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3 years ago

Which profession is likely to use a radiation shield?

Chemistry

2 answers:

Sidana [21]3 years ago

5 0

Answer:

astronomer

Explanation:

just took the test :)

pantera1 [17]3 years ago

3 0

Which profession is likely to use a radiation shield?
A.)dentist
B.)astronomer
C.)mathematician
D.)geologist

The answer is B: Astronomer

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What contribution did these scientists make to atomic models of the atom?

SashulF [63]

Answer:

Explanation:

Dalton's atomic theory proposed that all matter was composed of atoms, indivisible and indestructible building blocks. While all atoms of an element were identical, different elements had atoms of differing size and mass.

In 1897, J.J. Thomson discovered the electron by experimenting with a Crookes, or cathode ray, tube. He demonstrated that cathode rays were negatively charged. In addition, he also studied positively charged particles in neon gas.

Rutherford overturned Thomson's model in 1911 with his well-known gold foil experiment in which he demonstrated that the atom has a tiny and heavy nucleus. Rutherford designed an experiment to use the alpha particles emitted by a radioactive element as probes to the unseen world of atomic structure.

The Bohr model shows the atom as a small, positively charged nucleus surrounded by orbiting electrons. Bohr was the first to discover that electrons travel in separate orbits around the nucleus and that the number of electrons in the outer orbit determines the properties of an element.

7 0

3 years ago

The molarity (M) of an aqueous solution containing 22.5 g of sucrose (C12H22O11) in 35.5 mL of solution is ________.

Nonamiya [84]

Answer:

1.86 M

Explanation:

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

Mass of sucrose (C12H22O11) = 22.5 g

Volume of solution = 35.5 mL

Molarity of solution =?

Next, we shall determine the number of mole in 22.5 g of sucrose (C12H22O11). This can be obtained as follow:

Mass of sucrose (C12H22O11) = 22.5 g

Molar mass of C12H22O11 = (12×12) + (22×1) + (16×11)

= 144 + 22 + 176

= 342 g/mol

Mole of C12H22O11 =?

Mole = mass /Molar mass

Mole of C12H22O11 = 22.5 /342

Mole of sucrose (C12H22O11) = 0.066 mole

Next, we shall convert 35.5 mL to litres (L). This can be obtained as follow:

1000 mL = 1 L

Therefore,

35.5 mL = 35.5 mL × 1 L / 1000 mL

35.5 mL = 0.0355 L

Thus, 35.5 mL is equivalent to 0.0355 L.

Finally, we shall determine the molarity of the solution as follow:

Mole of sucrose (C12H22O11) = 0.066 mole

Volume of solution = 0.0355 L.

Molarity of solution =?

Molarity = mole /Volume

Molarity of solution = 0.066/0.0355

Molarity of solution = 1.86 M

Therefore, the molarity of the solution is 1.86 M.

8 0

3 years ago

The energy E of the electron in a hydrogen atom can be calculated from the Bohr formula: E = R_y/n^2 In this equation R_y stands

Katarina [22]

Answer:

The wavelength of the line in the emission line spectrum of hydrogen caused by the transition of the electron for the given energy levels is $5.23\times 10^{-5} m$

Explanation:

Given :

The energy E of the electron in a hydrogen atom can be calculated from the Bohr formula:

$E=\frac{R_y}{n^2}$

$R_y=2.18\times 10^{-18} J$ = Rydberg energy

n = principal quantum number of the orbital

Energy of 11th orbit = $E_{11}$

$E_{11}=\frac{2.18\times 10^{-18} J}{11^2}=1.80\times 10^{-20} J$

Energy of 10th orbit = $E_{10}$

$E_{10}=\frac{2.18\times 10^{-18} J}{10^2}=2.18\times 10^{-20} J$

Energy difference between both the levels will corresponds to the energy of the wavelength of the line which can be calculated by using Planck's equation.

$E'=E_{10}-E_{11}=2.18\times 10^{-20} J-1.80\times 10^{-20} J$

$=E'=0.38\times 10^{-20} J$

$\lambda =\frac{hc}{E'}$ (Planck's' equation)

$\lambda = \frac{6.626\times 10^{-34} Js\times 3\times 10^8 m/s}{0.38\times 10^{-20} J}$

$\lambda = 5.2310\times 10^{-5} m\approx 5.23\times 10^{-5} m$

The wavelength of the line in the emission line spectrum of hydrogen caused by the transition of the electron for the given energy levels is $5.23\times 10^{-5} m$

3 0

3 years ago

The functional groups in an organic compound can frequently be deduced from its infrared absorption spectrum. A compound, C5H10O

vova2212 [387]

Answer:The functional groups in an organic compound can frequently be deduced from its infrared absorption spectrum. A compound, C5H10O2, exhibits strong, broad absorption across the 2500-3200 cm^1 region and an intense absorption at 1715 cm'^-1. Relative absorption intensity: (s)=strong, (m)-medium, (w) weak. What functional class(cs) docs the compound belong to List only classes for which evidence is given here. Attach no significance to evidence not cited explicitly. Do not over-interpret exact absorption band positions. None of your inferences should depend on small differences like 10 to 20 cm^1. The functional class(es) of thla compound is(are) alkane (List only if no other functional class applies.) alkene terminal alkyne internal alkyne arene alcohol ether amine aldehyde or ketone carboxylic acid ester nitr

3 0

3 years ago

Large telescopes have revealed information about the motions of stars. They helped scientists realize that most stars and galaxi

fiasKO [112]

Answer:b

Explanation:sorry if I’m wrong

5 0

3 years ago