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otez555 [7]
3 years ago
13

What is POH of a solution containing hydrogen ion concentration of 0.008​

Chemistry
1 answer:
kotykmax [81]3 years ago
6 0

Given-

Hydrogen ion concentration pH = 0.008

Now,

we know that pH + pOH = 14

pOH = 14-0.008

pOH = 13.992

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A first-order reaction has a half-life of 16.7 s . How long does it take for the concentration of the reactant in the reaction t
andrew11 [14]

It takes 33.4 s for the concentration of A to fall to one-fourth of its original value.

A <em>half-life</em> is the time it takes for the concentration to fall to half its original value.

Assume the initial concentration is 1.00 mol/L. Then,

\text{1.00 mol/L }\stackrel{\text{1st half-life }}{\longrightarrow}\text{ 0.50 mol/L } \stackrel{\text{ 2nd half-life }}{\longrightarrow}\text{0.25 mol/L}\\

The concentration drops to one-fourth of its initial value in two half-lives.

∴ Time = 2 × 16.7 s = 33.4 s

5 0
2 years ago
I just need the right answer plz
astraxan [27]

Answer:

-0.5

Explanation:

5 0
2 years ago
How many grams of water will absorb a total of 2400 joules of energy when the temperature changes from 10.0°C to 30.0°C?​
Jobisdone [24]

Answer:

28.7 grams of water

Explanation:

Calorimetry problem:

Q = C . m . ΔT

2400 J = 4.18 J/g°C . m . (30°C - 10°C)

2400 J = 4.18 J/g°C . m . 20°C

2400J = 83.6 J/g . m

2400J / 83.6 g/J = m

28.7 g = m

8 0
3 years ago
How is atomic number and mass related to protons, neutrons, &amp; electrons?
artcher [175]
The atomic number is the numbers of protons

6 0
3 years ago
Read 2 more answers
What volume of 0.235 M H2SO4 is needed to titrate 40.0 mL of 0.0500 M Na2CO M NaCO3 solution?
Andreyy89

Answer: The volume of 0.235 M H_2SO_4 needed to titrate 40.0 mL of 0.0500 M Na_2CO_3 is 8.51 ml

Explanation:

To calculate the volume of acid, we use the equation given by neutralization reaction:

n_1M_1V_1=n_2M_2V_2

where,

n_1,M_1\text{ and }V_1 are the n-factor, molarity and volume of acid which is H_2SO_4

n_2,M_2\text{ and }V_2 are the n-factor, molarity and volume of base which is Na_2CO_3

We are given:

n_1=2\\M_1=0.235M\\V_1=?mL\\n_2=2\\M_2=0.0500M\\V_2=40.0mL

Putting values in above equation, we get:

2\times 0.235\times V_1=2\times 0.0500\times 40.0\\\\V_1=8.51mL

Thus volume of 0.235 M H_2SO_4 needed to titrate 40.0 mL of 0.0500 M Na_2CO_3 is 8.51 ml

5 0
3 years ago
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