PromptDescribe the general process by which new elements are synthesized.


Please help!!!!

Answers

Answer 1
Answer:

New elements are synthesized from the nuclear reaction of already existing elements. The change in nuclear composition of an atom leads to the conversion to new element.

What are elements ?

Elements are the basic things of anything in the world. Elements join together to form molecule and compounds. There are 118 known elements in periodic table.

Many more elements  discovered by scientists are accidental. The properties of each elements are determined based on their atomic number. The nuclear reactions mostly for radioactive elements make new elements with different nuclear composition.

New elements are produced as sub products that are resulted from several synthesis reactions involving the nuclear change. Radioactivity involving bombardment of charged particles such as alpha, beta or their emission also leads to formation of new elements.

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Answer 2
Answer:

Answer:

Describing Describe the general process by which new elements are synthesized.


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Four beakers containing potassium nitrate dissolved in water are allowed to evaporate to dryness. Beakers 1 through 4 contain 2.3, 1.91, 5.985, and 0.52 g of dry potassium nitrate respectively. How many moles of potassium nitrate were recovered after the water evaporated?

Answers

Explanation:

Molar mass of potassium nitrate will be calculated as follows.

   Molar mass KNO_(3) = molar mass of K + molar mass of N + 3 × molar mass of O

                             = 39.098 g/mol + 14.006 g/mol + 3 × 15.999 g/mol

                             = 102.102 g/mol

Now, adding the given amount of potassium nitrate present in each beaker as follows.

               (2.3 + 1.91 + 5.985 + 0.52) g

               = 10.715 g

Therefore, calculate number of moles as follows.

                   No. of moles = \frac{mass}{\text{molar mass}}

                                         = (10.715 g)/(102.102 g/mol)

                                         = 0.105 mol

Thus, we can conclude that 0.105 moles of potassium nitrate were recovered after the water evaporated.

Barium chloride + sodium phosphate 
Answer with double replacement

Answers

Answer:

Barium chloride + Sodium phosphate    →   barium phosphate + sodium chloride

Explanation:

Double replacement:

It is the reaction in which two compound exchange their ions and form new compounds.

AB + CD → AC +BD

Chemical equation:

BaCl₂ + Na₃PO₄         →     Ba₃(PO₄)₂ + NaCl

Balanced chemical equation:

3BaCl₂ + 2Na₃PO₄         →     Ba₃(PO₄)₂ + 6NaCl

The cation and anion of both reactants are exchanged with each other.

Ba²⁺ react with PO₄³⁻ and form Ba₃(PO₄)₂ while Cl⁻ react with Na⁺ and form sodium chloride.  

Molecular equation:

Barium chloride + Sodium phosphate    →   barium phosphate + sodium chloride

Final answer:

A double replacement reaction between barium chloride and sodium phosphate results in the formation of barium phosphate and sodium chloride.

Explanation:

The given chemical equation represents a double replacement reaction. The chemical reactants are barium chloride (BaCl2) and sodium phosphate (Na3PO4). In a double replacement reaction, the cations and anions of the two reactants switch places to form two new compounds.

So here is how the reaction would proceed: BaCl2 + Na3PO4 -> Ba3(PO4)2 + NaCl.

This translates to: Barium chloride reacts with sodium phosphate to form barium phosphate and sodium chloride.

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Benzene is a starting material in the synthesis of nylon fibers and polystyrene (styrofoam). Its specific heat capacity is 1.74 J/g·°C. If 16.7 kJ of energy is absorbed by a 225-g sample of benzene at 20.0°C, what is its final temperature?

Answers

Answer: The final temperature of the sample is 62.66°C

Explanation:

To calculate the amount of heat absorbed, we use the equation:

Q=mc\Delta T

where,

Q = heat absorbed = 16.7 kJ = 16700 J   (Conversion factor:  1 kJ = 1000 J)

m = Mass of the sample = 225 g

c = specific heat capacity of sample = 1.74J/g.^oC

\Delta T = change in temperature = T_2-T_1=(T_2-20.0)

Putting values in above equation, we get:

16700=225g* 1.74J/g.^oC* (T_2-20)^oC\n\nT_2=62.66^oC

Hence, the final temperature of the sample is 62.66°C

A student is heating a chemical in a beaker with a Bunsen burner.In a paragraph of at least 150 words, identify the safety equipment that should be used and the purpose of it for the given scenario.

