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guajiro [1.7K]
3 years ago
15

If a radioactive isotope of thorium (atomic number 90, mass number 232) emits 6 alpha particles and 4 beta particles during the

course of radioactive decay, what is the mass number of the stable daughter product?
Chemistry
1 answer:
kolbaska11 [484]3 years ago
5 0

Answer:

The mass number of the stable daughter product is 208

Explanation:

First thing's first, we have to write out the equation of the reaction. This is given as;

²³²₉₀Th → 6 ⁴₂α  +  4 ⁰₋₁ β + X

In order to obtain the identity of X, we have to obtain it's mass numbers and atomic number.

There is conservation of matter so we expect the mass number to remain the same in both the reactant and products.

Mass Number

Reactant = 232

Product = (6* 4 = 24) + (4 * 0 = 0) + x = 24 + x

since reactant = product

232 = 24 + x

x = 232 - 24 = 208

Atomic Number

Reactant = 90

Product = (6* 2 = 12) + (4 * -1 = -4) + x = 8 + x

since reactant = product

90 = 8 + x

x = 90 - 8 = 82

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5 0
3 years ago
calculate q for the following: 125.0 ml of 0.0500 m pb(no3)2 is mixed with 75.0 ml of 0.0200 m nacl at 25oc chegg
alexandr402 [8]

The value of Q for 125.0 ml of 0.0500 m Pb(NO3)2 is mixed with 75.0 ml of 0.0200 m NaCl at 25°C is 2.11 × 10^(-6).

Aa we know that, 125mL of 0.06M Pb(NO3)2 is mixed with 75.0 ml of 0.0200 m NaCl.

Given, T = 25°C.

<h3>Chemical equation:</h3>

Pb(NO3)2 + NaCl ---- NaNO3 + PbCl2

PbCl2 in aqueous solution split into following ions

PbCl2 ------ Pb(+2) + 2Cl-

Q = [Pb(+2)] [Cl-]^2

The Concentration of Pb(+2) ions and Cl- ions can be calculated as

[Pb(+2)] = 0.06 × 125/200

= 0.0375

[Cl-] = 0.02 × 75/200

= 0.0075

By substituting all the values, we get

[0.0375] [0.0075]^2

= 2.11 × 10^(-6).

Thus, we calculated that the value of Q for 125.0 ml of 0.0500 m Pb(NO3)2 is mixed with 75.0 ml of 0.0200 m NaCl at 25°C is 2.11 × 10^(-6).

learn more about Ions:

brainly.com/question/13692734

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6 0
2 years ago
Convert 7.34 miles into meters (1mile = 1.6 Km)
Lera25 [3.4K]

Answer:

11812.58 meters = 11.81258 Km

Explanation:

hope that helps

8 0
3 years ago
If 4.20 mol Al was mixed with 1.75 mol Fe2 O3 which reactant is the limiting reactant?
Dafna11 [192]
4.20 mol Al would react completely with 4.20 x (1/2) = 2.10 mol Fe2O3, but there is not that much Fe2O3 present, so Fe2O3 is the limiting reactant. (1.75 mol Fe2O3) x (2/1) x ( 55.8452 g Fe/mol) = 195 g Fe 3 MgO + 2 H3PO4 → Mg3(PO4)2 + 3 H2O (15.0 g MgO) / (40.3045 g MgO/mol) = 0.37217 mol MgO (18.5 g H3PO4) / (97.9953 g H3PO4/mol) = 0.18878 mol H3PO4 0.18878 mol H3PO4 would react completely with 0.18878 x (3/2) = 0.28317 mole of MgO, but there is more MgO present than that, so MgO is in excess and H3PO4 is the limiting reactant. Now we must consider why the problem tells us "17.6g of Mg3(PO4)2 is obtained". The first possibility is that it's just there for the sake of confusion -- in which case ignore it and proceed this way: ((0.37217 mol MgO initially) - (0.28317 mole MgO reacted)) x (40.3045 g MgO/mol) = 3.59 g MgO left over However, if the amount of magnesium phosphate obtained is given because the reaction was stopped before it was complete, the amount obtained governs the amount reacted and the amount left over, so proceed this way: (17.6g Mg3(PO4)2) / (262.8581 g Mg3(PO4)2/mol) x (3/1) = 0.20087 mol MgO reacted ((0.37217 mol MgO initially) - (0.20087 mole MgO reacted)) x (40.3045 g MgO/mol) = 6.90 MgO left over
7 0
3 years ago
If a single gold atom has a diameter of 2.9×x10−8 cm, how many atoms thick was rutherford's foil
ale4655 [162]

27,586

<h3>Further explanation</h3>

<u>Given:</u>

A single gold atom has a diameter of \boxed{ \ 2.9 \times 10^{-8} \ cm. \ }

From a reference, the Rutherford gold foil used in his scattering experiment had a thickness of approximately \boxed{ \ 8 \times 10^{-3} \ mm. \ }

<u>Question:</u>

How many atoms thick were Rutherford's foil?

<u>The Process:</u>

Convert thickness from mm to cm.

\boxed{ \ 8 \times 10^{-3} \ mm = 8 \times 10^{-3} \times 10{-1} \ cm \ } \rightarrow \boxed{ \ 8 \times 10^{-4} \ cm \}

The number of atoms is calculated from gold foil thickness divided by the atomic diameter.

\boxed{ \ = \frac{8 \times 10^{-4} \ cm}{2.9 \times 10^{-8} \ cm} \ }

\boxed{ \ =2.7586 \times 10^4 \ atoms \ }

Therefore, we get an atomic thickness of 27,586 atoms.

<u>Notes:</u>

  • In 1909-1910, Ernest Rutherford with two of his assistants, namely Hans Geiger and Ernest Marsden, conducted a series of experiments to find out more about the arrangement of atoms. They fired at a very thin gold plate with high-energy alpha particles.
  • One of their observations is that a small portion of alpha particles are reflected. This greatly surprised Rutherford. The reflected alpha particle must have hit something very dense in the atom. This fact is incompatible with the atomic model proposed by J.J. Thomson where the atoms are described as homogeneous in all parts with electrons and protons evenly distributed.
  • In 1911, Rutherford was able to explain the scattering of alpha rays by proposing ideas about atomic nuclei. According to him, most of the mass and positive charge of atoms are concentrated at the center of the atom, hereinafter referred to as the nucleus.
<h3>Learn more</h3>
  1. The energy density of the stored energy  brainly.com/question/9617400
  2. The theoretical density of platinum which has the FCC crystal structure. brainly.com/question/5048216
  3. Compound microscope brainly.com/question/4000241

Keywords: if a single gold atom, has a diameter of 2.9 x 10⁻⁸ cm, how many, atoms thick, Rutherford's foil, his scattering experiment

7 0
2 years ago
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