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olganol [36]
3 years ago
13

The process in which a large atom is split into two smaller atoms and releases energy is called

Chemistry
1 answer:
vazorg [7]3 years ago
8 0
Answer is: <span>nuclear fission.
</span>Fission is nuclear reaction<span> or a </span>radioactive decay<span> process in which the </span>nucleus of an atom splits, <span>resulting atoms are not the same </span>elements<span> as the original atom</span>. The fission reaction<span> releases large amount of </span>energy.
<span>Nuclear fission was discovered on 1938.</span>
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HNO3 + H2O &gt; H3O(aq) + NO3-(aq)
Kruka [31]
What is the question and are there answers to go with it
8 0
3 years ago
Reactants are substances that?
RSB [31]
A substance that undergoes change during a reaction, usually from coming in contact with another substance
4 0
2 years ago
How many Ne atoms are contained in 32.0 g of the element?
Svetlanka [38]
Mass atomic of Ne=20.18 u
Therefore:
molar mass=20.18 g/1 mol

1 mole=6.022*10²³ particles (atoms or molecules)

Then: 6.022*10²³ atoms are contained in 20.18g

Now, We can solve this problem by the three rule.

6.022*10²³ atoms-------------------20.18 g
x------------------------------------------32 g

x=(6.022*10²³ atoms * 32 g)/20.18 g=9.55*10²³ atoms.

Answer: 9.55*10²³ Ne atoms are contained in 32 g of the element.
3 0
3 years ago
A mysterious white powder could be powdered sugar (C12H22O11), cocaine (C17H21NO4), codeine (C18H21NO3), norfenefrine (C8H11NO2)
rodikova [14]

Norfenefrine (C₈H₁₁NO₂).

<h3>Further explanation</h3>

We will solve a case related to one of the colligative properties, namely freezing point depression.

The freezing point of the solution is the temperature at which the solution begins to freeze. The difference between the freezing point of the solvent and the freezing point of the solution is called freezing point depression.

\boxed{ \ \Delta T_f = T_f(solvent) - T_f(solution) \ } \rightarrow \boxed{ \ \Delta T_f = K_f \times molality \ }

<u>Given:</u>

A mysterious white powder could be,

  • powdered sugar (C₁₂H₂₂O₁₁) with a molar mass of 342.30 g/moles,
  • cocaine (C₁₇H₂₁NO₄) with a molar mass of 303.35 g/moles,
  • codeine (C₁₈H₂₁NO₃) with a molar mass of 299.36 g/moles,
  • norfenefrine (C₈H₁₁NO₂) with a molar mass of 153.18 g/moles, or
  • fructose (C₆H₁₂O₆) with a molar mass of 180.16 g/moles.

When 82 mg of the powder is dissolved in 1.50 mL of ethanol (density = 0.789 g/cm³, normal freezing point −114.6°C, Kf = 1.99°C/m), the freezing point is lowered to −115.5°C.

<u>Question: </u>What is the identity of the white powder?

<u>The Process:</u>

Let us identify the solute, the solvent, initial, and final temperatures.

  • The solute = the powder
  • The solvent = ethanol
  • The freezing point of the solvent = −114.6°C
  • The freezing point of the solution = −115.5°C

Prepare masses of solutes and solvents.

  • Mass of solute = 82 mg = 0.082 g
  • Mass of solvent = density x volume, i.e., \boxed{ \ 0.789 \ \frac{g}{cm^3} \times 1.50 \ cm^3 = 1.1835 \ g = 0.00118 \ kg  \ }

We must prepare the solvent mass unit in kg because the unit of molality is the mole of the solute divided by the mass of the solvent in kg.

The molality formula is as follows:

\boxed{ \ m = \frac{moles \ of \ solute}{kg \ of \ solvent} \ } \rightarrow \boxed{ \ m = \frac{mass \ of \ solute \ (g)}{molar \ mass \ of \ solute \times kg \ of \ solvent} \ }

Now we combine it with the formula of freezing point depression.

\boxed{ \ \Delta T_f =  K_f \times \frac{mass \ of \ solute \ (g)}{molar \ mass \ of \ solute \times kg \ of \ solvent} \ }

It is clear that we will determine the molar mass of the solute (denoted by Mr).

We enter all data into the formula.

\boxed{ \ -114.6^0C - (-115.5^0C) = 1.99 \frac{^0C}{m} \times \frac{0.082 \ g}{Mr \times 0.00118 \ kg} \ }

\boxed{ \ 0.9 = \frac{1.99 \times 0.082}{Mr \times 0.00118} \ }

\boxed{ \ Mr = \frac{0.16318}{0.9 \times 0.00118} \ }

We get \boxed{ \ Mr = 153.65 \ }

These results are very close to the molar mass of norfenefrine which is 153.18 g/mol. Thus the white powder is norfenefrine.

<h3>Learn more</h3>
  1. The molality and mole fraction of water brainly.com/question/10861444
  2. About the mass and density of ethylene glycol as an  antifreeze brainly.com/question/4053884
  3. About the solution as a homogeneous mixture  brainly.com/question/637791

Keywords: a mysterious white powder, sugar, cocaine, codeine, norfenefrine, fructose, the solute, the solvent, dissolved, ethanol, normal freezing point, the freezing point depression, the identity

7 0
3 years ago
Read 2 more answers
What is the mass of silver in 4.3g AgNO3?
gregori [183]
4.22 grams.
1. First find out how much AgNO3 weighs with one mole (107.87 g Ag + 14.007 g N + 48 g O = 169.89 grams)
2. Find the percent of Ag you have. So, (107.87 g/mol Ag)/(169.89 g/mol AgNO3)= 0.63 * 100 = 63%.
3. If you have 6.7 grams total, you know 63% of it is going to be silver, so just multiply 6.7 grams by .63 and you get 4.22 g Ag
5 0
3 years ago
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