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aivan3 [116]
2 years ago
12

Look at the periodic table and in the space below, type in the missing element name or symbol

Chemistry
1 answer:
Jet001 [13]2 years ago
7 0
Fe
Cobalt
Sodium
Sn
Phosphorus
Florine
I don’t know
Mg
I don’t know
Calcium
C
Pb
Silver
Zinc
Nickel
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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
What is an experimental control
attashe74 [19]

The experimental control is the standard used as a comparison for the experimental groups.


For example, you may be trying to find out how different types of disinfectants affect bacterial growth.  The control group would receive <em>no</em> disinfectant whereas the experimental groups would be the ones on which the disinfectants were tested.


Hope this makes sense!

3 0
3 years ago
Circle only those of the following reactions that are redox reactions:
Citrus2011 [14]

Answer: options B,D and F

Explanation:

Since redox reactions are those which involves both oxidation and reduction

In B , Cu is oxidized and S gets reduced

D, Na gets oxidized and hydrogen gets reduced

F, carbon gets oxidized and Oxygen gets reduced

In g, there is no change in oxidation no of s in both product and Reactants is same +4

Similarly in the case of Ag and Mg.

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The shape of the nucleus is maintained by
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B. the necular lamina
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What is a balanced equation for the neutralization of these two compounds? <br><br> HCl <br> Ca(OH)2
loris [4]

In order to balance this, you have to count each element where the elements in the reactants side and the product side should have equal number of molecules. The balanced reaction is as follows:


2HCl + Ca(OH)2 = CaCl2 + 2H2O

8 0
2 years ago
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