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3 years ago

Which radioisotopes have the same decay mode and have half-lives greater than

Chemistry

2 answers:

Damm [24]3 years ago

5 0

The answer is (3) I-131 and P-32. They both have β- decay mode and with half-lives greater than hour. You can get these half-lives and decay mode from table N selected radioisotopes.

AveGali [126]3 years ago

4 0

The correct answer is (3)

I-131 and P-32

The explanation:

according to attached table:

- we can see that the half life of p 32 is 14.28d (more than one hour)

- and the half life of I-131 is 8.021 d (more than one hour)

and They both have β- decay mode and with half-lives greater than hour.

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The key enzyme in the regulation of the citric acid cycle is:.

bixtya [17]

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2 years ago

Need help on this question asap pleasee

algol [13]

Answer:

9.0 moles of CaO

Explanation:

We have the reaction equation as follows;

Fe2O3 + Ca3(PO4)2 -------> 2FePO4 + 3CaO

Now we know from the equation that;

1 mole of iron III oxide yields 3 moles of CaO

Therefore;

3 moles of iron III oxide yields 3 * 3/1

= 9.0 moles of CaO

6 0

2 years ago

Particles of matter how have both potential and kinetic energy is that true or false?

guapka [62]

False: No,any particles of matter do not have any potential or kinetic energy.

5 0

3 years ago

1. The solubility of lead(II) chloride at some high temperature is 3.1 x 10-2 M. Find the Ksp of PbCl2 at this temperature.

solniwko [45]

Answer:

1) The solubility product of the lead(II) chloride is $1.2\times 10^{-4}$ .

2) The solubility of the aluminium hydroxide is $1.6\times 10^{-10} M$ .

3)The given statement is false.

Explanation:

Solubility of lead chloride = $S=3.1\times 10^-2M$

$PbCl_2(aq)\rightleftharpoons Pb^{2+}(aq)+2Cl^-(aq)$

S 2S

The solubility product of the lead(II) chloride = $K_{sp}$

$K_{sp}=[Pb^{2+}][Cl^-]^2$

$K_{sp}=S\times (2S)^2=4S^3=4\times (3.1\times 10^{-2})^3=1.2\times 10^{-4}$

The solubility product of the lead(II) chloride is $1.2\times 10^{-4}$ .

Concentration of aluminium nitrate = 0.000010 M

Concentration of aluminum ion = $1\timed 0.000010 M=0.000010 M$

Solubility of aluminium hydroxide in aluminum nitrate solution = $S$

$Al(OH)_3(aq)\rightleftharpoons Al^{3+}(aq)+3OH^-(aq)$

S 3S

The solubility product of the aluminium nitrate = $K_{sp}=1.0\times 10^{-33}$

$K_{sp}=[Al^{3+}][OH^-]^3$

$1.0\times 10^{-33}=(0.000010+S)\times (3S)^3$

$S=1.6\times 10^{-10} M$

The solubility of the aluminium hydroxide is $1.6\times 10^{-10} M$ .

$Molarity=\frac{Moles}{Volume (L)}$

Mass of NaCl= 3.5 mg = 0.0035 g

1 mg = 0.001 g

Moles of NaCl = $\frac{0.0035 g}{58.5 g/mol}=6.0\times 10^{-5} mol$

Volume of the solution = 0.250 L

$[NaCl]=\frac{6.0\times 10^{-5} mol}{0.250 L}=0.00024 M$

1 mole of NaCl gives 1 mole of sodium ion and 1 mole of chloride ions.

$[Cl^-]=[NaCl]=0.00024 M$

Moles of lead (II) nitrate = n

Volume of the solution = 0.250 L

Molarity lead(II) nitrate = 0.12 M

$n=0.12 M]\times 0.250 L=0.030 mol$

1 mole of lead nitrate gives 1 mole of lead (II) ion and 2 moles of nitrate ions.

$[Pb^{2+}]=[Pb(NO_2)_3]=0.030 M$

$PbCl_2(aq)\rightleftharpoons Pb^{2+}(aq)+2Cl^-(aq)$

Solubility of lead(II) chloride = $K_{sp}=1.2\times 10^{-4}$

Ionic product of the lead chloride in solution :

$Q_i=[Pb^{2+}][Cl^-]^2=0.030 M\times (0.00024 M)^2=1.7\times 10^{-9}$

$Q_i$ ( no precipitation)

The given statement is false.

3 0

3 years ago

What would you need to do to calculate the molality of 10 g of NaCl in 2 kg of

jasenka [17]

Answer:

O B. Convert the 10 g of NaCl to moles of NaCl.

Explanation:

The formula for finding the molality is m=moles of solute/kg of solvent. The solute for this question is NaCl and the solvent is water.

(10g NaCl)(1 mol NaCl/58.44g NaCl)=0.1711 mol NaCl

58.44 is the molar mass of NaCl

m=0.1711 mol NaCl/2 kg H2O

m=0.085557837

7 0

3 years ago