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3 years ago

Household bleach which has a ph of 12.6 has more hydroxide (oh-) ions than lye which has a ph of 13.

Chemistry

2 answers:

Elena-2011 [213]3 years ago

7 0

Answer:

Household bleach has more hydroxide ions than lye.

Explanation:

The pH of the solution id the negative logarithm of hydrogen ion concentration in an aqueous solution.

$pH=-\log[H^+]$

Higher the pH value lessor the hydrogen ion concentration in the solution. Higher will be the concentration of hydroxide ions
Lower the pH value higher the hydrogen ion concentration in the solution.Lower will be the concentration of hydroxide ions.

The pH of the household bleach = 12.6

$12.6=-\log[H^+]$

$[H^+]=2.511\times 10^{-13} M$

The pH of the lye= 13

$13=-\log[H^+]$

$[H^+]= 10^{-13} M$

Since , house hold bleach has the lower pH than lye which means that it will have higher concentration of hydroxide ions.

Arisa [49]3 years ago

6 0

Answer is: false, concentration of hydroxide ions is greater in lye than in household bleach.

1) pH = 12.6.

pH + pOH = 14.

pOH = 14 - 12.6.

pOH = 1.4.

pOH = -log[OH⁻] ⇒ [OH⁻] = 10∧(-pOH).

[OH⁻] = 10∧(-1.4) = 0.039 M; concentration of hydroxide ions.

2) pH = 13.

pOH = 14 - 13.

pOH = 1.

[OH⁻] = 10⁻¹ M = 0.1 M.

0.1 M > 0.039 M.

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Answer 1:

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Answer 3:

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3 years ago

How many hydrogen atoms are attached to each carbon adjacent to a double bond? nurition?

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That depends. there are 2 possible answers.
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4 years ago

A sample of gaseous arsine (AsH3) in a 460 mL flask at 332 Torr and 223 K, is heated to 437 K, at which temperature arsine decom

Taya2010 [7]

Answer:

28/95 = 29.,5 % of Arsine decomposed

Explanation:A sample of gaseous arsine (AsH3) in a 460 mL flask at 332 Torr and 223 K, is heated to 437 K, at which temperature arsine decom- poses to solid arsenic and hydrogen gas. The flask is then cooled to 273 K, at which tem- perature the pressure in the flask is 488 Torr. What percentage of arsine molecules have de- composed?

Answer in units of %.

initial pressure 332 Torr initial volume 0.46 L initial temperature 223K

final pressure 488 Torr final volume 0.46 L final 273 K

Torr is 1/760 atm 332 torr = 0.437 atm 488 Torr =0.642 atm

PV = nRT so n=RT/PV

INITIAL n= 0.082 X 223/(0.437)(0.46) = 91 moles

final n= 0.082 X 273 / (.437)(488) = 105 moles

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x moles of Arsine decomposed to make 1.5 moles of H2

the final number of moles was

(91 -X)+ 1.5 X = 105 moles

91 + 0.5 X = 105

0.5 X = 14

X =28

CHECK

if 28 moles of Arsine , then the container would have

91 --28 + 1.5(28) = 91 +14 =105 check

so 28/95 = 29.,5 % of Arsine decomposed

Your answer

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polyalchemVirtuoso

Answer:

Explanation:

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