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snow_tiger [21]
3 years ago
10

Lithium sulfide formula?

Chemistry
1 answer:
zavuch27 [327]3 years ago
6 0
BOLD MEANS THE NUMBER GOES DOWN...
Li2S

PS: BE CAREFUL IF HANDLING LITHIUM SULFIDE; IT PRODUCES HYDROGEN SULFIDE WHEN REACTED WITH WATER, MOISTURE OR STEAM!

Hope this helps xox :)
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What is the mass of 1.2 x 1023 atoms of arsenic?
Gre4nikov [31]

Answer:

14.93 g

Explanation:

First we <u>convert 1.2 x 10²³ atoms of arsenic (As) into moles</u>, using <em>Avogadro's number</em>:

  • 1.2 x 10²³ atoms ÷ 6.023x10²³ atoms/mol = 0.199 mol As

Then we can<u> calculate the mass of 0.199 moles of arsenic</u>, using its<em> molar mass</em>:

  • 0.199 mol * 74.92 g/mol = 14.93 g

Thus, 1.2x10²³ atoms of arsenic weigh 14.93 grams.

6 0
3 years ago
determine the ph of a buffer that is 0.55 M HNO2 and 0.75 M KNO2. tha value of Ka for HNO2 is 6.8*10^-4
Mariana [72]

Answer:

pH = 3.3

Explanation:

Buffer solutions minimize changes in pH when quantities of acid or base are added into the mix. The typical buffer composition is a weak electrolyte (wk acid or weak base) plus the salt of the weak electrolyte. On addition of acid or base to the buffer solution, the solution chemistry functions to remove the acid or base by reacting with the components of the buffer to shift the equilibrium of the weak electrolyte left or right to remove the excess hydronium ions or hydroxide ions is a way that results in very little change in pH of the system. One should note that buffer solutions do not prevent changes in pH but minimize changes in pH. If enough acid or base is added the buffer chemistry can be destroyed.

In this problem, the weak electrolyte is HNO₂(aq) and the salt is KNO₂(aq). In equation, the buffer solution is 0.55M HNO₂ ⇄ H⁺ + 0.75M KNO₂⁻ . The potassium ion is a spectator ion and does not enter into determination of the pH of the solution. The object is to determine the hydronium ion concentration (H⁺) and apply to the expression pH = -log[H⁺].

Solution using the I.C.E. table:

              HNO₂ ⇄    H⁺   +   KNO₂⁻

C(i)        0.55M       0M      0.75M

ΔC            -x            +x          +x

C(eq)  0.55M - x       x     0.75M + x    b/c [HNO₂] / Ka > 100, the x can be                                    

                                                             dropped giving ...

           ≅0.55M        x       ≅0.75M        

Ka = [H⁺][NO₂⁻]/[HNO₂] => [H⁺] = Ka · [HNO₂]/[NO₂⁻]

=> [H⁺] = 6.80x010⁻⁴(0.55) / (0.75) = 4.99 x 10⁻⁴M

pH = -log[H⁺] = -log(4.99 x 10⁻⁴) -(-3.3) = 3.3

Solution using the Henderson-Hasselbalch Equation:

pH = pKa + log[Base]/[Acid] = -log(Ka) + log[Base]/[Acid]

= -log(6.8 x 10⁻⁴) + log[(0.75M)/(0.55M)]

= -(-3.17) + 0.14 = 3.17 + 0.14 = 3.31 ≅ 3.3

3 0
3 years ago
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Vesna [10]
The molecule that contains the fewest number of Hydrogen atoms would be B. Al(OH)3. It only has 3 Hydrogen atoms.
7 0
3 years ago
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Calculate the percent activity of the radioactive isotope strontium-89 remaining after 5 half-lives.
maria [59]
The answer to this question would be: 3.125%

Half-life is the time needed for a radioactive molecule to decay half of its mass. In this case, the strontium-89 is already gone past 5 half lives. Then, the percentage of the mass left after 5 half-lives should be:
100%*(1/2^5)= 100%/32=3..125%
5 0
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Plant use sunlight as energy to convert carbon dioxide and water into glucose and oxygen. Which best describes the reaction?
Illusion [34]

Answer:

Its the process of photosynthesis

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