Which of the following equations is balanced?

A polar covalent bond will form between which two atoms?

Alex_Xolod [135]

Types of Bonds can be predicted by calculating the difference in electronegativity.

If, Electronegativity difference is,

Less than 0.4 then it is Non Polar Covalent

Between 0.4 and 1.7 then it is Polar Covalent

Greater than 1.7 then it is Ionic

For Be and F,

E.N of Fluorine = 3.98

E.N of Beryllium = 1.57

________

E.N Difference 2.41 (Ionic Bond)

For H and Cl,

E.N of Chorine = 3.16

E.N of Hydrogen = 2.20

________

E.N Difference 0.96 (Polar Covalent Bond)

For Na and O,

E.N of Oxygen = 3.44

E.N of Sodium = 0.93

________

E.N Difference 2.51 (Ionic Bond)

For F and F,

E.N of Fluorine = 3.98

________

E.N Difference 0.00 (Non-Polar Covalent Bond)

Result:

A polar covalent bond is formed between Hydrogen and Chlorine atoms.

5 0

3 years ago

5. Given the cell notation Cu(s)|Cu2+(aq)||Ag+(aq)|Ag(s), what is the half-reaction that occurs at the anode?

Nataly_w [17]

What is [H+] given that the measured cell potential is -0.464 V and the anode reduction ... What half-reaction occurs at the cathode during the electrolysis of molten ... PbO2(s) + 4H+(aq) + SO42-(aq) + 2e- → PbSO4(s) + 2H2O(l); E° = 1.69 V .... For the cell Cu(s)|Cu2+||Ag+|Ag(s), the standard cell potential is 0.46 V. A cell ... hopw this helps

3 0

3 years ago

Read 2 more answers

What must be the molarity of an aqueous solution of trimethylamine, (ch3)3n, if it has a ph = 11.20? (ch3)3n+h2o⇌(ch3)3nh++oh−kb

Stolb23 [73]

0.040 mol / dm³. (2 sig. fig.)

<h3>Explanation</h3>

$(\text{CH}_3)_3\text{N}$ in this question acts as a weak base. As seen in the equation in the question, $(\text{CH}_3)_3\text{N}$ produces $\text{OH}^{-}$ rather than $\text{H}^{+}$ when it dissolves in water. The concentration of $\text{OH}^{-}$ will likely be more useful than that of $\text{H}^{+}$ for the calculations here.

Finding the value of $[\text{OH}^{-}]$ from pH:

Assume that $\text{pK}_w = 14$ ,

$\begin{array}{ll}\text{pOH} = \text{pK}_w - \text{pH} \\ \phantom{\text{pOH}} = 14 - 11.20 &\text{True only under room temperature where }\text{pK}_w = 14 \\\phantom{\text{pOH}}= 2.80\end{array}$ .

$[\text{OH}^{-}] =10^{-\text{pOH}} =10^{-2.80} = 1.59\;\text{mol}\cdot\text{dm}^{-3}$ .

Solve for $[(\text{CH}_3)_3\text{N}]_\text{initial}$ :

$\dfrac{[\text{OH}^{-}]_\text{equilibrium}\cdot[(\text{CH}_3)_3\text{NH}^{+}]_\text{equilibrium}}{[(\text{CH}_3)_3\text{N}]_\text{equilibrium}} = \text{K}_b = 1.58\times 10^{-3}$

Note that water isn't part of this expression.

The value of Kb is quite small. The change in $(\text{CH}_3)_3\text{N}$ is nearly negligible once it dissolves. In other words,

$[(\text{CH}_3)_3\text{N}]_\text{initial} = [(\text{CH}_3)_3\text{N}]_\text{final}$ .

Also, for each mole of $\text{OH}^{-}$ produced, one mole of $(\text{CH}_3)_3\text{NH}^{+}$ was also produced. The solution started with a small amount of either species. As a result,

$[(\text{CH}_3)_3\text{NH}^{+}] = [\text{OH}^{-}] = 10^{-2.80} = 1.58\times 10^{-3}\;\text{mol}\cdot\text{dm}^{-3}$ .

$\dfrac{[\text{OH}^{-}]_\text{equilibrium}\cdot[(\text{CH}_3)_3\text{NH}^{+}]_\text{equilibrium}}{[(\text{CH}_3)_3\text{N}]_\textbf{initial}} = \text{K}_b = 1.58\times 10^{-3}$ ,

$[(\text{CH}_3)_3\text{N}]_\textbf{initial} =\dfrac{[\text{OH}^{-}]_\text{equilibrium}\cdot[(\text{CH}_3)_3\text{NH}^{+}]_\text{equilibrium}}{\text{K}_b}$ ,

$[(\text{CH}_3)_3\text{N}]_\text{initial} =\dfrac{(1.58\times10^{-3})^{2}}{6.3\times10^{-5}} = 0.040\;\text{mol}\cdot\text{dm}^{-3}$ .

8 0

3 years ago

I need help please please

inna [77]

Answer:

B should be the answer, and ur low-key valid lol

Explanation:

6 0

3 years ago

What are the hydrogen bonding atoms in melatonin?

nydimaria [60]

The hydrogen bonding atoms in melatonin are phospholipids and cholesterol.

<h3>What is Hydrogen bonding?</h3>

This is the type of interaction which involves a hydrogen atom being covalently bound to a more electronegative "donor".

Melatonin controls sleep-wake cycle and has phospholipids and cholesterol which have high affinity for hydrogen bonding.

Read more about Hydrogen bonding here brainly.com/question/12798212

#SPJ1

6 0

2 years ago