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Bogdan [553]
3 years ago
7

What type of elements are most likely to form more than one type of ion

Chemistry
2 answers:
Mamont248 [21]3 years ago
8 0

Answer: Naming Compounds and Writing Formulas

1. The elements most likely to form more than one type of ion are the____ Transition Metals

2. The name iron (ll) indicates that a compound contains___ Iron ions with a 2+ charge

3. Alkali metals, alkaline earth metals, and aluminum all form ions with positive charges equal to___ The group number

4. In the name carbon dioxide, the prefix of the second word indicates that molecule of carbon dioxide contains___Two oxygen atoms

Explanation:

ss7ja [257]3 years ago
7 0
Metals are the type of elements that are most likely to form more than one type of ion, for instance iron can form the ion of Fe^2+ or Fe^3+.
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What is the difference in mean solar time between 30 N 75 W and 30 N 90 W​
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Every 15 degrees is an hour

Explanation:

What is the time difference in mean solar time between 30° N 75° W and 30° N 90° W?

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3 years ago
Calculate the change in pH when 71.0 mL of a 0.760 M solution of NaOH is added to 1.00 L of a solution that is 1.0O M in sodium
Eddi Din [679]

Explanation:

It is known that pK_{a} value of acetic acid is 4.74. And, relation between pH and pK_{a} is as follows.

                    pH = pK_{a} + log \frac{[CH_{3}COOH]}{[CH_{3}COONa]}

                          = 4.74 + log \frac{1.00}{1.00}

So, number of moles of NaOH = Volume × Molarity

                                                   = 71.0 ml × 0.760 M

                                                    = 0.05396 mol

Also, moles of  CH_{3}COOH = moles of CH_{3}COONa

                                          = Molarity × Volume

                                          = 1.00 M × 1.00 L

                                          = 1.00 mol

Hence, addition of sodium acetate in NaOH will lead to the formation of acetic acid as follows.

            CH_{3}COONa + NaOH \rightarrow CH_{3}COOH

Initial :    1.00 mol                                  1.00 mol

NaoH addition:               0.05396 mol

Equilibrium : (1 - 0.05396 mol)    0           (1.00 + 0.05396 mol)

                    = 0.94604 mol                       = 1.05396 mol

As, pH = pK_{a} + log \frac{[CH_{3}COONa]}{[CH_{3}COOH]}

               = 4.74 +  log \frac{0.94604}{1.05396}

               = 4.69

Therefore, change in pH will be calculated as follows.

                         pH = 4.74 - 4.69

                               = 0.05

Thus, we can conclude that change in pH of the given solution is 0.05.

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