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Sergio039 [100]
1 year ago
10

For single bonds between similar types of atoms, how does the strength of the bond relate to the sizes of the atoms? Explain.

Chemistry
1 answer:
uysha [10]1 year ago
7 0

When comparing single bonds between atoms of comparable types, the stronger the bond is, the bigger the atom, the weaker it is.

The length of the X-H bond lengthens while the strength of the bond shortens with increasing halogen size (F-H strongest, I-H weakest). When comparing single bonds between atoms of similar sorts, the larger the atom, the weaker the bond. It can be explained by the fact that less energy is required to break the bond the bigger the atom's atomic size. The force of attraction from the nucleus to the outermost orbit will be less for iodine since it has a larger atom than the other elements in the group.

Learn more about single bonds here-

brainly.com/question/16626126

#SPJ4

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What is the mass of 2.5 mol of ca, which has a molar mass of g/mol
swat32
Atomic mass Ca = 40 a.m.u

1 mole Ca ----------- 40 g
2.5 mols Ca -------- ( mass Ca )

Mass Ca = 2.5 x 40 / 1

Mass Ca = 100 / 1

= 100 g of Ca

hope this helps!
3 0
3 years ago
What would happen if hydrogen and nitrogen combine?
Tanzania [10]
<span>At room temperature and atmospheric pressure, nothing happens when the two gasses are mixed. However, at high temperature and pressure (450C, 200atm), in the presence of an iron oxide catalyst, the production of ammonia is thermodynamically advantageous.</span>
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2 years ago
Refer to the graph. the graph shows the progress of digestion as carbohydrates, fats, and proteins pass through the human digest
nika2105 [10]
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8 0
3 years ago
Calculate the standard reaction enthalpy for the reaction NO2(g) → NO(g) + O(g) given +142.7 kJ/mol for the standard enthalpy of
bulgar [2K]

Answer:

The standard reaction enthalpy for the given reaction is 235.15 kJ/mol.

Explanation:

O_2(g) \rightarrow \frac{2}{3}O_3(g),\Delta H^o_{1}=142.7 kJ/mol..[1]

O_2(g) \rightarrow 2 O(g),\Delta H^o_{2}=498.4 kJ/mol..[2]

NO(g) + O_3(g)\rightarrow NO_2(g) + O_2(g) ,\Delta H^o_{3} = -200 kJ/mol..[3]

NO_2(g)\rightarrow NO(g) + O(g),\Delta H^o_{4}=?..[4]

Using Hess's law:

Hess’s law of constant heat summation states that the amount of heat absorbed or evolved in a given chemical equation remains the same whether the process occurs in one step or several steps.

2 × [4] = [2]- (3 ) × [1] - (2) × [3]

2\times \Delta H^o_{4}=\Delta H^o_{2} -3\times \Delta H^o_{1}-2\times \Delta H^o_{3}

2\times \Delta H^o_{4}=498.4 kJ/mol-3\times 142.7 kJ/mol-2\times -200 kJ/mol

2\times \Delta H^o_{4}=470.3 kJ/mol

\Delta H^o_{4}=\frac{470.3 kJ/mol}{2}=235.15 kJ/mol

The standard reaction enthalpy for the given reaction is 235.15 kJ/mol.

7 0
2 years ago
A student made an initial 1:5 dilution of protein lysate. Then 2mL of that was added to 8mL of water. Lastly, the student made a
Licemer1 [7]

Answer:

The final dilution is 1:400

Explanation:

Let's analyze what we are told: we have an initial 1:5 dilution of protein lysate. This means that the initial solution (stock solution) was diluted 5 times. Then, from this dilution the student prepared another dilution taking 2 mL of the first dilution in 8 mL of water. This is the same as saying we took 1 mL of first dilution in 4 mL of water (the ratio is the same), so we now have a second 1:4 dilution of the first dilution (1:5). Finally, the student made a third 1:20 dilution, this means that the second dilution was further diluted 20 times.

So, to calculate the final dilution of protein lysate, we have to multiply all the dilution factors of every dilution prepared: in this case we have a final dilution of 1:20, this means we have a factor dilution of 20. But it was previously diluted 4 times, so we have a factor dilution of 20×4 = 80. However, this dilution was also previously diluted 5 times, so the new dilution factor is 80 × 5 = 400

This means that the final dilution of the compound was diluted a total of 400 times compared to the initial concentration of stock solution.

3 0
2 years ago
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