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hram777 [196]
3 years ago
6

gThe atomic radii of a divalent cation and a monovalent anion are 0.085 nm and 0.125 nm, respectively. (a) Calculate the force o

f attraction between these two ions at their equilibrium interionic separation (i.e., when the ions just touch one another)./homework-help/electron-subshell-filled-rare-earth-series-elements-periodic-chapter-2-problem-14qp-solution-9781118324578-exc
Chemistry
1 answer:
frozen [14]3 years ago
6 0

Answer:

1.05 × 10⁻⁸ N

Explanation:

Using Coulomb's law

F force of attraction between them = k \frac{q^{2} }{r^{2} } = k \frac{z1z2q^{2} }{r^{2} }

K coulomb constant = 8.99 × 10⁹ N m²C⁻²

Z₁ = + 2 ( valency electron)

Z₂ = -1 ( valency electron)

r is the distance between then since they just touch one another = 0.085 nm + 0.125 nm = 0.21 nm = 0.21 × 10⁻⁹ m

q charge = 1.602 × 10⁻¹⁹ C of an electron

F = ( 2 × 1 × 8.99 × 10⁹ N m²C⁻² × (1.602 × 10⁻¹⁹ C)² / ( 0.21 × 10⁻⁹ m )² = 1046.34 × 10 ⁻¹¹ = 1.05 × 10⁻⁸ N

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bagirrra123 [75]
There are 1000 grams in a kg.
To convert g to kg, dovide by 1000.

3.5/1000= 0.0035 kg

Final answer: D
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Which of the following could be considered a scientific method? 01.01 the science of chemistry
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0.5 moles of sodium chloride is dissolved to make 0.05 liters of solution what is the molarity
Sunny_sXe [5.5K]
Molarity =  moles of solute / liters of solution

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hope this helps!
4 0
3 years ago
If 15.6 grams of copper (ii) chloride react with 20.2 grams of sodium nitrate how many grams of sodium chloride can be formed? W
olasank [31]

Answer:

- 13.56 g of sodium chloride are theoretically yielded.

- Limiting reactant is copper (II) chloride and excess reactant is sodium nitrate.

- 0.50 g of sodium nitrate remain when the reaction stops.

- 92.9 % is the percent yield.

Explanation:

Hello!

In this case, according to the question, it is possible to set up the following chemical reaction:

CuCl_2+2NaNO_3\rightarrow 2NaCl+Cu(NO_3)_2

Thus, we can first identify the limiting reactant by computing the yielded mass of sodium chloride, NaCl, by each reactant via stoichiometry:

m_{NaCl}^{by\ CuCl_2}=15.6gCuCl_2*\frac{1molCuCl_2}{134.45gCuCl_2} *\frac{2molNaCl}{1molCuCl_2} *\frac{58.44gNaCl}{1molNaCl} =13.56gNaCl\\\\m_{NaCl}^{by\ NaNO_3}=20.2gNaNO_3*\frac{1molNaNO_3}{84.99gNaNO_3} *\frac{2molNaCl}{2molNaNO_3} *\frac{58.44gNaCl}{1molNaCl} =13.89gNaCl

Thus, we infer that copper (II) chloride is the limiting reactant as it yields the fewest grams of sodium chloride product. Moreover the formed grams of this product are 13.56 g. Then, we take 13.56 g of sodium chloride to compute the consumed mass sodium nitrate as it is in excess:

m_{NaNO_3}^{by\ NaCl}=13.56gNaCl*\frac{1molNaCl}{58.44gNaCl}*\frac{2molNaNO_3}{2molNaCl} *\frac{84.99gNaNO_3}{1molNaNO_3}=19.72gNaNO_3

Therefore, the leftover of sodium nitrate is:

m_{NaNO_3}^{leftover}=20.2g-19.7g=0.5gNaNO_3

Finally, the percent yield is computed via:

Y=\frac{12.6g}{13.56g} *100\%\\\\Y=92.9\%

Best regards!

6 0
3 years ago
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