Chemistry, 17.10.2019 18:30 lalalaaldgydyhg8823

Determine the [oh−] , ph, and poh of a solution with a [h+] of 2.2×10−12 m at 25 °c. [oh−]= m ph= poh= determine the [h+] , ph, and poh of a solution with an [oh−] of 9.7×10−5 m at 25 °c. [h+]= m ph= poh= determine the [h+] , [oh−] , and poh of a solution with a ph of 10.30 at 25 °c. [h+]= m [oh−]= m poh= determine the [h+] , [oh−] , and ph of a solution with a poh of 9.06 at 25 °c. [h+]= m [oh−]= m ph=

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Answer from: mandy9386

For 1: The OH^- , pH and pOH of the solution is $4.57\times 10^{-3}M$ , 11.66 and 2.36 respectively.

For 2: The OH^- , pH and pOH of the solution is $1.023\times 10^{-10}M$ , 4.01 and 9.99 respectively.

For 3: The H^+ , OH^- and pOH of the solution is $5.012\times 10^{-11},1.99\times 10^{-5}M$ and 3.7 respectively.

For 4: The H^+ , OH^- and pOH of the solution is $8.71\times 10^{-10},1.07\times 10^{-5}M$ and 4.97 respectively.

Explanation:

To calculate the pH of the solution, we use the equation:

$pH=-\log[H^+]$ .....(1)

To calculate pOH of the solution, we use the equation:

pH+pOH=14 .....(2)

To calculate [OH^] of the solution, we use the equation:

$pOH=-\log[OH^-]$ ......(3)

For 1:

We are given:

$[H^+]=2.2\times 10^{-12}M$

Putting values in equation 1, we get:

$pH=-\log(2.2\times 10^{-12})\\\\pH=11.66$

Using equation 2, we get:

pOH=14-11.66=2.34

Now, using equation 3, we get:

$2.36=-\log[OH^-]$

$[OH^-]=10^{-2.36}=4.57\times 10^{-5}M$

Hence, the OH^- , pH and pOH of the solution is $4.57\times 10^{-3}M$ , 11.66 and 2.36 respectively.

For 2:

We are given:

$[H^+]=9.7\times 10^{-5}M$

Putting values in equation 1, we get:

$pH=-\log(9.7\times 10^{-5})\\\\pH=4.01$

Using equation 2, we get:

pOH=14-4.01=9.99

Now, using equation 3, we get:

$9.99=-\log[OH^-]$

$[OH^-]=10^{-9.99}=1.023\times 10^{-10}M$

Hence, the OH^- , pH and pOH of the solution is $1.023\times 10^{-10}M$ , 4.01 and 9.99 respectively.

For 3:

We are given:

pH = 10.30

Putting values in equation 1, we get:

$10.30=-\log[H^+]$

$[H^+]=10^{-10.30)=5.012\times 10^{-11}M$

Using equation 2, we get:

pOH=14-10.30=3.7

Now, using equation 3, we get:

$3.7=-\log[OH^-]$

$[OH^-]=10^{-3.7}=1.99\times 10^{-4}M$

Hence, the H^+ , OH^- and pOH of the solution is $5.012\times 10^{-11},1.99\times 10^{-5}M$ and 3.7 respectively.

For 4:

We are given:

pH = 9.06

Putting values in equation 1, we get:

$9.03=-\log[H^+]$

$[H^+]=10^{-9.30)=8.71\times 10^{-10}M$

Using equation 2, we get:

pOH=14-9.03=4.97

Now, using equation 3, we get:

$4.97=-\log[OH^-]$

$[OH^-]=10^{-4.97}=1.07\times 10^{-5}M$

Hence, the H^+ , OH^- and pOH of the solution is $8.71\times 10^{-10},1.07\times 10^{-5}M$ and 4.97 respectively.

Answer from: Quest

1.consider the following balanced equation and answer parts a through. show calculations for each part.

3 ag(s) + 4 hno3(aq) → 3 agno3(aq) + no(g) + 2 h2o(l)

a. calculate the number of moles of agno3 produced by the reaction of 3 moles of ag with excess hno3. if the number to the left of the compound represents the number of moles of the compound which are present, then 3 moles of ag react with 4 moles hno3 to produce 3 moles agno3 (aq).

so the answer to a would be 3 moles.

b. calculate the number of moles of agno3 produced by the reaction of 108 grams of ag with excess hno3.

i'm not sure how to plug in the grams of silver into an equation using moles to find how many moles are produced.

c. calculate the number of grams of agno3 produced by the reaction of 108 grams of ag with excess hno3

d. calculate the number of moles of no produced as a byproduct of the reaction of 108 grams of ag with excess hno3.

e. calculate the number of grams of no produced as a byproduct starting with 108 grams of ag and excess hno3.

explanation:

Answer from: Quest

indicate valleys, hills, steepness and gentleness of slopes.

explanation:

contour lines on a map indicate elevation, or depth. numeric notations on contour lines refer to the level of elevation, or height about sea level.