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bulgar [2K]
3 years ago
6

What is an example of a change of state. Help I need this answered asappp pls & tyyy

Chemistry
1 answer:
melamori03 [73]3 years ago
6 0

pretty sure its A

water being boiled in a pot shows a phase change from liquid water to gaseous water when it evaporates

all other choices are just changing its form or shape basically, solid to solid

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What is a chemical substance that cannot be broken down into another chemical substance?
Dimas [21]

Molecule is a chemical substance that cannot be broken down into another chemical substance.

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Porque la gasolina cintemos plomo?
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2 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
1. Decomposition reactions can be classified into three types. Pick the correct
Alex777 [14]

Answer:

Explanation:

Option A is the correct answer

3 0
3 years ago
17. A balloon contains 4.50 X 1022 atoms of helium gas. Calculate the mass of helium in<br> grams.
Rashid [163]

Answer:

1.08 grams

Explanation:

first, we need to find the number of moles

we divide 12.1/22.4=0.54 moles

we multiply the number of moles with the molecular mass 0.54x2=1.08g

4 0
3 years ago
Read 2 more answers
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