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mezya [45]
3 years ago
10

Which element is placed in the same period as ruthenium but has a higher atomic number than it

Chemistry
2 answers:
Blababa [14]3 years ago
8 0
Since it is in period five, any element that has a higher atomic number than it could be an answer, therefore, silver, bismuth, and osmium are possibilities.
Klio2033 [76]3 years ago
6 0

Answer:

The answer is Silver.

Explanation:

The complete question is:

<em>"Which element is placed in the same period as ruthenium but has a higher atomic number than it? bismuth osmium silver zirconium"</em>

In the periodic table the elements are arranged so that those with similar chemical properties are located close to each other.

The horizontal rows of the periodic table are called periods. The elements that make up the same row have different properties but similar masses: all the elements of a period have the same number of orbitals. In other words, the period that an element occupies coincides with its last electronic layer. In other words, an element with five electronic layers will be in the fifth period.

The atomic number is represented by the letter Z and is the number of protons that have their atoms in the nucleus. The atomic number of silver [47] is higher than that of ruthenium [44] because silver has higher number of protons more than ruthenium. Ruthenium and silver are in the same period and they have the same number of atomic orbital but have different atomic numbers.

So <u><em>the answer is Silver.</em></u>

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Answer:

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Explanation:

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What is the pH of a 0.050 M triethylamine, (C2H5)3N, solution? Kb for triethylamine is 5.3 ´ 10-4. Question options: 1) 11.69 2)
blsea [12.9K]
<span>the pH of a 0.050 M triethylamine, is 11.70
</span>
For triehtylamine, (C_{2}H_{5})_{3}N, the reaction will be
(C_{2}H_{5})_{3}N + H_{2}O ---\ \textgreater \ ( C_{2}H_{5})_{3}NH^{+} + OH^{-}

 and we know, pH = -log[H+] and pOH = -log[OH-]

Also, pOH + pH = 14

Now, the Kb value  = 5.3 x 10^-4

And kb =  \frac{( [( C_{2}H_{5})_{3}NH^{+} ]*  OH^{-} )}{[( C_{2}H_{5})_{3}N]}

 thus, [OH-] =(5.3 ^ 10-4) ^2 / 0.050

                    =0.00516 M 

Thus, pOH = 2.30 
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