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egoroff_w [7]
3 years ago
8

If the mass of a radioactive substance is 8 grams and it has a half-life of 4 hours, how much mass remains after 8 hours?

Chemistry
2 answers:
lisov135 [29]3 years ago
6 0

Answer: 2 grams will be remaining.

Explanation: Half life is the time in which the amount of radioactive substance remains half.

For example if half life of a radioactive substance is 3 hours and we have 16 grams of it then in two hours, 8 grams will be remaining. In next two hours, 4 grams will be remaining. Similarly, in next two hours, 2 grams and further in next two hours 1 gram of the substance will be remaining.

For the given problem, the half life of the radioactive substance is 4 hours. It means the initial amount of this substance will remain half in 4 hours.

Originally we have 8 grams of the radioactive substance. In first 4 hours, 4 grams of the substance will be remaining. Now, in next 4 hours that is in total 8 hours, 2 grams of the substance will be remaining.

These problems could also be solved mathematically using the formula:

\frac{N}{N_0}=(0.5)^n

where, N_0 is the initial amount of the substance, N is remaining amount and n is the number of half lives.

number of half lives(n) = \frac{time}{half life}

From given information, time is 8 hours and half life is 4 hours.

So, n = \frac{8}{4}  = 2

Initial amount is given as 8 grams and it asks to calculate the remaining amount. Let's plug in the values in the equation:

\frac{N}{8}=(0.5)^2

\frac{N}{8} = 0.25

N = 8(0.25)

N = 2

So, from both ways 2 grams of the radioactive substance will be remaining after 8 hours.

Alborosie3 years ago
3 0
The sample will lose half of its mass after 4 hours. The half life.
82=4
The sample will lose half of the remaining four after another half life.
42=2      Hope this helps! :)
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3 years ago
Law of conservation of mass to explain why a chemical equation must be balanced?
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Chemical equations must always balance due to the principles outlined in The Law Of Conservation of Matter. This scientific law states that matter cannot be created out of nothing nor can it be destroyed.
7 0
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A 0.43g samle of KHP required 24.11cm of NaOH for neutralization. Calculate the molarity of NaOH
expeople1 [14]

Answer:

0.083 M

Explanation:

We'll begin by calculating the number of mole in 0.43 g of KHP (potassium hydrogen phthalate, C₈H₅O₄K). This is can be obtained as follow:

Mass of C₈H₅O₄K = 0.43 g

Molar mass of C₈H₅O₄K = (8×12) + (5×1) + (16× 4) + 39

= 96 + 5 + 64 + 39 = 204 g/mol

Mole of C₈H₅O₄K =?

Mole = mass / molar mass

Mole of C₈H₅O₄K = 0.43 / 204

Mole of C₈H₅O₄K = 0.002 mole

Next, we shall determine the number of mole of NaOH required to react with 0.43 g (i.e 0.002 mole) of KHP. This can be obtained as follow:

C₈H₅O₄K + NaOH → C₈H₄O₄KNa + H₂O

From the balanced equation above,

1 mole of KHP reacted with 1 mole of NaOH.

Therefore, 0.002 mole of KHP will also react with 0.002 mole of NaOH.

Next, we shall convert 24.11 cm³ to L. This can be obtained as follow:

1000 cm³ = 1 L

Therefore,

24.11 cm³ = 24.11 cm³ × 1 L / 1000 cm³

24.11 cm³ = 0.02411 L

Finally, we shall determine the molarity of NaOH. This can be obtained as follow:

Mole of NaOH = 0.00 2 mole

Volume = 0.02411 L

Molarity of NaOH =?

Molarity = mole /Volume

Molarity of NaOH = 0.002 / 0.02411

Molarity of NaOH = 0.083 M

3 0
2 years ago
How does the A Hreaction relate to the A He of molecules involved in a reaction?
igor_vitrenko [27]

Answer:

B. ΔHreaction = ΔH°f reactants- ΔH°f products

Explanation:

<em>Using Hess's law, it is possible to sum ΔH of several related reactions to find ΔH of a particular reaction</em>.

Having in mind Hess's law, ΔH°f is defined as the change in enthalpy during the formation of 1 mole of substance from its constituent elements (That is, pure elements, mono or diatomics, that have a ΔH° = 0).

For example, in ΔH°f of H₂O, the equation is:

H₂(g) + 1/2O₂(g) → H₂O(g)

The constituent elements with ΔH°f = 0 are H₂(g) and O₂(g).

Now, using Hess's law, you can sum the ΔH°f of substance in a reaction as, for example:

NaOH + HCl → H₂O + NaCl. ΔHr

The ΔH°f for each substance in the reaction is:

NaOH: Na + 1/2H₂ + 1/2O₂ → NaOH <em>(1)</em>

HCl: 1/2H₂ + 1/2Cl₂ → HCl <em>(2)</em>

H₂O: H₂ + 1/2O₂ → H₂O <em>(3)</em>

NaCl: Na + 1/2Cl₂ → NaCl <em>(4)</em>

The algebraic sum of (3) + (4) is -(ΔH°f reactants):

H₂ + 1/2O₂ + Na + 1/2Cl₂ → NaCl + H₂O ΔH°f reactants

This reaction - {(1)+(2)} ΔH°f products

NaOH + HCl → H₂O + NaCl.

ΔHr = ΔH°f reactants- ΔH°f products

In the example, we obtain this relationship that can be expanded for all reactions. Thus, right answer is:

<h3>B. ΔHreaction = ΔH°f reactants- ΔH°f products</h3>

8 0
3 years ago
If 7.25g of chlorine is reacted with excess hydrogen and produces 7.25 g HCl. What is the percent yield?
lozanna [386]

Answer:

94.61 %

Explanation:

percent yield = (actual yield / theoretical yield) X 100%

The balanced equation for the reaction is:

H2 (g ) + Cl2(g) => 2 HCl (l)

So, the theoretical yield =  

 7.25g of chlorine X (2mol of Cl / 35.453 g of Cl) X (2mol Cl / 2mol of HCl) X  37.469g of Hcl / 2mol of Hcl = 0.409 x 18.735 = 7.663g of Hcl

Using this theoretical yield and the provided value for actual yield, the percent yield = (actual yield / theoretical yield) X 100%

= (7.25 g / 7.663g) X 100

= 94.61 %

   

3 0
3 years ago
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