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2 years ago

Complete the following radioactive decay problem. 210/84 Po --> 206/82 Pb + __/

Chemistry

1 answer:

vazorg [7]2 years ago

4 0

₈₄²¹⁰Po ⟶ ₈₂²⁰⁶Pb + $_2^4He$

Your equation is:

₈₄²¹⁰Po ⟶ ₈₂²⁰⁶Pb + ?

It becomes easier to balance the equation if we replace the "?" with an element symbol _x^yZ.

Then the equation becomes

₈₄²¹⁰Po ⟶ ₈₂²⁰⁶Pb +_x^yZ

The main point to remember in balancing nuclear equations is that the sums of the superscripts and the subscripts must be the same on each side of the equation.

Sum of superscripts: 210 = 206 + y, so y =4.

Sum of subscripts: 84 = 82 + x, so x = 2.

₈₄²¹⁰Po ⟶ ₈₂²⁰⁶Pb + $_2^4He$

We should recall that $_2^4He$ is an α particle.

Thus, the equation represents the α decay of polonium-210 to lead-206.

The nuclear equation is

₈₄²¹⁰Po ⟶ ₈₂²⁰⁶Pb + $_2^4He$

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What volume of 0.100 M Na3PO4 is required to precipitate all the lead(ii) ions from 150.0 mL of 0.250 M Pb(NO3)2?

My name is Ann [436]

First, we write the reaction equation:

3Pb(NO₃)₂ + 2Na₃PO₄ → 6NaNO₃ + 3Pb₃(PO₄)₂

Moles of Pb ions present:
moles = concentration x volume
= 0.15 x 0.25
= 0.0375

From the equation,
moles Pb : moles Na₃PO₄
= 3 : 2
Moles of Na₃PO₄:
2/3 x 0.0375
= 0.025

volume = moles / concentration
= 0.025 / 0.1
= 0.25 L
= 250 ml

7 0

2 years ago

Read 2 more answers

Aluminum–lithium (Al–Li) alloys have been developed by the aircraft industry to reduce the weight and improve the performance of

gtnhenbr [62]

Answer:

The concentration of Li (in wt%) is 3,47g/mol

Explanation:

To obtain the 2,42g/cm³ of density:

2,42g/cm³ = 2,71g/cm³X + 0,534g/cm³Y (1)

Where X is molar fraction of Al and Y is molar fraction of Li.

X + Y = 1 (2)

Replacing (2) in (1):

Y = 0,13

Thus, X = 0,87

The weight of Al and Li is:

0,87*26,98g/mol = 23,4726 g of aluminium

0,13*6,941g/mol = 0,84383 g of lithium

The concentration of Li (in wt%) is:

0,84383g/(0,84383g+23,4726g) ×100= 3,47%

6 0

1 year ago

If the mass percentage composition of a compound is 72.1% Mn and 27.9% O, its empirical formula is

mojhsa [17]

Answer:

MnO- Manganese Oxide

Explanation:

Empirical formula: This is the formula that shows the ratio of elements

present in a

compound.

How to determine Empirical formula

1. First arrange the symbols of the elements present in the compound

alphabetically to determine the real empirical formula. Although, there

are exceptions to this rule, E.g H2So4

2. Divide the percentage composition by the mass number.

3. Then divide through by the smallest number.

4. The resulting answer is the ratio attached to the elements present in

a compound.

Mn O

% composition 72.1 27.9

Divide by mass number 54.94 16

1.31 1.74

Divide by the smallest number 1.31 1.31

1 1.3

The resulting ratio is 1:1

Hence the Empirical formula is MnO, Manganese oxide

8 0

2 years ago

Consider isotopes ions protons and electrons. how many of these did dalton not discuss in his atomic theory?

MAVERICK [17]

Answer: He did not discuss about any of these.

Explanation: Dalton proposed some of the postulates for his atomic theory. They are:

1) Matter is made up of atoms which are not divisible.

2) Atoms of different elements combine in a fixed ratio to form compounds.

3) The atomic properties of given element are same including mass. This states that all the atoms of an element have same mass but the atoms of different elements have different masses.

4) No atoms are either created or destroyed during a chemical reaction.

5) Atoms of an element are identical in mass, size and other chemical and physical properties.

As it is visible from the postulates, he only discussed only about the atoms but not subatomic particles or isotopes.

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2 years ago

Carbon dioxide readily absorbs radiation with an energy of 4.67 x 10-20 J. What is the wavelength and frequency of this radiatio

Tanya [424]

Answer:

ν = 7.04 × 10¹³ s⁻¹

λ = 426 nm

It falls in the visible range

Explanation:

The relation between the energy of the radiation and its frequency is given by Planck-Einstein equation:

E = h × ν

where,

E is the energy

h is the Planck constant (6.63 × 10⁻³⁴ J.s)

ν is the frequency

Then, we can find frequency,

$\nu = \frac{E}{h}= \frac{4.67 \times 10^{-20}J }{6.63 \times 10^{-34}J.s} = 7.04 \times 10^{13} s^{-1}$

Frequency and wavelength are related through the following equation:

c = λ × ν

where,

c is the speed of light (3.00 × 10⁸ m/s)

λ is the wavelength

$\lambda = \frac{c}{\nu } =\frac{3.00 \times 10^{8} m/s }{7.04 \times 10^{13} s^{-1} } =4.26 \times 10^{-6}m.\frac{10^{9}nm }{1m} = 426 nm$

A 426 nm wavelength falls in the visible range (≈380-740 nm)

7 0

2 years ago

Complete the following radioactive decay problem. 210/84 Po --> 206/82 Pb + ____/____ __

Complete the following radioactive decay problem. 210/84 Po --> 206/82 Pb + __/