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3 years ago

What are three common types of radioactivity given off by unstable atoms? How are they

Chemistry

1 answer:

Kisachek [45]3 years ago

4 0

Answer:

<h3>ALPHA DECAY</h3>

Alpha decay or α-decay is a type of radioactive decay in which an atomic nucleus emits an alpha particle and thereby transforms or 'decays' into a different atomic nucleus, with a mass number that is reduced by four and an atomic number that is reduced by two.

<h3>BETA DECAY</h3>

Beta decay is one process that unstable atoms can use to become more stable

in beta DECAY parent nuclide has proton number increased by 1 and mass number is same

<h3>GAMMA DECAY</h3>

Gamma decay is one type of radioactive decay that a nucleus can undergo. What separates this type of decay process from alpha or beta decay is that no particles are ejected from the nucleus when it undergoes this type of decay. Instead, a high energy form of electromagnetic radiation - a gamma ray photon - is released. Gamma rays are simply photons that have extremely high energies which are highly ionizing.[1] As well, gamma radiation is unique in the sense that undergoing gamma decay does not change the structure or composition of the atom. Instead, it only changes the energy of the atom s

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Calculate the molality for each of the following solutions. Then, calculate the freezing-point depression ΔTF = iKFcm produced b

zlopas [31]

Answer:

a) Cm= 3.9 m ; ΔTf= 14.51 ºC

b) Cm= 0.21 m ; ΔTf= 0.79ºC

Explanation:

In order to solve the problems, we have to remember that the molality (m) of a solution is equal to moles of solute in 1 kg of solvent.

m= mol solute/kg solvent

a) In this case we have molarity, which is moles of solute in 1 liter of solution. We have to know how many kg of solvent (water) we have in 1 L of solution.

3.2 M NaCl= 3.2 mol NaCl/ 1 L solution

1 L solution= 1000 ml solution x 1.00 g/ml= 1000 g

A solution is composed by solute (NaCl) + solvent, so:

1000 g solution = g NaCl + g solvent

g NaCl= 3.2 mol NaCl x 58.44 g/mol= 187 g NaCl

g solvent= 1000 g - 187 g NaCl= 813 g= 0.813 kg

Cm= 3.2 g NaCl/0.813 kg solvent= 3.9 m

NaCl is an electrolyte and it dissociates in water in two ions: Na⁺ anc Cl⁻, si the van't Hoff factor (i) is 2.

ΔTf= i x KF x Cm= 2 x 1.86ºC/m x 3.9 m= 14.51ºC

b) In this case we have 24 g of solute in 1.5 L of solvent. We have to convert the liters of solvent to kg, and to convert the mass of solute to mol by using the molecular weight of KCl (74.55 g/mol):

24 g KCl x 1 mol KCl/74.55 g= 0.32 mol

1.5 L solvent= 1500 g solvent x 1.00 g/ml= 1500 g = 1.5 kg

Cm= 0.32 g KCl/1.5 kg solvent= 0.21 m

KCl is an electrolyte and when it dissolves in water, it dissociates in 2 ions: K⁺ and Cl⁻. For this, van't Hoff factor (i) is equal to 2.

ΔTf= i x KF x Cm= 2 x 1.86ºC x 0.21 m= 0.79ºC

7 0

3 years ago

100 PIONTSSSSS HELP ASAP

valina [46]

Left Panel

Short answer A

Solution

Since you have been given choices, my sloppy numbers will do, but it anyone is going to see this, YOU SHOULD CLEAN THEM UP WITH THE NUMBERS THAT COME FROM YOUR PERIODIC TABLE.

Equation

Sodium Phosphate + Calcium Chloride ===> Sodium Chloride + Calcium Phosphate.

Na3PO4 + CaCl2 ===> NaCl + Ca3(PO4)2

Step One

Balance the Equation

2Na2PO4 + 3CaCl2 ==> 6NaCl + Ca3(PO4)2

Step Two

Find the molar mass of CaCl2

Ca = 40

2Cl = 71

Molar Mass = 40 + 71 = 111 grams/mole

Step Three

Find the number of moles of CaCl2

Given mass = 379.4

Molar Mass = 111

moles = given Mass / molar Mass

moles of CaCl2 = 379.4/111 = 3.418 moles

Step Four

Find the number of moles of Ca3(PO4)2 needed.

This requires that you use the balance numbers from the balanced equation.

