1.Tritium, of hydrogen-3, is prepared by bombarding lithium-6 with neutrons. A 0.250-mg sample of tritium decays at the rate of 8.94 x 1010 disintegrations per second. What is the decay constant (in per sec) of tritium, whose atomic mass is 3.02 amu

The molarity of KOH is 0.2678 moles/L.

           Molarity(M) = number of moles/ volume of solutions in liter.

      Number of moles(n) = mass of the substance/Molar mass

The mass of Oxalic acid is 0.546g.

The molar mass of Oxalic acid [tex]H_{2}C_{2}O_{4}[/tex] is 90g (1*2+12*2+16*4=90)

The number of moles of Oxalic acid is,

n = 0.546/90

n = 0.006 moles

So, the number of moles of Oxalic acid is 0.006 moles.

The volume of KOH is 22.40mL = 22.40*1000 = 0.0224L

Therefore the molarity of KOH is,

M = number of moles/ volume of solutions in liter

M = 0.006/0.0224

M = 0.2678 moles/L

Therefore, the molarity of KOH is 0.2678 moles/L

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Taking into account the Charles' law, the temperature of the gas if the volume drops to 75.0 L is 246 K or -27 C.

Charles' law establishes the relationship between the temperature and the volume of a gas when the pressure is constant. This law says that for a given sum of gas at constant pressure, as the temperature increases, the volume of the gas increases and as the temperature decreases, the volume of the gas decreases. That is, the volume is directly proportional to the temperature of the gas.

Mathematically, Charles' law is a law that says that when the amount of gas and pressure remain constant, the ratio between volume and temperature will always have the same value:

[tex]\frac{V}{T}=k[/tex]

Considering an initial state 1 and a final state 2, it is fulfilled:

[tex]\frac{V1}{T1}=\frac{V2}{T2}[/tex]

In this case, you know:

Replacing in Charles's law:

[tex]\frac{100 L}{328 K}=\frac{75 L}{T2}[/tex]

Solving:

[tex]T2x\frac{100 L}{328 K}=75 L[/tex]

[tex]T2=\frac{75 L}{\frac{100 L}{328 K}}[/tex]

T2= 246 K= -27 C

Finally, the temperature of the gas if the volume drops to 75.0 L is 246 K or -27 C.

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Three examples of environmental, industrial and bio-chemistry are listed below:

Chemistry can be defined as a branch of science which is concerned with the substances matter is composed of, their properties and reactions,

Chemistry also deals with the use of such reactions to form new substances.

In conclusion, Three examples of environmental, industrial and bio-chemistry are listed anove

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In a Base catalyst elimination, the base extracts a proton and the leaving group leaves simultaneously. In a Concentrated process, the leaving group leaves  the base abstracts a proton.

Just as there are two mechanisms of substitution (SN2 and S N1), there are two

mechanisms of elimination (E2 and E1).

E2 mechanism — bimolecular elimination

E1 mechanism — unimolecular elimination

The E2 and E1 mechanisms differ in the timing of bond cleavage and bond formation, analogous to the SN2 and S N1 mechanisms.

E2 and S N2 reactions have some features in common, as do E1 and SN1 reactions.

Base removes a proton from the β-carbon atom,

while the halogen atom leaves from the α-carbon resulting in the formation of a π-bond. Such eliminations are also called β-eliminatione liminations are also called β elimination reactions

In a base-catalyzed elimination reaction, the base takes away a proton from the β position of the leaving group (the group that goes away with the bonding pair of electrons) and a double bond is formed. When the base is bulky it prefers to take away the proton from the less hindered position.

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The dehydration-condensation reaction occurs between 'ethanol' and 'butanoic acid' giving an ester called as "ethyl butanoate".

Esters are produced when carboxylic acids are heated with alcohols in the presence of an acid catalyst.

The catalyst is usually concentrated sulphuric acid.

Dry hydrogen chloride gas is used in some cases, but these tend to involve aromatic esters (ones containing a benzene ring).

Esters are represented by the formula RCOOR', where R and R' are hydrocarbon groups.

