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LUCKY_DIMON [66]
3 years ago
15

What term defines the specific amount of time required for half of a radioactive substance to become stable?

Chemistry
2 answers:
Genrish500 [490]3 years ago
6 0
A. half-life that is the correct answer because i"ve seen this question before. 

goldenfox [79]3 years ago
5 0
Half life is the specific amount of time required for half of a radioactive substance to become stable.

You might be interested in
A 20.0 mL 0.100 M solution of lactic acid is titrated with 0.100 M NaOH.
yan [13]

Answer:

(a) See explanation below

(b) 0.002 mol

(c) (i) pH = 2.4

(ii) pH = 3.4

(iii) pH = 3.9

(iv) pH = 8.3

(v) pH = 12.0

Explanation:

(a) A buffer solution exits after addition of 5 mL of NaOH  since after reaction we will have  both the conjugate base lactate anion and unreacted weak  lactic acid present in solution.

Lets call lactic acid HA, and A⁻ the lactate conjugate base. The reaction is:

HA + NaOH ⇒ A⁻ + H₂O

Some unreacted HA will remain in solution, and since HA is a weak acid , we will have the followin equilibrium:

HA  + H₂O ⇆ H₃O⁺ + A⁻

Since we are going to have unreacted acid, and some conjugate base, the buffer has the capacity of maintaining the pH in a narrow range if we add acid or base within certain limits.

An added acid will be consumed by the conjugate base A⁻ , thus keeping the pH more or less equal:

A⁻ + H⁺ ⇄ HA

On the contrary, if we add extra base it will be consumed by the unreacted lactic acid, again maintaining the pH more or less constant.

H₃O⁺ + B ⇆ BH⁺

b) Again letting HA stand for lactic acid:

mol HA =  (20.0 mL x  1 L/1000 mL) x 0.100 mol/L = 0.002 mol

c)

i) After 0.00 mL of NaOH have been added

In this case we just have to determine the pH of a weak acid, and we know for a monopric acid:

pH = - log [H₃O⁺] where  [H₃O⁺] = √( Ka [HA])

Ka for lactic acid = 1.4 x 10⁻⁴  ( from reference tables)

[H₃O⁺] = √( Ka [HA]) = √(1.4 x 10⁻⁴ x 0.100) = 3.7 x 10⁻³

pH = - log(3.7 x 10⁻³) = 2.4

ii) After 5.00 mL of NaOH have been added ( 5x 10⁻³ L x 0.1 = 0.005 mol NaOH)

Now we have a buffer solution and must use the Henderson-Hasselbach equation.

                            HA          +         NaOH          ⇒   A⁻ + H₂O

before rxn         0.002                  0.0005                0

after rxn    0.002-0.0005                  0                  0.0005

                        0.0015

Using Henderson-Hasselbach equation :

pH = pKa + log [A⁻]/[HA]

pKa HA = -log (1.4 x 10⁻⁴) = 3.85

pH = 3.85 + log(0.0005/0.0015)

pH = 3.4

iii) After 10.0 mL of NaOH have been ( 0.010 L x 0.1 mol/L = 0.001 mol)

                             HA          +         NaOH          ⇒   A⁻ + H₂O

before rxn         0.002                  0.001               0

after rxn        0.002-0.001                  0                  0.001

                        0.001

pH = 3.85 + log(0.001/0.001)  = 3.85

iv) After 20.0 mL of NaOH have been added ( 0.002 mol )

                            HA          +         NaOH          ⇒   A⁻ + H₂O

before rxn         0.002                  0.002                 0

after rxn                 0                         0                   0.002

We are at the neutralization point and  we do not have a buffer anymore, instead we just have  a weak base A⁻ to which we can determine its pOH as follows:

pOH = √Kb x [A⁻]

We need to determine the concentration of the weak base which is the mol per volume in liters.

At this stage of the titration we added 20 mL of lactic acid and 20 mL of NaOH, hence the volume of solution is 40 mL (0.04 L).

The molarity of A⁻ is then

[A⁻] = 0.002 mol / 0.04 L = 0.05 M

Kb is equal to

Ka x Kb = Kw ⇒ Kb = 10⁻¹⁴/ 1.4 x 10⁻⁴ = 7.1 x 10⁻¹¹

pOH is then:

[OH⁻] = √Kb x [A⁻]  = √( 7.1 x 10⁻¹¹ x 0.05) = 1.88 x 10⁻⁶

pOH = - log (  1.88 x 10⁻⁶ ) = 5.7

pH = 14 - pOH = 14 - 5.7 = 8.3

v) After 25.0 mL of NaOH have been added (

                            HA          +         NaOH          ⇒   A⁻ + H₂O

before rxn           0.002                  0.0025              0

after rxn                0                         0.0005              0.0005

Now here what we have is  the strong base sodium hydroxide and A⁻ but the strong base NaOH will predominate and drive the pH over the weak base A⁻.

So we treat this part as the determination of the pH of a strong base.

V= (20 mL + 25 mL) x 1 L /1000 mL = 0.045 L

[OH⁻] = 0.0005 mol / 0.045 L = 0.011 M

pOH = - log (0.011) = 2

pH = 14 - 1.95 = 12

7 0
2 years ago
scandium47 has a half-life of 35s. suppose you have a 45g sample of scadium 47 how much of the sample remains unchanged after 14
Mars2501 [29]

 The much  of the sample that would remain  unchanged  after 140 seconds is 2.813 g

Explanation

Half life  is time taken for the quantity  to reduce  to half its original value.

if the half life  for Scandium  is 35 sec, then the number  of half life in 140 seconds

=140 sec/ 35 s = 4 half life

Therefore 45 g after first half life = 45 x1/2 =22.5 g

               22.5 g after second half life = 22.5 x 1/2 =11.25 g

            11.25 g after third half life = 11.25 x 1/2 = 5.625 g

             5.625 after  fourth half life = 5.625 x 1/2 = 2.813

therefore 2.813 g  of Scandium 47 remains  unchanged.

4 0
2 years ago
Which way would CCI2F2 orient in an electric field?
ArbitrLikvidat [17]

Answer:

See explanation

Explanation:

In looking at molecules to determine whether they are polar or not we have to look at two things basically;

i) presence of polar bonds

ii) geometry of the molecule

Now, we know that CCI2F2 is a tetrahedral molecule, but the molecule is not symmetrical. It has four polar bonds that are not all the same hence the molecule is polar.

In an electric field, polar molecules orient themselves in such a way that  the positive ends of the molecule are being attracted to the negative plate while the negative ends of the molecules are attracted to the positive plate.

So the positive ends of CCI2F2  are oriented towards the negative plate of the field while the negative ends of CCI2F2 are oriented towards the positive ends of the field.

3 0
2 years ago
Will bromine react with sodium
denis-greek [22]
Yes. bromine and sodium iodide can react to form sodium bromine and free iodine
4 0
2 years ago
Qué tipo de reacción es el siguiente? CH4 + 02<br><br> CO2 + H2O
Tatiana [17]

Answer:

Combustion reaction

Explanation:

Let's consider the following balanced equation.

CH₄ + 2 O₂  ⇒ CO₂ + 2 H₂O

This reaction is known as a combustion reaction, in which a compound reacts with oxygen to form a compound of carbon and water.

  • If the product is carbon dioxide, the combustion is complete.
  • If the product is carbon monoxide or carbon, the combustion is incomplete.
7 0
3 years ago
Read 2 more answers
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