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Rufina [12.5K]
4 years ago
9

Which is one common use of neutron activation analysis?

Chemistry
1 answer:
artcher [175]4 years ago
3 0
Neutron activation analysis is a nuclear process that is used to determine the concentration of elements in various types of materials. The technique uses excitation of neurons to produce gamma ray emission by the treated material. Neutron activation analysis has many applications; one of them is that it is used  in forensic medicine to establish evidences in criminal cases. 
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Deep sea divers use a mixture of helium and oxygen to breathe. Assume that a diver is going to a depth of 150 feet where the tot
Svetlanka [38]

Answer:

4.525% is the percentage by volume of oxygen in the gas mixture.

Explanation:

Total pressure of the mixture = p = 4.42 atm

Partial pressure of the oxygen = p_1=0.20 atm

Partial pressure of the helium = p_2

p_1=p\times \chi_1 (Dalton law of partial pressure)

0.20 atm=4.42 atm\times \chi_1

\chi_1=\frac{0.20 atm}{4.42 atm}=0.04525

\chi_2=1-\chi_1=1-0.04525=0.95475

chi_1+chi_2=1

n_1=0.04525 mol,n_2=0.95475 mol

According Avogadro law:

Moles\propto Volume (At temperature and pressure)

Volume occupied by oxygen gas  =V_1

Total moles of gases = n = 1 mol

Total Volume of the gases = V

\frac{n_1}{V_1}=\frac{n}{V}

\frac{V_1}{V}=\frac{n_1}{n}=\frac{0.04525 mol}{1 mol}

Percent by volume of oxygen in the gas mixture:

\frac{V_1}{V}\times 100=\frac{0.04525 mol}{1 mol}\times 100=4.525\%

6 0
4 years ago
How many Liters of a 4.5 M HCL solution can be prepared by using 250.0 mL of a 12.0 M HCl solution?
netineya [11]

Answer:

0.667 L

Explanation:

From the question given above, the following data were obtained:

Initial volume (V₁) = 250 mL

Initial concentration (C₁) = 12 M

Final concentration (C₂) = 4.5 M

Final volume (V₂) =?

The final volume of the solution can be obtained by using the dilution formula as illustrated below:

C₁V₁ = C₂V₂

12 × 250 = 4.5 × V₂

3000 = 4.5 × V₂

Divide both side by 4.5

V₂ = 3000 / 4.5

V₂ = 667 mL

Finally, we shall convert 667 mL to L. This can be obtained as follow:

1000 mL = 1 L

Therefore,

667 mL = 667 mL × 1 L / 1000 mL

667 mL = 0.667 L

Thus, the volume of the solution prepared is 0.667 L

3 0
3 years ago
Calculate the formula mass of calcium iodate, ca(io3)2, in atomic mass units (amu or u).
VARVARA [1.3K]
<span>389.88094 amu First we look up the atomic mass of all elements contained in calcium iodate using the periodic table: Ca: 40.078 I: 126.90447 O: 15.999 As an intermediate step we calculate the molecular mass of the ion IO3: 126.90447 + 3*15.999 = 174.90147 Then we calculate the mass of one calcium atom and 2 iodate ions: 2*174.90147 + 40.078 = 389.88094 amu</span>
7 0
3 years ago
3. Discuss the difference between molarity and molality, state the units of each, state the symbol for each, and give an example
dangina [55]

<span>Molality(m) or molal concentration is a measure of concentration and it refers to amount of substance in a specified amount of mass of the solvent. Used unit for molality is mol/kg which is also sometimes denoted as 1 molal. It is equal to the moles of solute (the substance being dissolved) divided by the kilograms of solvent (the substance used to dissolve).</span>

Molarity(M) or molar concentration is also a measure of concentration and represents the amount of substance per unit volume of solution(number of moles per litre of solution. Used unit for molarity is mol/L or M. A solution with a concentration of 1 mol/L is equivalent to 1 molar (1 M).

Molality is preferred when the temperature of the solution varies, because it does not depend on temperature, (neither number of moles of solute nor mass of solvent will be affected by changes of temperature), while molarity changes as temperature changes(volume of solution changes as temperature changes).


4 0
4 years ago
What are five general properties of aqueous bases? b. Name some common substances that have one or more of these properties.
ella [17]

Answer:

Explanation:(1)base are slippery to touch e.g sodium hydroxide NaOH(aq)..

(2) They can be corrosive e.g pottasium hydroxide KOH(aq) and sodium hydroxide NOH(aq)

(3) They can act as electrolytes e.g NaOH(aq)

(4) they react to acids to form salt and water

Na0H(s)+HCL(aq)>>>NaCL(a)+H20(l)

(5)they dissolve in water to form hydroxyl ion

KoH(aq) >>>>>k+ + 0H-

4 0
3 years ago
Read 2 more answers
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