The chemical reaction occurring in the formation of the ester is known as an esterification reaction.

The reaction that occurs between ethanol and butanoic acid:

Ethanol reacts with butanoic acid in the presence of an acid-catalyst and gives ethyl butanoate as ester and water as products. If an acid reacts with a base, salt and water are the products. But ester is not a salt.

i) The carbon skeletal notation for the given reaction is shown below.

ii) The ester is named "ethyl butanoate".

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The best reactant is: iPrCl

The best reagent is: Mg

The best solvent is: CH3CH2OCH2CH3

A chemical molecule with the general formula RMgX, where X is a halogen and R is typically an alkyl or aryl group, is known as a Grignard reagent or Grignard compound. Methylmagnesium chloride (ClMgCH3) and phenylmagnesium bromide (C6H5)-Mg-Br, respectively, are two prominent examples. They fall within the category of organomagnesium compounds.

Typically, magnesium ribbon is needed for the reaction that results in the synthesis of Grignard reagents. A magnesium oxide passivating coating protects the whole magnesium, preventing interactions with organic halides. As an initiator, Grignard's prepared reagent can also be applied.

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Answer: The concentration of ammonia is 0.0286M.

Explanation:

The diluted equation is presented as: [tex]M_{s} V_{s} =M_{d} V_{d}[/tex].

~M= Molarity (labeled as M)

~V= Volume (labeled in L)

~s= stock solution (what you started with)

~d= diluted solution (what you now have)

Now that we have our definitions, we can knowingly plug the information into the formula shown above.

[tex]0.117M*24.4mL=M_{d} *100mL[/tex]

We have the information plugged in, but the volume must be measured into liters, not milliliters. We can fix that by converting it. To convert milliliters to liters, you divide the milliliters by 1000.

[tex]24.4mL/1000=0.0244L[/tex]     [tex]100mL/1000=0.1L[/tex]

Now that the volumes are in liters, let's plug that into the equation.

[tex]0.117M*0.0244L=M_{d} *0.1L[/tex]

We are now ready to solve the concentration with this equation!

[tex]M_{d} =\frac{0.117M*0.0244L}{0.1L}[/tex]      [tex]M_{d} =0.0286M[/tex]

With all of the steps shown above, we can conclude that the new concentration of ammonia is 0.0286M.

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The number of mole of ethanol present in the sample is 0.005 mole

The mole of a substance is related to it's mass and molar mass according to the following equation

Mole = mass / molar mass

With the above formula, we can obtain the mole of ethanol

Mole = mass / molar mass

Mole of ethanol = 0.2301 / 46

Mole of ethanol = 0.005 mole

Complete question

A sample of ethanol (C2H6O) has a mass of 0.2301 g. Complete combustion of this sample causes the temperature of a bomb calorimeter to increase by 1.33°C. The calorimeter has a mass of 2.000 kg and a specific heat of 2.45 J/g•°C. How many moles of ethanol are present in the sample?

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Taking into account the reaction stoichiometry, 3 moles of NH₃ will be formed.

In first place, the balanced reaction is:

Li₃N + 3 H₂O → NH₃ + 3 LiOH

By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles of each compound participate in the reaction:

The limiting reagent is one that is consumed first in its entirety, determining the amount of product in the reaction. When the limiting reagent is finished, the chemical reaction will stop.

To determine the limiting reagent, it is possible to use a simple rule of three as follows: if by stoichiometry 1 mole of Li₃N reacts with 3 moles of H₂O, 3.20 moles of Li₃N reacts with how many moles of H₂O?

[tex]moles of H_{2} O=\frac{3.2 moles of Li_{3}Nx 3 moles of H_{2} O}{1 mole of Li_{3}N}[/tex]

moles of H₂O= 9.6 moles

But 9.6 moles of H₂O are not available, 9 moles are available. Since you have less moles than you need to react with 3.20 moles of Li₃N, H₂O will be the limiting reagent.

Considering the limiting reagent, the following rule of three can be applied: if by reaction stoichiometry 3 moles of H₂O form 1 mole of NH₃, 9 moles of H₂O form how many moles of NH₃?

