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GalinKa [24]
3 years ago
6

James exclaims to his chemistry teacher that a new element has been discovered that fits between Nickel and Copper on the period

ic table.
Mrs. Nogaki calls this fake news and explains to him why this is not possible.

Write 2 or more sentences of what you would tell James about why this new element is not possible.
Chemistry
2 answers:
myrzilka [38]3 years ago
8 0

Aye you have the same class as me bruh I need help on some chemistry qustions

ehidna [41]3 years ago
7 0

Answer:

See below  

Step-by-step explanation:

Nickel and copper are next to each other in the Periodic Table.

Ni has 28 protons, and Cu has 29.

You can't split a proton, so there's <em>no room</em> for another element between Ni and Cu.

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3 years ago
Based on these data, what is the average atomic mass of element b? 10.01 10.51 10.81 11.01
Lelechka [254]
The right answer is C. 10.81
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During a drought of 3 years would you expect to see adaptations in organism populations?
TiliK225 [7]

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yes

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6 0
3 years ago
50.0 mL of 0.200 M HNO2 is titrated to its equivalence point with 1.00 M NaOH. What is the pH at the equivalence point?
yKpoI14uk [10]

Answer:

8.279

Explanation:

The pH can be determined by hydrolysis of a conjugate base of weak acid at the equivalence point.

At the equivalence point, we have

$n_{NaOH}=n_{HNO_2}$

           = 25.00 x 0.200

           = 5.00 m-mol

           = 0.005 mol

Volume of the base that is added to reach the equivalence point is

$\frac{0.005}{1.00} \times 1000= 5.00 \ mL$

Number of moles of $NO^-_{2}=n_{HNO_2}$

                                           = 0.005 mol

Volume at the equivalence point is 25 + 5 = 30.00 mL

Therefore, concentration of $NO^-_{2}= \frac{5}{30}$

                                                        = 0.167 M

Now the ICE table :

            $NO^-_2 + H_2O \rightarrow HNO_3 + OH^-$

I (M)       0.167                   0            0

C (M)         -x                      +x          +x

E (M)      0.167-x                  x           x

Now, the value of the base dissociation constant is ,

$K_w=K_a \times K_b$            $(K_w \text{ is the ionic product of water })$

$K_b =\frac{K_w}{K_a}$

$K_b =\frac{1 \times 10^{-14}}{4.6 \times 10^{-4}}$

    = $2.174 \times 10^{-11}$

Base ionization constant, $K_b = \frac{\left[HNO_2\right] \left[OH^- \right]}{\left[NO^-_2 \right]}$

$2.174 \times 10^{-11}=\frac{x^2}{0.167 -x}$

$x= 1.9054 \times 10^{-6}$

So, $[OH^-]=1.9054 \times 10^{-6 } \ M$

pOH =- $\log[OH^-]$

       = $- \log(1.9054 \times 10^{-6} \ M)$

        =5.72

Now, since pH + pOH = 14

           pH = 14.00 - 5.72

                = 8.279

Therefore the ph is 8.279 at the end of the titration.

8 0
3 years ago
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