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Nuetrik [128]
3 years ago
9

Which statement correctly describes potassium iodide, KI? There is a one-to-one ratio of potassium ions to iodide ions. Potassiu

m gains electrons and iodine loses electrons during the reaction. The lattice is held together by potassium anions and iodide cations.
Chemistry
2 answers:
Natali [406]3 years ago
7 0

Answer:There is a one-to-one ratio of potassium ions to iodide ions.

Explanation:

elena-14-01-66 [18.8K]3 years ago
5 0

The answer is there is a one-to-one ratio of potassium ions to iodide ions.

Explanation :

- (K) belongs to Alkali metals in group (1A) that contains (1) electron in the outermost energy level, whereas, (I) is from halogens in group (7A) that contains (7) electron in the outermost energy level.

- To achieve stability, both atoms tend to reach the nearest noble state (outermost level occupies 8 electrons). Therefore, (K) loses its outer electron and gives it to (I) which now has a completely filled outer level and an ionic bond is formed between the two.

- The valency (number of electrons lost, gained or shared) of both atoms is equal ”monovalent” which means one-to-one ratio..


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What is the concentration of NaCI in an aqueous solution that contains 0.032 grams of NaCI in 600. Grams of the solution
jolli1 [7]

Answer:

0.00032 Grams of NaCl per 1 gram of the solution

Explanation:

8 0
3 years ago
What observation would you look for to tell if a reaction was endothermic?
horsena [70]

Exothermic reactions release heat into their surroundings and endothermic reactions absorb heat and are cool. I believe the correct answer is D.

5 0
2 years ago
A certain liquid sample has a volume of 14.7 mL and a mass of 22.8 grams. Calculate the density.
ira [324]
M = 22.8 g
V = 14.7 mL
ρ - ?

ρ = m/V
ρ = 22.8/14.7 = 1.55 g/mL

4 0
3 years ago
A solution is made by combining 500 mL of 0.10 M HF (Ka=7.2 x 10^-4) with 300 mL of 0.15 M NaF. What is the pH of the resulting
n200080 [17]

Answer:

b) 3.10

Explanation:

HF ⇄ H + + F

Using Henderson-Hasselbalch Equation:

pH = pKa + log [A-]/[HA].

Where;

pKa = Dissociation constant = -log Ka

Hence, pKa of HF = -log 7.2 x 10^-4 = 3.14266

[A-] = concentration of conjugate base after dissociation = moles of base/total volume

          = 0.15 x 0.3/0.8

               = 0.05625 M

[HA] = concentration of the acid = moles of acid/total volume

             = 0.10 x 0.5/0.8

                    = 0.0625 M

Note: <em>Total volume = 500 + 300 = 800 mL = 0.8 dm3</em>

pH = 3.14266 + log [0.05625/0.0625]

      = 3.14267 + (-0.04575749056)

           = 3.09691250944

<em>From all the available options below:</em>

<em>a) 2.97 </em>

<em>b) 3.10 </em>

<em>c) 3.19 </em>

<em>d) 3.22 </em>

<em>e) 3.32</em>

The correct option is b.

4 0
3 years ago
Using the reaction below determine the amount of Sulfur proceduced in grams (this is a limiting reaction question) 2 H2S + SO2 -
marusya05 [52]

Answer:

<u>= 2.2 g pf S. produced</u>

Explanation:

Balanced Reaction equation:

2H_{2} S + SO_{2} →  3S + 2H_{2} O

1 mole of H2S - 34.1g

? moles - 3.2g

= 3.2/34.1 =<u> 0.09 moles of H2S</u>

Also,

1 mole of S02 - 64.07 g

? moles - 4.42g

= 4.42/64.07 <u>= 0.069 moles of SO2</u>

<u />

<em>Meaning SO2 is the limiting reagent</em>

Finally, 3 moles of S -  32g of sulphur

0.069 mole = ? g of Sulphur

= 0.069 x 32

<u>= 2.2 g pf S.</u>

7 0
2 years ago
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