[tex]molesof NH_{3} =\frac{9 moles H_{2} Ox1 mole of NH_{3}}{3 moles H_{2} O}[/tex]

moles of NH₃= 3 moles

Finally, 3 moles of NH₃ will be formed.

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A key element to a successful response to a chemical, biological, or radiological incident is the rapid identification of the hazard and immediate control over all responding units.

Rapid techniques commonly used for species identification consist of species-precise primer pairs and probes, randomly amplified polymorphic DNA (RAPD), amplified fragment period polymorphisms (AFLP), restriction fragment length polymorphisms (RFLP), and karyotyping.

Elements favoring fast identity kits are reduced time in getting ready, inoculating, incubating, and cleaning up; also a convenience, decreased area for trying out and garage, ease of operation, and saving of cloth and cash are many motives for the food industry to use rapid identification structures for ordinary.

A urine lifestyle test can discover micro organism or yeast causing urinary tract contamination (UTI). If microorganisms multiply, an antibiotic sensitivity take look at can discover the antibiotic most in all likelihood to kill those specific bacteria. Your healthcare issuer may additionally order a urine tradition if you get persistent or difficult-to-deal with UTIs.

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Answer and Explanation:

To calculate this, simply use an equality of proportions:

50 particles  :  1 billion particles = x tons  :  1.49 million tons

The value of x can be solved for by multiplying the ratio on the left by the value 1.49 million / 1 billion:

So, x = (50)(0.00149) tons = 0.0745 tons

Then, do the same thing for the larger concentration:

x = (200)(0.00149) tons = 0.298 tons

The chemical formula will be  [tex]C_4H_9O_2[/tex].

From the model, there are a total of 4 carbon atoms, 9 hydrogen atoms, and 2 oxygen atoms.

All the atoms are bonded to each other accordingly.

Thus, the only plausible chemical formula for the compound is [tex]C_4H_9O_2[/tex]

All other available options fall short in one way or the other as far as the number of atoms of each element in the compound is concerned.

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B

A mixture can be separated by a physical means because there is no chemical bond between the constituents

The strongest type of intermolecular force between solute and solvent in CH3OCH3(g) in H2O(l) is H-bond and ion-induced dipole.

Hydrogen bonding is a special type of dipole-dipole attraction between molecules, not a covalent bond to a hydrogen atom.

It comes about as a result of the attraction between a hydrogen atom that is covalently attached to a very electronegative atom, like an N, O, or F atom, and another extremely electronegative atom.

Hydrogen is covalently joined to the more electronegative oxygen atom in water molecules (H2O). Therefore, the dipole-dipole interactions between the hydrogen atom of one water molecule and the oxygen atom of another H2O molecule are what cause hydrogen bonding to form in water molecules.

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Answer: C. 1.40×10^24

Explanation:

To do this, we must convert this from grams to moles, then from moles to molecules. For the first step, you will need to find the molar mass of HCl on the periodic table, which is 36.461gHCl.

[tex]85.0gHCl*\frac{1molHCl}{36.461gHCl} =2.33molHCl[/tex]

Now that we have the moles, we must convert that to molecules by using Avogadro's number.

[tex]2.33molHCl*\frac{6.022*10^{24} }{1molHCl} =1.40*10^{24}[/tex]

Now we have our answer. The best answer choice here is C. 1.40×10²⁴.

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Answer:

-Unknown

Explanation:

Questions is not seen properly bro

The given question is based on the concept of conversion factor

1 KL = 10³ L, 1cL = 10⁻³ L, 1dL = 10⁻² L, 1 nL = 10⁻⁹ L

A conversion factor is an expression for the relationship between units that is used to change the units of a measured quantity without changing the value.

A conversion ratio (or unit factor) always equals one (1), where the numerator and the denominator have the same value expressed in different units.

