 M−^1sec^−1) = 166145 J/mol = 166 kJ/mol => choice b
6 0
3 years ago
I need help on the first and second one please help
svetoff [14.1K]
The hypothesis would be that more green bugs would be found than bright red bugs because the green bugs are more camouflaged asunder the green plants than the bright red bugs.
4 0
4 years ago
which equation is setup correctly to determine the volume of a 1.5 mol sample of oxygen gas at 22 degrees Celsius and 100 kPa
rjkz [21]
Hello!

We have the following data:


v (volume) = ? (in L)
n (number of mols) = 1,5 mol
T (temperature) = 22 ºC 
First let's convert the temperature on the Kelvin scale, let's see:
TK = TºC + 273,15
TK = 22 + 273,15
TK = 295,15

P (pressure) = 100 kPa → P = 100000 Pa → P ≈ 0,987 atm
R (gas constant) = 0,082 atm.L / mol.K

<span>We apply the data above to the Clapeyron equation (gas equation), let's see:

</span>P*V = n*R*T

0,987*V = 1,5*0,082*295,15

0,987V = 36,30345

V =  \dfrac{36,30345}{0,987}

\boxed{\boxed{V \approx 36,78\:L}} \end{array}}\qquad\checkmark

I hope this helps. =)
4 0
4 years ago
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