Assume that you want to construct a voltaic cell that uses the following half reactions:

slader Consider the following reactions: A. uranium-238 emits an alpha particle; B. plutonium-239 emits an alpha parti- cle; C.

pishuonlain [190]

Answer:

For A: The equation is $_{92}^{238}\textrm{U}\rightarrow _{90}^{234}\textrm{Th}+_2^4\alpha$

For B: The equation is $_{94}^{239}\textrm{Pu}\rightarrow _{92}^{235}\textrm{U}+_2^4\alpha$

For C: The equation is $_{90}^{239}\textrm{Th}\rightarrow _{91}^{235}\textrm{Pa}+_{-1}^0\beta$

Explanation:

Alpha decay process is the process in which nucleus of an atom disintegrates into two particles. The first one which is the alpha particle consists of two protons and two neutrons. This is also known as helium nucleus. The second particle is the daughter nuclei which is the original nucleus minus the alpha particle released.

$_Z^A\textrm{X}\rightarrow _{Z-2}^{A-4}\textrm{Y}+_2^4\alpha$

Beta decay process is defined as the process the neutrons get converted into an electron and a proton. The released electron is known as the beta particle. In this process, the atomic number of the daughter nuclei gets increased by a factor of 1 but the mass number remains the same.

$_Z^A\textrm{X}\rightarrow _{Z+1}^A\textrm{Y}+_{-1}^0\beta$

For A: Uranium-238 emits an alpha particle

The nuclear equation for this process follows:

$_{92}^{238}\textrm{U}\rightarrow _{90}^{234}\textrm{Th}+_2^4\alpha$

For B: Plutonium-239 emits an alpha particle

The nuclear equation for this process follows:

$_{94}^{239}\textrm{Pu}\rightarrow _{92}^{235}\textrm{U}+_2^4\alpha$

For C: Thorium-239 emits a beta particle

The nuclear equation for this process follows:

$_{90}^{239}\textrm{Th}\rightarrow _{91}^{235}\textrm{Pa}+_{-1}^0\beta$

6 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

The valences of metal x,y and z are 1,2 and 3 respectively. What are the formulae of their;a) hydroxides, b) sulphates, c) hydro

Rina8888 [55]

Answer:

See answer below

Explanation:

AS we know that the valence for those metals X, Y, and Z are 1, 2 and 3, we can determine the formula of each compound.

1. Hydroxides.

An hydroxide is formed when an oxyde of a metal reacts with water. When this happens, the general molecular formula is:

Meₐ(OH)ₙ

Where:

a: valence or charge of the hydroxide (Which is -1)

n: valence of the metal.

Following this, the formula for X, Y and Z would be:

XOH

Y(OH)₂

Z(OH)₃

2. Sulphates

Sulphates follow a similar rule of hydroxide in the general molecular formula, but instead of having a charge of -1, it has a charge of -2 so:

Mₐ(SO₄)ₙ

So, following the rule:

X₂SO₄

Y₂(SO₄)₂ ------> YSO₄

Z₂(SO₄)₃

3. Hydrogens

Following the same rule as the previous, hydrogens works with a charge of -1, so:

MₐHₙ

Then:

XH

YH₂

ZH₃

4. Carbonates.

This follows the same rule as sulphates, with the same charge so:

Mₐ(CO₃)ₙ

Then:

X₂CO₃

YCO₃

Z₂(CO₃)₃

5. Nitrates

Follow the same rule as the hydroxides, with the same charge of -1.

Mₐ(NO₃)ₙ

Then:

XNO₃

Y(NO₃)₂

Z(NO₃)₂

6. Phosphates

In the case of phosphates, these have a charge of -3 so:

Mₐ(PO₄)ₙ

Then:

X₃PO₄

Y₃(PO₄)₂

Z₃(PO₄)₃ ----> ZPO₄

Hope this helps

6 0

3 years ago

Strontium-90 is a radioisotope that will decrease in mass by one-half every 29 years. How many years will it take for a 10.0-gra

Lelechka [254]

Answer: It will take 29 years for a 10.0-gram sample of strontium-90 to decay to 5.00 grams

Explanation:

Radioactive decay process is a type of process in which a less stable nuclei decomposes to a stable nuclei by releasing some radiations or particles like alpha, beta particles or gamma-radiations. The radioactive decay follows first order kinetics.

Half life is the amount of time taken by a radioactive material to decay to half of its original value.

Half life is represented by $t_{\frac{1}{2}$

$t_{\frac{1}{2}=\frac{0.693}{\lambda}$

$\lambda$ = rate constant

Given : Strontium-90 decreases in mass by one-half every 29 years , that is half life of Strontium-90 is 29 years.

As half life is independent of initial concentration, it will take 29 years for a 10.0-gram sample of strontium-90 to decay to 5.00 grams as the amount gets half.

7 0

3 years ago

Please help Explain why a molecule that has polar bonds can be a polar molecule.

Zepler [3.9K]

Answer:

Explanation:

Well, obviously a molecule with polar bonds can be polar in itself. It's like saying I am an atheltic person who can just reach the basketball rim with my head and also I can dunk.

But if the question is how can a molecule that in non-polar have polar bonds, well, its because the polar bonds' dipole cancels each other out. It's like a tight rope. If a person pulls in one direction, it intuitively, the rope would go in that direction. However, if a person pulls in the other direction with the same amount of force, the rope stays still. This is the same case. Although molecules can have different electronegativities, the pull of electrons in one direction is cancelled out by a pull in the opposite direction, making the net dipole 0.

This is common for main VSERP shaped molecules like linear, trigonal planar, tetrahedral, trigonal bipyramidal, and octahedral.

8 0

3 years ago