Hence, The complete filled table is given as follows ;

1 KL = 10³ L,

1cL = 10⁻³ L,

1dL = 10⁻² L,

1 nL = 10⁻⁹ L

Where,

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If a naturally occurring sample of an unidentified element is found to contain three isotopes (A, B, and C) and consists of 90.5% isotope A (mass number 20), 0.3% isotope B (mass number 21), and 9.3% isotope C (mass number 22), the atomic weight of the element measured from the sample will be greater than 21 amu.

To calculate the average atomic mass, multiply the fraction through the mass number for every isotope, then add them together. Whenever we do mass calculations concerning elements or compounds (combos of elements), we usually use average atomic loads.

For instance, carbon-14 is a radioactive isotope of carbon that has six protons and 8 neutrons in its nucleus. We name it carbon-14 because the overall range of protons and neutrons within the nucleus also called the mass number, provides up to fourteen (6+8=14).

Together, the quantity of protons and the range of neutrons determine an detail's mass variety. Due to the fact, that an element's isotopes have barely unique mass numbers, the atomic mass is calculated by obtaining the suggested mass numbers for its isotopes.

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Answer:

Explanation:

Equation with given

P2O5      ====> 2P^5 + 5O-2

                          0.620     x

Solution

What you are going to do first is find the molar amount of oxygen. Use a proportion to do this

2/0.620 = 5/x                    Cross Multiply

2x = 0.620 * 5                   Divide both sides by 2

2x/2 = 0.620 *5/2

x = 0.310 * 5                      Combine

x = 1.55                              moles of oxygen.

Now comes the answer. How many atoms are there in 1.55 moles of oxygen?

1 mole of anything contains 6.02 * 10^23 items of  that mole

1.55 moles of oxygen contains x  atoms of oxygen

1/1.55 = 6.02 * 10^23 / x                  Cross multiply

1*x = 1.55 * 6.02*10^23                    Combine

x = 9.331 e 23

Answer

9.33e23

[tex]V^{2+}[/tex]is the symbol of the species with 23 protons and 21 electrons.

Hence, option F is correct.

Electrons are a type of subatomic particle with a negative charge. Protons are a type of subatomic particle with a positive charge.

If it has 23 protons, then its proton number (also know as its atomic number) is 23. You can look that symbol up in your periodic table.

If it has 21 electrons then it has a net charge of 23(+) - 21. So we know the ionisation state of the species, and we can record that as a superscript after the symbol for the species.

Hence, [tex]V^{2+}[/tex] is the symbol of the species with 23 protons and 21 electrons.

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Answer:

Option B

Explanation:

Therefore our answer must be Option B

Explanation:

sulphur has a low melting point.

sulphur has a low boiling point thus it can be extracted by pumping hot water which turns it ito solution form

Answer:

37

Explanation:

( 98.6 - 32 ) × 5(100c) ÷ 9(180f) = 37

Answer: [tex]340gAgNO_{3}[/tex].

Explanation:

You will need to create a stoichiometric equation with the information given. The formula is in the problem and so are the molar masses. I will add and plug in the data for you below.

[tex]24gMg*\frac{1molMg}{24gMg} *\frac{2molAgNO_{3} }{1molMg} *\frac{170gAgNO_{3}}{1molAgNO_{3}} =340gAgNO_{3}[/tex]

I plugged the numbers into my scientific calculator, which is how I got the answer. I also added the letters in the formula above because you have to be able to cancel them out so you can find your missing data.

Therefore, [tex]340gAgNO_{3}[/tex] would react with 24gMg.

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[tex]340g AgNO_3[/tex] would react with 24 grams of Mg.

A mole is defined as 6.02214076 × [tex]10^{23}[/tex] of some chemical unit, be it atoms, molecules, ions, or others. The mole is a convenient unit to use because of the great number of atoms, molecules, or others in any substance.

Create a stoichiometric equation with the information given.

[tex]24g Mg X \frac{1 mol of Mg}{24g Mg} X \frac{2 mol of AgNO_3}{1 mol of Mg} X \frac{170g AgNO_3 }{1 mol AgNO_3} = 340g AgNO_3[/tex]

[tex]340g AgNO_3[/tex]

Hence, [tex]340g AgNO_3[/tex] would react with 24 grams of Mg.

