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lapo4ka [179]
1 year ago
12

Define bond energy using the H¬Cl bond as an example. When this bond breaks, is energy absorbed or released? Is the accompanying

\Delta H value positive or negative? How do the magnitude and sign of this \Delta H value relate to the value that accompanies H¬Cl bond formation?
Chemistry
1 answer:
Rainbow [258]1 year ago
4 0

In HCl-HCl, the hydrogen-chlorine link is a polar covalent bond. It is produced when two atoms share an electron pair.

When atoms with various electronegativities share electrons in a covalent link, the result is a polar covalent bond. Think about the molecule of hydrogen chloride (HCl). In order to generate an inert gas electron configuration, each atom of HCl needs an additional electron. Despite having a stronger electronegativity than hydrogen, the chlorine atom cannot remove an electron from hydrogen due to its inability to attract electrons. As a result, a polar covalent bond in hydrogen chloride has an unbalanced distribution of bonding electrons.

Learn more about electronegativity here-

brainly.com/question/17762711

#SPJ4

You might be interested in
Which statement about the orinoco river is true.
WITCHER [35]
Hello,

I think the answer is B) It is second longest river in south america 

Hope this helps!!

~Girlygir101~
3 0
3 years ago
Given the amount of camphor (200mg) we are using in this experiment, please determine how many mg of sodium borohydride to use i
swat32

Answer:

Explanation:

From the given information:

Camphor may be reduced as readily in the presence of sodium borohydride(NaHB4). The resulting compound which is stereoselective requires 1 mole of sodium borohydride (NaHB4) to reduce 1 mole of camphor in this reaction. The reaction is shown below.

Through the reduction process of camphor, the reducing agent can reach the carbonyl face with a one-carbon linkage. The product stereoisomer is known as borneol.

If the molecular weight of camphor = 152.24 g/mol

and it mass = 200 mg

The its no of moles = 200 mg/ 152.24 g/mol

= 1.3137 mmol

Now the amount of the required mmol for NaBH4 to be consumed in the reaction = 5.2 × 1.3137 mmol

= 6.831 mmol

since the molar mass of NaBH4 = 37.83 g/mol

Then, using the same formula:

No of moles = mass/molar mass

mass = No of moles × molar mass

mass = 6.831 mmol × 37.83 g/mol

mass of NaBH4 used = 258.42 mg  

7 0
2 years ago
Calculate the cell potential for the reaction as written at 25.00 °C 25.00 °C , given that
Taya2010 [7]

Answer:

E = 2.02 V

Explanation:

In order to do this, we need to apply the Nernst equation which is:

E = E° - RT/nF lnQ

The value of RT/F can be simplified to just 0.059 because we are doing this experiment at 25 °C, and R and F are constants. so we need the value of Q which in this case is:

Q = [Mg²⁺] / [Ni²⁺]

We already have the concentrations, so, all we have left is the standard reduction potential, which are:

E° Mg = -2.38 V

E° Ni = -0.25 V

According to the overall reaction:

Mg(s) + Ni²⁺(aq) -------> Mg²⁺(aq) + Ni(s)

we can see that one element is reducting and the other is oxidizing, so we need to write the semi equation of reduction for each element:

Mg(s) ---------> Mg²⁺ + 2e⁻     E° = 2.38 V       oxidizing (Value of E° inverted)

Ni²⁺ + 2e⁻ -----------> Ni(s)      E° = -0.25 V     reducting

------------------------------------------------------------

Mg(s) + Ni²⁺(aq) -------> Mg²⁺(aq) + Ni(s)      E° = 2.13 V

We have the value of the standard potential, now we need to replace all given data into the nernst equation to solve for the cell potential:

E = 2.13 - 0.059/2 ln(0.757/0.0160)

E = 2.13 - 0.0295 ln(47.3125)

E = 2.13 - 0.11

E = 2.02 V

This is the cell potential

3 0
3 years ago
A compound made of two elements, iridium (Ir) and oxygen (O), was produced in a lab by heating iridium while exposed to air. The
natima [27]
1) Compund Ir (x) O(y)

2) Mass of iridium = mass of crucible and iridium - mass of crucible = 39.52 g - 38.26 g = 1.26 g

3) Mass of iridium oxide = mass of crucible and iridium oxide - mass of crucible = 39.73g - 38.26g = 1.47g

4) Mass of oxygen = mass of iridum oxide - mass of iridium = 1.47g - 1.26g = 0.21g

5) Convert grams to moles

moles of iridium = mass of iridium / molar mass of iridium = 1.26 g / 192.17 g/mol = 0.00656 moles

moles of oxygen = mass of oxygen / molar mass of oxygen = 0.21 g / 15.999 g/mol = 0.0131

6) Find the proportion of moles

Divide by the least of the number of moles, i.e. 0.00656

Ir: 0.00656 / 0.00656 = 1

O: 0.0131 / 0.00656 = 2

=> Empirical formula = Ir O2 (where 2 is the superscript for O)

Answer: Ir O2
5 0
3 years ago
Read 2 more answers
Consider the generic reaction: 2 A(g) + B(g) → 2 C(g). If a flask initially contains 1.0 atm of A and 1.0 atm of B, what is the
irina1246 [14]

Answer:

b. 1.5 atm.

Explanation:

Hello!

In this case, since the undergoing chemical reaction suggests that two moles of A react with one moles of B to produce two moles of C, for the final pressure we can write:

P=P_A+P_B+P_C

Now, if we introduce the stoichiometry, and the change in the pressure x we can write:

P=1.0-2x+1.0-x+2x

Nevertheless, since the reaction goes to completion, all A is consumed and there is a leftover of B, and that consumed A is:

x=\frac{1.0atm}{2}=0.5atm

Thus, the final pressure is:

P=1.0-2(0.5)+1.0-(0.5)+2(0.5)\\\\P=1.5atm

Therefore the answer is b. 1.5 atm.

Best regards!

3 0
2 years ago
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