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4 years ago

If the H+ in a solution is 1x10^-1 mol/L what is the OH-

1 answer:

8 0

PH scale is used to determine how acidic or basic a solution is.
we have been given the hydrogen ion concentration. Using this we can calculate pH,
pH = - log[H⁺]
pH = - log (1 x 10⁻¹ M)
pH = 1
using pH can calculate pOH
pH + pOH = 14
pOH = 14 - 1
pOH = 13
using pOH we can calculate the hydroxide ion concentration
pOH = - log [OH⁻]
[OH⁻] = antilog(-pOH)
[OH⁻] = 10⁻¹³ M
hydroxide ion concentration is 10⁻¹³ M

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Cobalt-60 is a strong gamma emitter that has a half- life of 5.26 yr. The co balt-60 in a radiotherapy unit must be replaced whe

Alenkasestr [34]

<u>Answer:</u> The sample of Cobalt-60 isotope must be replaced in January 2027

<u>Explanation:</u>

The equation used to calculate rate constant from given half life for first order kinetics:

$t_{1/2}=\frac{0.693}{k}$

where,

$t_{1/2}$ = half life of the reaction = 5.26 years

Putting values in above equation, we get:

$k=\frac{0.693}{5.26yrs}=0.132yrs^{-1}$

Rate law expression for first order kinetics is given by the equation:

$k=\frac{2.303}{t}\log\frac{[A_o]}{[A]}$

where,

k = rate constant = $0.132yr^{-1}$

t = time taken for decay process = ? yr

$[A_o]$ = initial amount of the sample = 100 grams

[A] = amount left after decay process = (100 - 75) = 25 grams

Putting values in above equation, we get:

$0.132=\frac{2.303}{t}\log\frac{100}{25}\\\\t=10.5yrs$

The original sample was purchased in June 2016

As, June is the 6th month of the year, which means the time period will be $2016+\frac{6}{12}=2016.5$

Adding the time in the original time period, we get:

$2016.5+10.5=2027$

Hence, the sample of Cobalt-60 isotope must be replaced in January 2027

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3 years ago

For hydrocyanic acid, HCN, Ka = 4.9 × 10-10. Calculate the pH of 0.20 M NaCN. What is the concentration of HCN in the solution?

gogolik [260]

By using the ICE table :

initial 0.2 M 0 0
change -X + X +X
Equ (0.2 -X) X X

when Ka = (X) (X) / (0.2-X)
so by substitution:

4.9x10^-10 = X^2 / (0.2-X) by solving this equation for X
∴X ≈ 10^-6
∴[HCN] = 10^-6

and PH = -㏒[H+]
= -㏒ 10^-6
= 6

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3 years ago

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nikdorinn [45]

Answer: 0.72 litres of water is wasted in one day.

Explanation:

First you need to find out how many minutes are in a day. Do this by multiplying the number of minutes in an hour (60) by the number of hours in a day (24). 24 x 60 = 1440. If the faucet is dripping at 5 drops per minute, then multiply 5 by the number of minutes in a day (1440) to see how many drops drip in one day. 5 x 1440 = 7200. Now we need to figure out how many mL fo water that is. if 10 drops is 1 mL, then we need to divide the total number of drops (7200) by 10. 7200 divided by 10 is 720. That means 720 mL of water is dripping per day. Finally, we must convert mL to litres. There are 1000 mL in one litre, so divide 720 by 1000. The final answer is 0.72

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3 years ago

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3 years ago

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Vinil7 [7]

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3 years ago