Answer:
pOH = 1.3, pH = 12.7
Explanation:
Since NaOH is a strong base, it will completely ionize; further, since it completely ionizes, our hydroxide concentration (a product of the ionization) will be the same as the given concentration of NaOH.
NaOH -> Na⁺ + OH⁻, [OH⁻] = 5.0 x 10^-2 M
pOH is the negative log of the hydroxide concentration, so plug our hydroxide concentration in:
pOH = -log[OH⁻] = -log[5.0 x 10^-2 M] = 1.3
Since pH + pOH = 14, we can plug in pOH and solve for pH:
pH + 1.3 = 14
pH = 14 - 1.3 = 12.7
Thus, our pOH = 1.3 and pH = 12.7.
Answer : The speed in miles per hour is 22 mile/hr.
The speed in yard per min is 26617.8 yard/min
Explanation :
The conversion used for meters to miles is:
The conversion used for second to hour is:
The conversion used for meter per second to mile per hour is:
As we are given the speed of 407.0 meter per second. Now we have to determine the speed in miles per hour.
So, 
Therefore, the speed in miles per hour is 22 mile/hr.
The conversion used for meter to yard
1m = 1.09 yard
The conversion used for second to hour is:
The conversion used for meter per second to mile per hour is:
As we are given the speed of 407.0 meter per second. Now we have to determine the speed in yards per min
So, 
Therefore, the speed in yard per min is 26617.8 yard/min
Sulfur forms compounds in oxidation states −2 (sulfide, S2−), +4 (sulfite, SO32−), and +6 (sulfate, SO42−). I don't know what type of ion but hope this helps!! :)
Answer:
785
Explanation:
Molecular. X. W
Weight
8000-16000 0.05 0.03
16000-24000. 0.017. 0.08
24000-32000. 0.22. 0.18
32000-40000. 0.25. 0.35
40000-48000. 0.22. 0.27
48000-56000. 0.09. 0.09
Mean weight X*M. W*M
12000. 600. 240
20000. 3200. 2000
28000. 6720. 5600
36000. 10080. 10800
44000. 8800. 11880
52000. 3640 3640
Total=33040g\mol 36240
Note before repeat molecular weight m= 3*12.01+6*1.008=
42.08g/mol
Degree of polymerization= total W*M/w=33040/42.08 =785
Answer:
The answers are in the explanation
Explanation:
A buffer is the mixture of a weak acid with its conjugate base or vice versa. Thus:
<em>1)</em> Mixing 100.0 mL of 0.1 M HF with 100.0 mL of 0.05 M mol KF. <em>Will </em>result in a buffer because HF is a weak acid and KF is its conjugate base.
<em>2)</em> Mixing 100.0 mL of 0.1 M NH₃ with 100.0 mL of 0.1 M NH₄Br. <em>Will not </em>result in a buffer because NH₃ is a strong base.
<em>3) </em>Mixing 100.0 mL of 0.1 M HCN with 100.0 mL of 0.05 M KOH. <em>Will </em>result in a buffer because HCN is a weak acid and its reaction with KOH will produce CN⁻ that is its conjugate base.
<em>4)</em> Mixing 100.0 mL of 0.1 M HCl with 100.0 mL of 0.1 M KCl <em>Will not </em>result in a buffer because HCl is a strong acid.
<em>5)</em> Mixing 100.0 mL of 0.1 M HCN with 100.0 mL of 0.1 M KOH <em>Will not </em>result in a buffer because each HCN will react with KOH producing CN⁻, that means that you will have just CN⁻ (Conjugate base) without HCN (Weak acid).
I hope it helps!
