Why are the atomic masses of elements usually decimal numbers? Provide an answer using 3 – 4 sentences in your own words.

Answers

Answer 1

Answer:

Explanation: An element does not exist alone in the nature. It exist in the form of isotopes of its own. Atomic mass of that element is the average atomic masses of the isotopes.

Average atomic mass of the element is defined as the sum of atomic masses of isotopes each multiplied with their respective natural fractional abundance.

Mathematically,

$\text{Average atomic mass}=\sum_(i=1)^n\text{(atomic mass of an isotopes)}_i* \text {(fractional abundance)}_i$

This is the reason, why an element does not have a whole number atomic mass, but exist in decimal numbers.

Answer 2

Answer:
The atomic masses of elements are usually written in decimal numbers because they are non-terminating. The numbers usually are rounded to the nearest thousandths place. The atomic masses are more exact in decimal numbers.

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Which radioisotope emits alpha particles?
(1) Fe-53 (3) Au-198
(2) Sr-90 (4) Pu-239

Answers

Answer : Option 4) Pu-239

Explanation : Plutonium-239 is a radioisotope of the element plutonium which emits alpha particles. The decaying process of Pu-239 emits alpha particles. It is widely used as a nuclear reactor fuel. Pu-239 is also used as a primary fissile isotope for the production of nuclear weapons.

The answer is (4) Pu-239. The decay mode of Pu-239 is α decay which means it will emit alpha particles. The other three have the decay mode of β decay.

A solution is highly concentrated if there is:a large volume of solution

a large amount of solvent and a small amount of solute

a large amount of solute and a small amount of solvent

a high density value for the solution

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TO be saturated you need an amount of solvent that will not dissolve anymore into the solution. So, the best answer from the list is C

If 4.00 g of NaCl react with 10.00 g of AgNO3, what is the excess reactant?

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Answer : The excess reagent is, $NaCl$

Explanation :

First we have to calculate the moles of $NaCl$ and $AgNO_3$ .

$\text{Moles of }NaCl=\frac{\text{Mass of }NaCl}{\text{Molar mass of }NaCl}=(4g)/(58.4g/mole)=0.068moles$

$\text{Moles of }AgNO_3=\frac{\text{Mass of }AgNO_3}{\text{Molar mass of }AgNO_3}=(10g)/(169.9g/mole)=0.058moles$

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

$NaCl+AgNO_3\rightarrow AgCl+NaNO_3$

From the balanced reaction we conclude that

As, 1 mole of $NaCl$ react with 1 mole of $AgNO_3$

So, 0.068 moles of $NaCl$ react with 0.068 moles of $AgNO_3$

From this we conclude that, the moles of $AgNO_3$ are less than the NaCl. So, $AgNO_3$ is a limiting reagent because it limits the formation of products and $NaCl$ is an excess reagent.

Hence, the excess reagent is, $NaCl$

The reaction formula of this is NaCl + AgNO3 = NaNO3 + AgCl. The mole number of NaCl is 4/58.5=0.068 mol. The mole number of AgNO3 is 10/170=0.059 mol. So the NaCl is excess.

Calculate the atmospheric pressure if the boiling point of water is 81 degrees Celsius.

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Answer : The atmospheric pressure of water if boiling at 81 °C will be 369 mmHg.

Explanation : The vapor pressure of boiling water can be calculated using the Antoine Equation:

log P = A - $[(B)/( C + T) ]$ ,

where P is the pressure in mmHg,

A = 8.07131,

B = 1730.63

C = 233.426

and T = temperature in °C (81°C)

Substituting the given values, we get,

Log P = 8.07131− $[(1730.63)/(233.426 + 81)]$

Therefore, Log P =8.07131 − 5.5041 = 2.567.

Now, P = $(10^(2.567) )$ =369 mmHg.

Hence, the atmospheric pressure will be 369 mmHg.

To calculate the atmospheric pressure at a specific boiling point of water, we use the steam table or the Antoine equation. From steam table via interpolation, at 81 C, the pressure where saturation occurs is the atmospheric pressure. The atmospheric pressure is 369.7 mm Hg or 0.4864 atm.

A finales del siglo XIX, los estudios y demostraciones que habían sobre los fenómenos eléctricos llevaban a sospechar que los átomos eran divisibles. ¿Cuáles fueron las experiencias realizadas por J.J Thomson para confirmar este hecho y qué postulado surgió de estas experiencias?

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Answer:

Joseph John Thomson studied the properties and the effects of the cathodic rays. Thomson's experiments involved the passage of electricity through a high-vacuum cathode-ray tube composed by a positive electrode (anode) and a negatively charged cathode. This experiment was crucial to describe the nature of the electric discharge.

Explanation:

Michael Faraday placed two electrodes in water solution and then he observed how electricity forces can separate elements in the solution.

William Crookes studied the forces that drive electricity phenomena by passing electricity through a gas in a sealed tube (cathode ray tubes).

Wilhelm Roentgen discovered the electromagnetic radiation in the X-ray spectrum

Finally, Ernest Rutherford was an apprentice of JJ Thompson. He designed an experiment involving alpha particles that were emitted by a radioactive element. This experiment showed that atoms have tiny and heavy nucleus.

How many grams of sucrose, C12H22O11, are in a 94.2 mL sample of a 2.6 M sucrose solution?

Answers

get the mass # of each element since its asking for grams
C- 12.011x12
H-1.008x22
O- 15.999x11
add all of those together to get the mass
put 2.6 moles on the top - so its the mass x 2.6 then divide by 94.2 Ml

Molar (M) means mol of solute in volume of solution (expressed by liters):

$2.6(mol)/(L)\cdot 94.2\cdot 10^(-3)\ L = 0.245\ mol$

The molecular weight of sucrose is: 12·12 + 22·1 + 11·16 = 342 g/mol.

$0.245\ mol\ C_(12)H_(22)O_(11)\cdot (342\ g)/(1\ mol) = \bf 83.79\ g\ C_(12)H_(22)O_(11)$

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