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ivolga24 [154]
3 years ago
5

The ions S-2, Cl-1, K+, Ca+2, Sc+3 have the same total number of electrons as the noble gas argon. How would you expect the radi

i of these ions to vary?
Chemistry
1 answer:
bearhunter [10]3 years ago
4 0

Answer:

Cl^- <S^2-<Sc^3+ <Ca^2+<K^+

Explanation:

We know that ionic radius of ions decreases from right to left in the periodic table.  This is because, ionic radii decreases with increase in nuclear charge. This explains why; Sc^3+ <Ca^2+<K^+.

Secondly, even though Cl^- is isoelectronic with S^2-, the size of the nuclear charge in Cl^-  is larger compared to that of  S^2- . Hence Cl^-  is smaller than S^2- in ionic radius owing to increased nuclear attraction in Cl^-.

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A 50.0-mL volume of 0.15 M HBr is titrated with 0.25 M KOH. Calculate the pH after the addition of 13.0 mL of KOH. Express your
Mazyrski [523]

Answer:

pH= 1.17

Explanation:

The neutralization reaction between HBr (acid) and KOH (base) is given by the following equation:

HBr(aq) + KOH(aq) → KBr(aq) + H₂O(l)

According to this equation, 1 mol of HBr reacts with 1 mol of KOH. Then, the moles can be expressed as the product between the molarity of the acid/base solution (M) and the volume in liters (V). So, we calculate the moles of acid and base:

<u>Acid</u>:

M(HBr) = 0.15 M = 0.15 mol/L

V(HBr) = 50.0 mL x 1 L/1000 mL = 0.05 L

moles of HBr = M(HBr) x V(HBr) = 0.15 mol/L x 0.05 L = 7.5 x 10⁻³ moles HBr

<u>Base</u>:

M(KOH) = 0.25 M = 0.25 mol/L

V(HBr) = 13.0 mL x 1 L/1000 mL = 0.013 L

moles of HBr = M(HBr) x V(HBr) = 0.25 mol/L x 0.013 L = 3.25 x 10⁻³ moles KOH

Now, we have: 7.5 x 10⁻³ moles HBr > 3.25 x 10⁻³ moles KOH

HBr is a strong acid and KOH is a strong base, so they are completely dissociated in water: the acid produces H⁺ ions and the base produces OH⁻ ions. So, the difference between the moles of HBr and the moles of KOH is equal to the moles of remaining H⁺ ions after neutralization:

moles of H⁺ = 7.5 x 10⁻³ moles HBr - 3.25 x 10⁻³ moles KOH = 4.25 x 10⁻³ moles H⁺

From the definition of pH:

pH = -log [H⁺]

The concentration of H⁺ ions is calculated from the moles of H⁺ divided into the total volume:

total volume = V(HBr) + V(KOH) = 0.05 L + 0.013 L = 0.063 L

[H⁺] = (moles of H⁺)/(total volume) = 4.25 x 10⁻³ moles/0.063 L = 0.067 M

Finally, we calculate the pH after neutralization:

pH = -log [H⁺] = -log (0.067) = 1.17

3 0
3 years ago
Nitrogen and hydrogen react to produce ammonia.
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6 0
3 years ago
In an experiment, 107.9 grams of H2SO, is produced when 196.2 grams
Ipatiy [6.2K]

Answer:

54.99% yield

Explanation:

percent yield is just the amount you obtained over the amount expected times 100%.

(experimental value/theoretical value) x 100%

= (107.9 g/196.2 g) x 100%

=54.99% yield

5 0
3 years ago
What is one major purpose for an animal to produce unusual sounds, display bright
bezimeni [28]
I would say courtship, we this is when an animal would be looking for a mate. They would want to make themselves different and bright colors could attract a possible mate for them. I hope this helps!
7 0
3 years ago
Consider the reaction 2CO * O2 —&gt; 2 CO2 what is the percent yield of carbon dioxide (MW= 44g/mol) of the reaction of 10g of c
Arturiano [62]

Answer:

Y = 62.5%

Explanation:

Hello there!

In this case, for the given chemical reaction whereby carbon dioxide is produced in excess oxygen, it is firstly necessary to calculate the theoretical yield of the former throughout the reacted 10 grams of carbon monoxide:

m_{CO_2}^{theoretical}=10gCO*\frac{1molCO}{28gCO}*\frac{2molCO_2}{2molCO}  *\frac{44gCO_2}{1molCO_2}\\\\ m_{CO_2}^{theoretical}=16gCO_2

Finally, given the actual yield of the CO2-product, we can calculate the percent yield as shown below:

Y=\frac{10g}{16g} *100\%\\\\Y=62.5\%

Best regards!

8 0
3 years ago
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