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GarryVolchara [31]
2 years ago
11

One way to identify the type of radioactive decay produced in a reaction is to pass the emission through an electric field. Desc

ribe the type of radioactive emission produced from the decay of uranium-
238 to thorium-234 and its reaction to the electric field
Chemistry
1 answer:
Sergeeva-Olga [200]2 years ago
4 0

Decay of Uranium-238 produces gamma radiaton whereas decay of thorium-234 releases beta radiation.

<h3>What type of radiation is produced?</h3>

In the decay of U-238, two gamma rays of different energies are emitted in addition to the alpha particle while on the other hand, in the decay of thorium-234 , beta rays are emitted.

So we can conclude that decay of Uranium-238 produces gamma radiaton whereas decay of thorium-234 releases beta radiation.

Learn more about decay here: brainly.com/question/25537936

You might be interested in
230g sample of a compound contains 136.6g carbon, 26.4g hydrogen, and 31.8g nitrogen. What is masspercentif oxygen
natulia [17]

Answer:

15.3 %

Explanation:

Step 1: Given data

  • Mass of the sample (ms): 230 g
  • Mass of carbon (mC); 136.6 g
  • Mass of hydrogen (mH): 26.4 g
  • Mass of nitrogen (mN): 31.8 g

Step 2: Calculate the mass of oxygen (mO)

The mass of the sample is equal to the sum of the masses of all the elements.

ms = mC + mH + mN + mO

mO = ms - mC - mH - mN

mO = 230 g - 136.6 g - 26.4 g - 31.8 g

mO = 35.2 g

Step 3: Calculate the mass percent of oxygen

%O = (mO / ms) × 100% = (35.2 g / 230 g) × 100% = 15.3 %

6 0
3 years ago
Gallium is a metallic element in Group III. It has similar properties to aluminium.
Liula [17]

Answer:

Gallium has Orthorhombic and forms ionic bonds with metals

Explanation:

The structure of gallium is Orthorhombic and it has a high neutron capture cross section. It is very stable in dry air but oxidizes in moist air. It reacts with oxygen to form Ga2O3. It has a valence of III. Its electronic configuration is  

Ar 3d10 4s2 4p1

It is found in the 3A group of periodic table and it mostly forms ionic bonds with metal.  

3 0
3 years ago
The following calculations must be handwritten in your notebook. – Acetic Acid ■ Hydrogen ion concentration ■ Ka ■ % Error – Ace
Nikitich [7]

Answer:

Explanation:

1) Acetic acid

Concentration is given as 0.103 M

The average pH of this solution = 2.96

we know that pH = - log [H+] therefore [H+] = 10-pH

[H+] = 10-2.96

= 1.1 x 10-3 M = 0.0011 M

Consider the equilibrium

CH3COOH ⇄CH3COO- + H+

Initial 0.103 0 0

Change -x +x +x

equlibrium 0.103 -x x x

Ka = x2 / 0.103 - x

Here the initial concentration of CH3COOH = 0.103 M

the equilibrium concentration of H+ = x = 0.0011 M

Therefore the equilibrium conc of acetic acid = 0.103 - 0.0011 = 0.1019 M

Therefore Ka = 0.0011 x 0.0011 / 0.1019 = 1.187 x 10-5

2) Acetic acid + NaOH

pH measured = 4.48 , therefore [H+} = 10-4.48 = 3.3 x 10-5

Volume and conc of acetic acid = 10 mL of 0.103 M

= 10 mL x 0.103 mmol / mL

= 1.03 mmol

Volume and conc of NaOH added = 4 mL of 0.0992 M

= 4 x 0.0992 mmol

= 0.397 mmol

Consider the equation

CH3COOH + NaOH -----------> CH3COONa + H2O

Initial 1.03 0.397 0

Final 0.633 0 0.397

0.397 mmole of NaOH will convert 0.397 mmole of acetic acid to sodium acetate.

