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patriot [66]
4 years ago
10

The amount of atomic particles released by a radioactive material in a specific time is determined by

Chemistry
2 answers:
shepuryov [24]4 years ago
8 0
The amount of atomic particles released by a radioactive material in a specific time is determined by the strong and the weak nuclear force. It is these forces that holds the nucleus of the atom and the force that needs to be overcome with to allow radioactive decay.
Georgia [21]4 years ago
6 0

The best answer is A. Strong and weak nuclear forces

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A concentration cell is constructed by using the same half-reaction for both the cathode and anode. What is the value of standar
Akimi4 [234]

Solution :

A cell that is concentrated is constructed by the same half reaction for the anode as well as he cathode.

We know,

In a standard cell,

the reduction half cell reaction is :

$Ag^+(aq)+e^- \rightarrow Ag(s) E^0 = -0.80 \ V$

The oxidation half ell reaction :

$Ag(s) \rightarrow Ag^+(aq) + e^- \ E^0= +0.80 \ V$

Thus the complete reaction of the cell is :

$Ag^+(aq)+ Ag(s) \rightarrow Ag^+(aq)+Ag(s)$

$E^0 $ cell = $E_R - E_L = 0.00  \ \text{volts}$

7 0
3 years ago
Aspirin (C9H8O4) is an acid which can be titrated with a base to determine purity. If an aspirin tablet weighing 0.615 g is titr
Iteru [2.4K]

Answer:

67.4 % of C₉H₈O₄

Explanation:

To make titrations problems we know, that in the endpoint:

mmoles of acid = mmoles of base

mmoles = M . volume so:

mmoles of acid = 20.52 mL . 0.1121 M

mmoles of acid = mg of acid / PM (mg /mmoles)

Let's determine the PM of aspirin:

12.017 g/m . 9 + 1.00078 g/m . 8 + 15.9994 g/m . 4 = 180.1568 mg/mmol

mass (mg) = (20.52 mL . 0.1121 M) . 180.1568 mg/mmol

mass (mg) = 414.4 mg

We convert the mass to g → 414.4 mg . 1g / 1000mg = 0.4144 g

We determine the % → (0.4144 g / 0.615 g) . 100 = 67.4 %

6 0
3 years ago
There are 2 gasses, A, B. They weigh 2.46g and 0.5g respectively, and the Volume of A is 3 times the volume of B. A has a molecu
dangina [55]

Answer:

B

Explanation:

molecular mass of B is 28

8 0
2 years ago
Read 2 more answers
Did I get these correctly?
Luda [366]
Those are both correct! great job, keep up the good work (-:
4 0
3 years ago
To demonstrate the formation of iron (iii) chloride from iron fillings
Pavel [41]

Iron (iii) chloride is obtained by vapor condensation from the reaction between chlorine gas and iron fillings.

<h3>How can iron (iii) chloride be formed from iron fillings?</h3>

Iron (ii) chloride can be formed from iron fillings in the laboratory as follows:

  • Iron fillings + Cl₂ → FeCl₃

Chlorine gas is introduced into a reaction vessel containing iron fillings and the iron (iii) chloride vapor formed is obtained by condensation.

In conclusion, iron (iii) chloride is formed by the the direct combination of iron fillings and chlorine gas.

Learn more about iron (iii) chloride at: brainly.com/question/14653649

#SPJ1

5 0
2 years ago
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