For every 3 moles of CaCl2 you have, you get 1 mole of Ca3(PO4)2

n_moles of Ca3(PO4)2 = 3.418 / 3 = 1.13933 moles

Step Five

Find the molar mass of Ca3(PO4)2

From the periodic table,

3Ca = 3 * 40 = 120

2 P = 2 * 31 = 62

8 O = 8 * 16 =128

Molar Mass = 120 + 62 + 128= 310 grams per mole.

Step Six

1 mole of Ca3(PO4)2 has a molar mass of 310 gram

1.13933 moles of Ca3(PO4)2 = x

x = 1.13933 moles * 310 grams /mole

x = 353.2 grams. As you can see, even with my rounding I'm only out 0.3 of a gram. DON'T FORGET TO PUT THIS TO THE PROPER SIG DIGS IF SOMEONE ELSE IS GOING TO SEE IT.

Middle Panel

Short Answer C

Equation

2HCl + Mg ===> H2 + MgCl2

The object of the first part of the game is to find the number of moles of H2.

Step One

Find the moles of HCl

1 mole HCl = 35.5 + 1 = 36.5

n = given mass divided by molar mass

n = 49 grams / 36.5 = 1.34 moles.

The balanced equation tells you that for ever mole of H2 produced, you need 2 moles of HCl. That's what the balance numbers are for.

So the number of moles of H2 is 1.34 / 2 = 0.671 moles of H2.

Now we come to Part II. We have to use an new friend of yours that I have seen only once before from you.

Find V using PV = nRT

R is going to be in kPa so the value of R = 8.314

V = ???

n = 0.671 moles

T = 25 + 273 = 298oK

P = 101.3 kPa

101.3 * V= 0.671*8.314 * 298

V = 0.671 * 8.314 * 298 / 101.3

V = 16.4

The answer is C and again, I have rounded almost everything except R, although it can go out to 8 places.

Right Panel

I can't see the panel. I don't know what the problem is. Never mind I got it. I'm going to be a little skimpy on this one since I've done two like it and they are long.

LiOH + HBr ===> LiBr + H2O and the equation is balanced.

You have to figure out the moles of LiOH and HBr. Use the LOWEST number of moles

n_LiOH = given mass / molar mass = 117/(7 + 16 + 1) = 117 / 24 = 4.875 moles

n_HBr = given mass / molar mass = 141/(1 + 80) = 141 / 81 = 1.741 moles

HBr is the lower number. That's all the LiBr you are going to get is 1.741. There is no adjustment to be made from the balance equation.

n = given mass / molar mass multiply both sides by the molar mass

n * Molar mass (LiBr) = n * (7 + 80) = 1.741 * 87 = 151 grams of

The answer is C

6 0

3 years ago

How many moles are in 6.80 x 10^23 atoms of gold, Au?

frosja888 [35]

Answer:

1.13 moles Au

Explanation:

Moles Au = 6.80x10²³atoms / 6.023x10²³atoms/mole = 1.13 moles Au

8 0

3 years ago

What would happen if the Earth was tilted more than its current tilt of 23.5°? HELP ASAP!!

Oduvanchick [21]

Answer:

Wouldn't the Earth's atmosphere be moving too fast that it eventually breaks out?

Explanation:

Do NOT trust me.

7 0

3 years ago

Consider the decomposition of a metal oxide to its elements, where M represents a generic metal. M 3 O 4 ( s ) − ⇀ ↽ − 3 M ( s )

Citrus2011 [14]

Answer:

a) ΔGrxn = 6.7 kJ/mol

b) K = 0.066

c) PO2 = 0.16 atm

Explanation:

a) The reaction is:

M₂O₃ = 2M + 3/2O₂

The expression for Gibbs energy is:

ΔGrxn = ∑Gproducts - ∑Greactants

Where

M₂O₃ = -6.7 kJ/mol

M = 0

O₂ = 0

$deltaG_{rxn} =((2*0)+(3/2*0))-(1*(-6.7))=6.7kJ/mol$

b) To calculate the constant we have the following expression:

$lnK=-\frac{deltaG_{rxn} }{RT}$

Where

ΔGrxn = 6.7 kJ/mol = 6700 J/mol

T = 298 K

R = 8.314 J/mol K

$lnK=-\frac{6700}{8.314*298} =-2.704\\K=0.066$

c) The equilibrium pressure of O₂ over M is:

$K=P_{O2} ^{3/2} \\P_{O2}=K^{2/3} =0.066^{2/3} =0.16atm$

3 0

3 years ago