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Vadim26 [7]
3 years ago
15

If the half-life of hydrogen-3 is 11.8 years, after two half-lives the radioactivity of a sample will be reduced to one-half of

the original amount. If the half-life of hydrogen-3 is 11.8 years, after two half-lives the radioactivity of a sample will be reduced to one-half of the original amount. True False
Chemistry
1 answer:
maw [93]3 years ago
4 0

Answer:

False

Explanation:

Half life is the time period at which the concentration of the radioactive substance in decay reduced to half.

<u>Thus, if the hydrogen-3 has gone 2 half lives, it means that it has first reduced to its half and then again the half of what it was, i.e. 1/4</u>

Thus, after two successive half-lives, the concentration must be 1/4 of the initial concentration and hence, the statement is false.

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A buffer solution contains 0.345 M acetic acid and 0.377 M sodium acetate . If 0.0613 moles of potassium hydroxide are added to
melamori03 [73]

Answer:

pH = 5.54

Explanation:

The pH of a buffer solution is given by the <em>Henderson-Hasselbach (H-H) equation</em>:

  • pH = pKa + log\frac{[CH_3COO^-]}{[CH_3COOH]}

For acetic acid, pKa = 4.75.

We <u>calculate the original number of moles for acetic acid and acetate</u>, using the <em>given concentrations and volume</em>:

  • CH₃COO⁻ ⇒ 0.377 M * 0.250 L = 0.0942 mol CH₃COO⁻
  • CH₃COOH ⇒ 0.345 M * 0.250 L = 0.0862 mol CH₃COOH

The number of CH₃COO⁻ moles will increase with the added moles of KOH while the number of CH₃COOH moles will decrease by the same amount.

Now we use the H-H equation to <u>calculate the new pH</u>, by using the <em>new concentrations</em>:

  • pH = 4.75 + log\frac{(0.0942+0.0613)mol/0.250L}{(0.0862-0.0613)mol/0.250L} = 5.54
6 0
3 years ago
What monatomic ions do selenium (z = 34) and phosphorus (z = 15) form? (type your answer using the format [cu]+ for cu+ and [cu]
Alex787 [66]
The answer:
for the monoatomic <span>selenium ions
</span>    -the ion charge of selenium is 2-, so the answer is [Se]2+ 
as for the monoatomic phosphorus ions
    -the ion charge of phosphorus is 3-, so the answer is [P]3- 
6 0
3 years ago
Which of the following is not found in a nucleotide?
Firlakuza [10]

Amino acid is not found in a nucleotide.

8 0
3 years ago
the equilibrium concentration of hydroxide ion in a saturated iron(ii) hydroxide solution is 1.2 x 10^-5 M at a certain temperat
Anit [1.1K]

From the calculation as shpwn in the procedure below, the equilibrium constant of the substance is 6.9 * 10^-15.

<h3>What is equilibrium constant?</h3>

The equilibrium constant for the solubility of aa solid in solution is called the solubility product Ksp. The Ksp shows the extent to which a solid is dissolved in solution.

Given that;

Fe(OH)2 ⇄Fe^2+ + 2(OH)^-

Ksp = s(2s)^2

We have s as 1.2 x 10^-5 M

So

Ksp = 4s^3

Ksp = 4( 1.2 x 10^-5 )^3

Ksp = 6.9 * 10^-15

Learn more about Ksp:brainly.com/question/27132799

#SPJ1

8 0
2 years ago
It takes 60 mL of 0.20 M of sodium hydroxide (NaOH) to neutralize 25 mL of carbonic acid (H2CO3) for the following chemical reac
irina1246 [14]

If It takes 60mL of 0.20M of sodium hydroxide (NaOH) to neutralize 25 mL of carbonic acid (H_2CO_3), then the concentration of the carbonic acid is 0.24M

The reaction between NaOH solution and H_2CO_3 is written below

2NaOH + H_2CO_3 \rightarrow Na_2CO_3 + 2H_2O

Volume of NaOH, V_B = 60 ml

Volume of H_2CO_3, V_A=25 ml

Molarity of H_2CO_3, C_A=?

Molarity of NaOH, C_B=0.20M

Number of moles of H_2CO_3, n_A=1

Number of moles of NaOH, n_B=2

The mathematical equation for neutralization reaction is:

\frac{C_AV_A}{C_BV_B} =\frac{n_A}{n_B}

Substitute  C_B=0.2 M,  n_A=1,  n_B=2,  V_B = 60ml, and  V_A=25 ml into the equation above in order to solve for C_A

\frac{C_A \times 25}{0.2 \times 60}=\frac{1}{2}  \\\\50C_A=12\\\\C_B=\frac{12}{50} \\\\C_B=0.24M

Therefore, the concentration of the carbonic acid is 0.24M

Learn more here: brainly.com/question/25943090

3 0
2 years ago
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