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Olin [163]
3 years ago
7

Which professional most likely uses x-ray technology

Chemistry
1 answer:
Aleks [24]3 years ago
5 0

Answer:

The professionals that most likely use X-ray technology are a dentist and an airport-security scanner.

A dentist would scan your teeth using X-ray technology to see whether there are any cavities within the teeth themselves which he couldn't otherwise see with the naked eye. An airport-security scanner would use this technology to scan your luggage to check whether there are any weapons and other illegal products.

Explanation:

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A certain half-reaction has a standard reduction potential E0red = +0.13V . An engineer proposes using this half-reaction at the
Ivan

Answer:

a. 1.23 V

b. No maximum

Explanation:

Required:

a. Is there a minimum standard reduction potential that the half-reaction used at the cathode of this cell can have?

b. Is there a maximum standard reduction potential that the half-reaction used at the cathode of this cell can have?

The standard cell potential (E°cell) is the difference between the standard reduction potential of the cathode and the standard reduction potential of the anode.

E°cell = E°red, cat - E°red, an

If E°cell must be at least 1.10 V (E°cell > 1.10 V),

E°red, cat - E°red, an > 1.10 V

E°red, cat - 0.13V > 1.10 V

E°red, cat > 1.23 V

The minimum standard reduction potential is 1.23 V while there is no maximum standard reduction potential.

4 0
3 years ago
What is the molar mass, in grams, of a mole of an element equal to?
insens350 [35]

Answer:

B. the atomic weight of the element

Explanation:

since it talks about mass i think weight

5 0
4 years ago
Ok so the question is Sort the processes based on the type of energy transfer they involve. condensation
Lesechka [4]

Maybe evaporation???????????????

7 0
3 years ago
Read 2 more answers
What is the outcome of hotspots? Question 13 options: Formation of islands Constant volcanic eruptions Divergent boundary intera
miss Akunina [59]

Answer:

Constant volcanic eruptions Hope this helps!

6 0
3 years ago
Butane (C4H10) has a heat of vaporization of 22.44 kJ/mol and a normal boiling point of -0.4 ∘C. A 250 mL sealed flask contains
erastova [34]

Given that:

  • The heat of vaporization = 22.44 kJ/mol = 22440 J/mol
  • normal boiling point which is the initial temperature = 0.4° C = (273 + (-0.4))K  = 272.6 K
  • volume  = 250 mL = 0.250 L
  • Mass of butane = 0.8 g
  • the final temperature = -22° C = (273 + (-22)) K = 251 K

The first step is to determine the vapor pressure at the final temperature of 251K by using the Clausius-Clapeyron equation. This is following by using the ideal gas equation to determine the numbers of moles of butane gas. After that, the mass of butane present in the liquid is determined by using the relation for the number of moles.

Using Clausius-Clapeyron Equation:

\mathbf{In (\dfrac{P_2}{P_1} )= -\dfrac{\Delta H_{vap}}{R}(\dfrac{1}{T_2} - \dfrac{1}{T_1})}

where;

P1 and P2 correspond to the temperature at T1 and T2.

∴

replacing the values into the given equation, we have;

\mathbf{In \dfrac{P_2}{1\  atm} = -\dfrac{22440 \ J/mol}{8.314 \ J/mol.K}(\dfrac{1}{251 \ K} - \dfrac{1}{272.6 \ K})}

\mathbf{In \dfrac{P_2}{1\  atm} =-(0.852053785)}

\mathbf{P_2=0.427 \ atm}

As such, at -22° C; the vapor pressure = 0.427 atm

Now, using the ideal gas equation:

PV = nRT

where:

  • P = Pressure
  • V = volume
  • n = number of moles of butane
  • R = universal gas constant
  • T = temperature

∴

Making (n) the subject of the formula:

\mathbf{n = \dfrac{PV}{RT}}

\mathbf{n = \dfrac{0.427 atm \times 0.250 L}{(0.08206 \ L.atm/k.mol) \times 251}}

\mathbf{n =0.00518 mol}

We all know that the standard molecular weight of butane = 58.12 g/mol

∴

Using the relation for the number of moles which is:

\mathbf{number \  of \  moles = \dfrac{mass}{molar mass}}

mass = 0.00518 mole × 58.12 g/mol

mass = 0.301 g

∴

The mass of butane in the flask = 0.301 g

But the mass of the butane present as a liquid in the flask is

= 0.8 g - 0.301 g

= 0.499 g

In conclusion, the mass of the butane present as a liquid in the flask is 0.499 g

Learn more about vapourization here:

brainly.com/question/17039550?referrer=searchResults

7 0
3 years ago
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