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RSB [31]
3 years ago
8

Can someone help please?

Chemistry
2 answers:
weeeeeb [17]3 years ago
8 0

Answer: 63.55

Explanation:

The attached chart shows the calculation.  The orange line points to the % abundance of Cu-65, which is the difference of 100% less the % Abundance of Cu-65 (since there are only 2 isotopes).

The weighted average contribution of each istope is calculated (blue arrow) and then summed to find the atomic mass of copper.

attashe74 [19]3 years ago
4 0
<h3>Take the weighted average of the individual isotopes.</h3><h3 /><h3>Explanation:</h3><h3>63</h3><h3>C</h3><h3>u</h3><h3> has </h3><h3>69.2</h3><h3>%</h3><h3> abundance.</h3><h3 /><h3>65</h3><h3>C</h3><h3>u</h3><h3> has </h3><h3>30.8</h3><h3>%</h3><h3> abundance.</h3><h3 /><h3>So, the weighted average is </h3><h3>62.93</h3><h3>×</h3><h3>69.2</h3><h3>%</h3><h3> </h3><h3>+</h3><h3> </h3><h3>64.93</h3><h3>×</h3><h3>30.8</h3><h3>%</h3><h3> </h3><h3>=</h3><h3> </h3><h3>63.55</h3><h3> </h3><h3>amu</h3><h3> .</h3><h3 /><h3>If we look at the Periodic Table, copper metal (a mixture of isotopes but </h3><h3>63</h3><h3>C</h3><h3>u</h3><h3> and </h3><h3>65</h3><h3>C</h3><h3>u</h3><h3> predominate) has an approximate atomic mass of </h3><h3>63.55</h3><h3> </h3><h3>g</h3><h3>⋅</h3><h3>m</h3><h3>o</h3><h3>l</h3><h3>−</h3><h3>1</h3><h3> , so we know we are right.</h3>

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A 0.43g samle of KHP required 24.11cm of NaOH for neutralization. Calculate the molarity of NaOH
expeople1 [14]

Answer:

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Explanation:

We'll begin by calculating the number of mole in 0.43 g of KHP (potassium hydrogen phthalate, C₈H₅O₄K). This is can be obtained as follow:

Mass of C₈H₅O₄K = 0.43 g

Molar mass of C₈H₅O₄K = (8×12) + (5×1) + (16× 4) + 39

= 96 + 5 + 64 + 39 = 204 g/mol

Mole of C₈H₅O₄K =?

Mole = mass / molar mass

Mole of C₈H₅O₄K = 0.43 / 204

Mole of C₈H₅O₄K = 0.002 mole

Next, we shall determine the number of mole of NaOH required to react with 0.43 g (i.e 0.002 mole) of KHP. This can be obtained as follow:

C₈H₅O₄K + NaOH → C₈H₄O₄KNa + H₂O

From the balanced equation above,

1 mole of KHP reacted with 1 mole of NaOH.

Therefore, 0.002 mole of KHP will also react with 0.002 mole of NaOH.

Next, we shall convert 24.11 cm³ to L. This can be obtained as follow:

1000 cm³ = 1 L

Therefore,

24.11 cm³ = 24.11 cm³ × 1 L / 1000 cm³

24.11 cm³ = 0.02411 L

Finally, we shall determine the molarity of NaOH. This can be obtained as follow:

Mole of NaOH = 0.00 2 mole

Volume = 0.02411 L

Molarity of NaOH =?

Molarity = mole /Volume

Molarity of NaOH = 0.002 / 0.02411

Molarity of NaOH = 0.083 M

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