Thus the final moles of acetic acid and sodium acetate in the solution are 0.633 and 0.397

therefore [salt] / [acid] = 0.397 / 0.633 = 0.627

By Hendersen equation pH = pKa + log[salt / acid]

pH = pKa + log 0.627 = pKa - 0.203

or pKa = pH + 0.203 = 4.48 + 0.203 [ since the measured pH = 4.48]

= 4.683

Ka = 10-4.683 = 2.07 x 10-5

3) Phosphate salts:

(i) mass of NaH2PO4 taken = 0.613 g

molar mass of NaH2PO4 = 120

therefore moles = 0.613 / 120 = 0.0051 mole

= 5.1 mmol

The volume is 30 mL therefore concentration = 5.1 /30 mmol/mL

= 0.17 M

consider the equilibrium

H2PO4-⇄ HPO42- + H+

Initial 0.17 0 0

Change -x +x + x

equilibrium 0.17-x x x

Ka = x2 / 0.17-x = 6.2 x 10-8 [ Ka is given]

neglect x in the denominator as it is very small x2 = 0.17 x 6.2 x 10-8

x =  1.03 x 10-4

Thus the equilirium conc of H+ = 1.03 x 10-4 therefore pH = - log 1.03 x 10-4 = 3.99

(ii) Mass of Na2HPO4.7H2O =0.601 g

therefore no of moles = 0.601 / 268.07 = 0.00224 mole

= 2.24 mmol

The volume = 30 mL , therefore conc = 2.24 / 30 mmol/ml

= 0.075 M

consider the equilibrium

HPO42- ⇄ PO43- + H+

Initial 0.075 0 0

Change -x +x + x

equilibrium 0.075-x x x

Ka = x2 / 0.075-x = 4.8 x 10-13 [ Ka is given]

neglect x in the denominator as it is very small x2 = 0.075 x 4.8 x 10-13

x =  1.9 x 10-7

Thus the equilirium conce of H+ = 1.9 x 10-7 therefore pH = - log 1.9 x 10-7 = 6.7

(iii) Mass of Na3PO4.12H2O taken = 0.208 g

moles of trisodiumphosphate 0.208/ 380 = 0.00055 moles

= 0.55 mmol

Volume = 10 mL therefore conc = 0.55/10 = 0.055 mmol/mL

= 0.055 M

Consider the equilibrium reaction

PO43- + H2O  ⇄ HPO42- + OH-

initial 0.055 0 0

Change -x +x +x

equilibrium 0.055-x x x

Kb = x2/ 0.055 -x = 0.0208 [Kb = Kw / Ka = 10-14 / 4.8 x 10-13 = 0.0208]

x2 + 0.0208x - 0.001144 = 0 Solving this equation we get x = 0.025

That is the conce of OH- ion = 0.025M

Therefore pH = 14 - pOH = 14 - 1.6 =12.4

3 0
2 years ago
An aqueous potassium iodate (KIO3) solution is made by dissolving 553 grams of KIO3 in sufficient water so that the final volume
liberstina [14]

Answer:

M KIO3 = 1.254 mol/L

Explanation:

  • molarity (M) [=] mol/L

∴ w KIO3 = 553 g

∴ mm KIO3 = 214.001 g/mol

∴ volumen sln = 2.10 L

⇒ mol KIO3 = (553 g)×(mol/210.001 g) = 2.633 mol

⇒ M KIO3 = (2.633 mol KIO3 / (2.10 L sln)

⇒ M KIO3 = 1.254 mol/L

3 0
3 years ago
Read 2 more answers
in example 5.11 of the text the molar volume of n2 at STP is given as 22.42 L/mol how is this number calculatd how does the mola
Valentin [98]

Answer:

V = 22.42 L/mol

N₂ and H₂ Same molar Volume at STP

Explanation:

Data Given:

molar volume of N₂ at STP = 22.42 L/mol

Calculation of molar volume of N₂ at STP  = ?

Comparison of molar volume of H₂ and N₂ = ?

Solution:

Molar Volume of Gas:

The volume occupied by 1 mole of any gas at standard temperature and pressure and it is always equal to 22.42 L/ mol

Molar volume can be calculated by using ideal gas formula  

                               PV = nRT

Rearrange the equation for Volume

                            V = nRT / P . . . . . . . . . (1)

where

P = pressure

V = Volume

T= Temperature

n = Number of moles

R = ideal gas constant

Standard values

P = 1 atm

T = 273 K

n = 1 mole

R = 0.08206 L.atm / mol. K

Now put the value in formula (1) to calculate volume for 1 mole of N₂

                   V = 1 x 273 K x 0.08206 L.atm / mol. K / 1 atm

                   V = 22.42 L/mol

Now if we look for the above calculation it will be the same for H₂ or any gas. so if we compare the molar volume of 1 mole N₂ and H₂ it will be the same at STP.

6 0
3 years ago
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