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Answer:

1s22s22p6: Neon (Ne)

1s22s22p63s23p3: Phosphorous (P)

1s22s22p63s23p64s1: Potassium (K)

1s22s22p63s23p64s2(im not sure what 308 is supposed to be): Calcium (Ca)

1s22s22p63s23p64s23d104p65s24d3: there is no pure element that ends 4d3 that I know of so this can either be Zirconium(Zr) if it ends in 4d2 or Niobium (Nb) if it ends in 4d4

Explanation:

you can look at the periodic table and the trends to find the rough idea of where the electron configuration ends, there are helpful articles and images on these, i attached an image that may help. After that you can look at the atomic number to find the number of electrons for a pure element and use the electron subshell pattern thing to find the exact number

Answer:

the volume of the gas at 2.50 atm pressure is equal to 1.44 Litres.

Explanation:

-31.2 kJ is the work  done during a reaction in which the internal volume contracts from 83 L to 13 L against an outside pressure of 4.4 atm.

It is represented by the equation

W = PΔV

where,

W = Work done

P = Pressure

ΔV = Change in volume (V₂ - V₁)

Now put the values in above formula we get

W = PΔV

    = 4.4 atm × (13 L - 83 L)

    = 4.4 atm × (-70 L)

    = -308 atm L

Now change L atm into J

= -308 × 101.3 J     [L × atm = 101.3 J]

= -31200.4 J

= -31200.4 × 10⁻³ kJ

= -31.2 kJ

Thus from the above conclusion we can say that -31.2 kJ is the work  done during a reaction in which the internal volume contracts from 83 L to 13 L against an outside pressure of 4.4 atm.

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Answer: 2.11 g

Explanation:

[tex]$$Equation: $2 \mathrm{AgC}_{2} \mathrm{H}_{3} \mathrm{O}_{2}(a q)+\mathrm{Cu}(s) \longrightarrow \mathrm{Cu}\left(\mathrm{C}_{2} \mathrm{H}_{3} \mathrm{O}_{2}\right)_{2}(a q)+2 \mathrm{Ag}(s)$[/tex]

[tex]$$No of moles of $\mathrm{Ag}$ produced from $\mathrm{Cu}=2.500 \mathrm{~g} \times \frac{1 \mathrm{~mol} \mathrm{Cu}}{63.55 \mathrm{~g}} \times \frac{2 \mathrm{~mol} \mathrm{Ag}}{1 \mathrm{~mol} \mathrm{Cu}}$ $=0.07868 \mathrm{~mol} \mathrm{Ag}$[/tex]

[tex]$$No of moles of $\mathrm{Ag}$ produced from $\mathrm{AgC}_{2} \mathrm{H}_{3} \mathrm{O}_{2}=200.0 \mathrm{~mL} \times \frac{1 \mathrm{~L}}{1000 \mathrm{~mL}} \times \frac{0.150 \mathrm{~mol} \mathrm{AgC_{2 } \mathrm { H } _ { 3 } \mathrm { O } _ { 2 }}}{1 \mathrm{~L}}$ $=0.0300 \mathrm{~mol} \mathrm{Ag}$[/tex]

[tex]Since $\mathrm{AgC}_{2} \mathrm{H}_{3} \mathrm{O}_{2}$ limits the production of $\mathrm{Ag}, \mathrm{AgC}_{2} \mathrm{H}_{3} \mathrm{O}_{2}$ acts as the limiting reactant \\\\No of moles of Ag formed $=0.0300 \mathrm{~mol}$\\Theoretical yield of $\mathrm{Ag}=0.0300 \mathrm{~mol} \mathrm{Ag} \times \frac{108 \mathrm{~g}}{1 \mathrm{~mol} \mathrm{Ag}}=3.24 \mathrm{~g}$\\Actual yield of silver $=65.0 \%(3.24 \mathrm{~g})$$$=\frac{65.0}{100}(3.24 \mathrm{~g})=2.106 \mathrm{~g} \approx 2.11 \mathrm{~g}$$[/tex]