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ELEN [110]
3 years ago
13

How did uranium become present in Earth's upper crust? Choose one: A. It was delivered by comets that crashed into Earth's surfa

ce. B. It developed naturally in the upper crust. C. It rose upward as it was carried within magma. D. It does not exist in the upper crust. Uranium only exists in the mantle.
Chemistry
1 answer:
Luda [366]3 years ago
5 0

Answer:

<em>Option A. It was delivered by comets that crashed into Earth's surface.</em>

Explanation:        

<em><u>Uranium (U) is a chemical element with atomic number 92.</u></em>        

<em />

<em>For many years, a large number of scientists have been studying the abundance and origin of the isotopes of uranium in Earth</em>. <u>According to some theories, the Earth's uranium was produced in one or more supernovae</u> (an explosive brightening of a star), in wich, the main process consists in the rapid capture of neutrons by seed nuclei at great rates. <u>Another theory proposes that uranium is created during the merger of two neutron stars</u> (neutron stars are very dense), because, when such dense bodies come closer together the gravitational force cause them to merge, producing huge amounts of hevy metals like uranium.                  

<u><em>Many analyses have been made of the uranium in rocks of the Earth. These measurements shows that the abundance of uranium is bigger in the crust and upper mantle of the Earth</em></u>.    

So, knowing that Earth's uranium was produced through one of these processes, <u><em>the best answer is option A, the uranium was delivered by comets that crashed into Earth's surface.</em></u>    

Have a nice day!                            

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olganol [36]

Answer:

Ocean, lakes and rivers. Are all liquids.

Explanation:

Ocean, lakes and rivers. Are all liquids. Snow starts off as a liquid, evaporates into a gas and camoes back as snow.

4 0
3 years ago
A chemist titrates 130.0mL of a 0.4248 M lidocaine (C14H21NONH) solution with 0.4429 M HBr solution at 25 degree C . Calculate t
jeka57 [31]

Answer:

pH = 3.36

Explanation:

Lidocaine is a weak base to be titrated with the strong acid HBr, therefore at equivalence point we wil have the protonated lidocaine weak conjugate acid of lidocaine which will drive the pH.

Thus to solve the question we will need to calculate the concentration of this weak acid at equivalence point.

Molarity = mol /V ∴ mol = V x M

mol lidocaine = (130 mL/1000 mL/L) x 0.4248 mol/L = 0.0552 mol

The volume of 0.4429 M HBr required to neutralize this 0.0552 mol is

0.0552 mol x  (1L / 0.4429mol) = 0.125 L

Total volume at equivalence is  initial volume lidocaine + volume HBr added

0 .130 L +0.125 L = 0.255L

and the concentration of protonated lidocaine at the end of the titration will be

0.0552 mol / 0.255 L = 0.22M

Now to calculate the pH we setup our customary ICE table for  weak acids for the equilibria:

protonated lidocaine + H₂O   ⇆  lidocaine + H₃O⁺

                      protonated lidocaine          lidocaine        H₃O⁺

Initial(M)               0.22                                       0                  0

Change                   -x                                      +x                 +x

Equilibrium          0.22 - x                                  x                    x

We know for this equilibrium

Ka = [Lidocaine] [H₃O⁺] / [protonaded Lidocaine] =  x² / ( 0.22 - x )

The Ka can be calculated from the given pKb for lidocaine

Kb = antilog( - 7.94 ) = 1.15 x 10⁻⁸

Ka = Kw / Kb = 10⁻¹⁴ / 1.15 x 10⁻⁸  = 8.71 x 10⁻⁷

Since Ka is very small we can make the approximation 0.22  - x  ≈ 0.22

and solve for x. The pH  will then  be the negative log of this value.

8.71 x 10⁻⁷  = x² / 0.22 ⇒ x = √(/ 8.71 X 10⁻⁷ x 0.22) = 4.38 x 10⁻⁴

( Indeed our approximation checks since 4.38 x 10⁻⁴ is just 0.2 % of 0.22 )

pH = - log ( 4.4x 10⁻⁴) = 3.36

3 0
3 years ago
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Vsevolod [243]

Answer:

11,000 cm

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Step 1: Given data

Width of the field (w): 17 meters

Length of the field (l): 38 meters

Step 2: Calculate the perimeter of the field

The field is a rectangle. We can find its perimeter (P) by adding its sides.

P = 2 × w + 2 × l = 2 × 17 m + 2 × 38 m = 110 m

Step 3: Convert the perimeter to centimeters

We will use the relationship 1 m = 100 cm.

110 m × (100 cm/1 m) = 11,000 cm

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Answer:

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