Answers

When a student is warming a chemical in a container using a special burner, it is very important to focus on safety by using the right safety tools.

What is the safety equipment

First, the student needs to wear the right safety clothes like a lab coat, gloves, and goggles to protect themselves from getting splashed or hurt by chemicals. A lab coat stops chemicals from touching the skin, gloves keep the hands safe, and safety goggles protect the eyes from chemicals

and hot things.

Furthermore, using a fume hood is necessary to make sure there is enough fresh air circulating and to remove any dangerous fumes or gases that might be released while heating things up.

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Answer:The student should be wearing a lab coat or maybe an apron to prevent chemicals from spilling or exploding onto their clothes, I do recommend a lab coat better though because it can protect your skin better. Next, make sure while messing with chemicals you are always wearing goggles, if you are not wearing them there is a chance that after touching chemicals you could touch your eyes. And that brings me to washing your hands straight away after messing with chemicals. You could also wear gloves and just take them off when you're done but if you don't have clean hands afterward you could always put the chemicals all over your skin. But in case you do touch your eyes there is always an emergency eyewash station somewhere in the lab room. And if you are to get Chemicals on your skin, in your hair, on your clothes, or to be on fire, there shall be a shower somewhere to get rid of that. But if you read the instructions or listen closely to the teacher you shall have no problem.

Explanation:

I kinda got off topic

A rock contains 0.37 mg of Pb-206 and 0.95 mg of U-238. Approximately how many U-238 atoms were in the rock when it was formed billions of years ago? (The half-life for 238U  206Pb is 4.5  109 yr.)

Answers

Half-life of a radioactive element

The Half-Life of a radioactive element osbtime taken for half the nucleus of the atom of the element to decay

Calculating the original amount of U-238

  • The number of moles present in each element is first determined:
  • Number of moles = mass/molar mass

For Pb-206:

mass = 0.37 mg = 0.37 * 10⁻³ g

molar mass = 206 g/mol

number of moles = 0.37 * 10⁻³ g/206 g/mol

number of moles of Pb-206 = 1.79 * 10⁻⁶ moles

For U-238:

mass = 0.95 mg = 0.95 * 10⁻³ g

molar mass = 238 g/mol

number of moles = 0.95 * 10⁻³ g/238 g/mol

number of moles = 3.99 * 10⁻⁶ moles

Assuming that all the Pb-206 were formed from U-238

  • Initial moles of U-238 = Moles of present U-238 + molesw of present Pb-208

Initial moles of U-238 = 3.99 * 10⁻⁶ moles + 1.79 * 10⁻⁶ moles

Initial moles of U-238 = 5.78 * 10⁻⁶ moles

One mole of U-238 contains = 6.02 * 10²³ atoms

5.78 * 10⁻⁶ moles of U-238 will contain 6.02 * 10²³ * 5.78 * 10⁻⁶ atoms

Number of atoms of U-238 initially present = 3.48 * 10¹⁸ atoms

Therefore, the number of atoms of U-238 initially present is 3.48 * 10¹⁸ atoms

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The voltage generated by the zinc concentration cell described by, zn(s)|zn2 (aq, 0.100 m)||zn2 (aq, ? m)|zn(s),is 16.0 mv at 25 °c. calculate the concentration of the zn2 (aq) ion at the cathode.

Answers

The concentration cell is:
Zn(s) \ Zn²⁺(aq,0.100 M) // Zn²⁺(aq, x M) \ Zn(s)
voltage = 16 mV x (1V / 10³ mV) = 16 x 10⁻³ V
- In the cell notation, the concentration on the left is that of the anode and that on the right is that of the cathode.
- Oxidation takes place at the anode and reduction takes place at the cathode.
so [Zn²⁺]oxidation = 0.100 M
[Zn²⁺] reduction = x M
From Nernst equation:
Ecell = -0.0592 / n log [Zn²⁺] oxidation / [Zn²⁺]reduction
Number of electrons, n = 2. Substitute and solve for x:
16 x 10⁻³ V = - 0.0592 / 2 log (0.100 /x)
log 0.100 / x = - 0.54
0.100 / x = 0.288
x = 0.347
So the concentration of Zn²⁺ at the cathode = 0.406