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Dima020 [189]
3 years ago
9

The half life of a certain radioactive element is 800 years. How old is an object if only 12.5% of radioactive atoms in it remai

ns?
Chemistry
1 answer:
Levart [38]3 years ago
6 0

Given,  half life of a certain radioactive element = 800 years.

Amount of substance remaining at time t = 12.5%

Lets consider the initial amount of the radioactive substance  = 100%

Using the half life equation:

A = A₀(1/2)^t/t₁/₂

where A₀ is the amount of radioactive substance at time zero and A is the amount of radioactive substance at time t, and t₁/₂ is the half-life of the radioactive substance.

Plugging the given data into the half life equation we have,

12.5 = 100 . (1/2)^t/800

12.5/100 = (1/2)^t/800

0.125 = (0.5)^t/800

(0.5)^3 = (0.5)^t/800

3 = t/800

t = 2400 years

Thus the object is 2400 years  old.



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Aluminum chloride can be formed from its elements:
saul85 [17]

<u>Answer:</u> The \Deltas H^o_{formation} for the reaction is -1406.8 kJ.

<u>Explanation:</u>

Hess’s law of constant heat summation states that the amount of heat absorbed or evolved in a given chemical equation remains the same whether the process occurs in one step or several steps.

According to this law, the chemical equation is treated as ordinary algebraic expressions and can be added or subtracted to yield the required equation. This means that the enthalpy change of the overall reaction is equal to the sum of the enthalpy changes of the intermediate reactions.

The chemical reaction for the formation reaction of AlCl_3 is:

2Al(s)+3Cl_2(g)\rightarrow 2AlCl_3 (s)    \Delta H^o_{formation}=?

The intermediate balanced chemical reaction are:

(1) HCl(g)\rightarrow HCl(aq.)    \Delta H_1=-74.8kJ    ( ×  6)

(2) H_2(g)+Cl_2(g)\rightarrow 2HCl(g)    \Delta H_2=-185kJ     ( ×  3)

(3) AlCl_3(aq.)\rightarrow AlCl_3(s)    \Delta H_3=+323kJ     ( ×  2)

(4) 2Al(s)+6HCl(aq.)\rightarrow 2AlCl_3(aq.)+3H_2(g)    \Delta H_4=-1049kJ

The expression for enthalpy of formation of AlCl_3 is,

\Delta H^o_{formation}=[6\times \Delta H_1]+[3\times \Delta H_2]+[2\times \Delta H_3]+[1\times \Delta H_4]

Putting values in above equation, we get:

\Delta H^o_{formation}=[(-74.8\times 6)+(-185\times 3)+(323\times 2)+(-1049\times 1)]=-1406.8kJ

Hence, the \Deltas H^o_{formation} for the reaction is -1406.8 kJ.

6 0
2 years ago
What is the molar mass of cholesterol if 0.00105 mol weighs 0.406 g?
goblinko [34]
<h3>Answer: 386.67 g/mol </h3>

Explanation:

Molar Mass = Mass ÷ Mole

                   =  0.406 g ÷ 0.00105 mol

                   =  386.67 g/mol

∴ molar mass of cholesterol = 386.67 g/mol

5 0
2 years ago
Calculate a reasonable amount (mass in g) of your unknown acid to use for a titration. You will want about 30 mL of titrant to g
Vinil7 [7]

Answer:

"0.60 g" is the appropriate solution.

Explanation:

The given values are:

Volume of base,

= 30 ml

Molarity of base,

= 0.05 m

Molar mass of acid,

= 400 g/mol

As we know,

⇒  Molarity=\frac{Number \ of \ moles \ of \ base}{Number \ of \ solution}

On substituting the values, we get

⇒           0.05=\frac{Number \ of \ moles \ of \ base}{30\times 10^{-3}}

⇒  Number \ of \ moles \ of \ base=0.05\times 30\times 10^{-3}

⇒                                             =1.5\times 10^{-3}  

hence,

⇒  Moles \ of \ acid=\frac{Mass \ of \ acid}{Molar \ mass \ of \ acid}

On substituting the values, we get

⇒  1.5\times 10^{-3}=\frac{Mass \ of \ acid}{400}

⇒  Mass \ of \ acid=1.5\times 10^{-3}\times 400

⇒                         =0.60 \ g

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11Alexandr11 [23.1K]

Answer:

1.0

Explanation:

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HCl(aq) → H⁺(aq) + Cl⁻(aq)

Then, the concentration of H⁺ will be equal to the initial concentration of the acid, i.e., 0.10 M.

We can calculate the pH using the following expression.

pH = -log [H⁺] = -log 0.10 = 1.0

3 0
3 years ago
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aivan3 [116]
Newton's third law of motion states that for every action force, there is an equal and opposite reaction force so that means that the wall is pushing you with the same amount of force that you put on it.
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3 years ago
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