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Kaylis [27]
3 years ago
9

Un isótopo de un elemento metálico tiene un número de masa de 65 y tiene 35 neutrones en su núcleo. El catión derivado de dicho

isótopo tiene 28 electrones. Identifique el elemento, y la carga correspondiente del ión.
Chemistry
1 answer:
poizon [28]3 years ago
8 0
According to Data,
                             Mass Number  =  65

                             # of Neutrons  =  35

As we know,
                             Mass Number  =  # of protons + # of Neutrons
Solving for # of protons,
                                # of protons  =  Mass Number - # of Neutrons
Putting values,
                                # of Protons  =  65 -35

                                # of protons  =  30
As, 

# of protons is the atomic number, so element with atomic number 30 is Zinc.

The cation derived from it has 28 electrons,
So, Zinc having 30 electron will be left with two extra protons, and each proton has +1 charge.

Result:
                                                    Zn
²⁺
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jek_recluse [69]

Answer:

C. A stretch of raised land, often built to help control flooding and the flow of a river.

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Embankments are made typically of construction adjacent to the river from spoil excavated from its bank.

4 0
3 years ago
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One method for determining the amount of corn in early Native American diets is the stable isotope ratio analysis (SIRA) techniq
frozen [14]

Answer:

a. i. 8.447 × 10⁻³ T ii.  27.14 cm

b. i. 2.14 cm ii. It is easily detectable.

Explanation:

a.

i. What strength of magnetic field is required?

Since the magnetic force F = Bqv equals the centripetal force F' = mv²/r on the C12 charge, we have

F = F'

Bqv = mv²/r

B = mv/re where B = strength of magnetic field, m = mass of C12 isotope = 1.99 × 10⁻²⁶ kg, v = speed of C 12 isotope = 8.50 km/s = 8.50 × 10³ m/s, q = charge on C 12 isotope = e = electron charge = 1.602 × 10⁻¹⁹ C (since the isotope loses one electron)and r = radius of semicircle = 25.0 cm/2 = 12.5 cm = 12.5 × 10⁻² m

So,

B = mv/rq

B = 1.99 × 10⁻²⁶ kg × 8.50 × 10³ m/s ÷ (12.5 × 10⁻² m × 1.602 × 10⁻¹⁹ C)

B = 16.915 × 10⁻²³ kgm/s ÷ (20.025 × 10⁻²¹ mC)

B = 0.8447 × 10⁻² kg/sC)

B = 8.447 × 10⁻³ T

(ii) What is the diameter of the 13C semicircle?

Since the magnetic force F = Bq'v equals the centripetal force F' = mv²/r' on the C13 charge, we have

F = F'

Bq'v = mv²/r'

r' = mv/Be where r = radius of semicircle, B = strength of magnetic field = 8.447 × 10⁻³ T, m = mass of C12 isotope = 2.16 × 10⁻²⁶ kg, v = speed of C 12 isotope = 8.50 km/s = 8.50 × 10³ m/s, q' = charge on C 13 isotope = e = electron charge = 1.602 × 10⁻¹⁹ C (since the isotope loses one electron) and  = d/2 = 12.5 cm = 12.5 × 10⁻² m

So, r' = mv/Be

r' = 2.16 × 10⁻²⁶ kg × 8.50 × 10³ m/s ÷ (8.447 × 10⁻³ T × 1.602 × 10⁻¹⁹ C)

r' = 18.36 × 10⁻²³ kgm/s ÷ 13.5321 × 10⁻²² TC)

r' = 1.357 × 10⁻¹ kgm/TC)

r' = 0.1357 m

r' = 13.57 cm

Since diameter d' = 2r', d' = 2(13.57 cm) = 27.14 cm

b.

i. What is the separation of the C12 and C13 ions at the detector at the end of the semicircle?

Since the diameter of the C12 isotope is 25.0 cm and that of the C 13 isotope is 27.14 cm, their separation at the end of the semicircle is 27.14 cm - 25.0 cm = 2.14 cm

ii. Is this distance large enough to be easily observed?

This distance of 2.14 cm easily detectable since it is in the centimeter range.

7 0
3 years ago
Who ever answer this is genius ????????? <br><br><br>let's see?​
Flura [38]

Can you do plsssssssss

help me plssssssssssss meee

3 0
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How many moles of MgCl2 are there in 319 g of the compound?
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Hey there!:

Molar mass MgCl2 = 95.2110 g/mol

So:

1 mole MgCl2 -------------- 95.2110 g

moles MgCl2 ---------------- 319 g

moles MgCl2 = 319 * 1 / 95.2110

moles MgCl2 = 319 / 95.2110

=> 3.350 moles of MgCl2


Hope that helps!

3 0
3 years ago
Why are hydrogen, lithium, and sodium classified as reactive elements?
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<span>The alkali metals and hydrogen are reactive because they have only one electron to give in order to complete their valence shell. It is easier to give that one electron so when given the opportunity they will. This means they will react with anything polar or willing to take an electron.</span>
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