In the nuclear transmutation represented by 168o(p, \alpha) 137n, the emitted particle is ________.

Answers

Answer 1

Answer: Answer is: the emitted particle is an alpha particle.

Nuclear reaction: ¹⁶O + p⁺→ ¹³N + α (alpha particle).
Alpha decay is radioactive decay in which an atomic nucleus emitsan alpha particle (helium nucleus) and transforms into an atomwith an atomic number that is reduced by two and massnumber that is reduced by four.
When oxygen-16 gain one proton, atomic mass is 17, but when lose alpha particleatomic mass reduces by four to 13.

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PLEASEEEEEEEE ASAPPPPPWhat is the density of a piece of cardboard that has a mass of 250 g and volume of 46 mL? *

Answers

HEELP ME HELP ME HELP ME HELP ME HELP ME

Answer:

5.4347826087

Explanation:

The formula for density is Mass/Volume so you would do 250/46 to get the answer of 5.4347826087 grams per milliliter

Which of the following is an example of scientific inference

Answers

give me more details pleaseee i can't see anything and i wanna help

Pharmacists sometimes measure medicines in the unit of "grains," where 1 gr = 65 mg. The label on a bottle of aspirin reads "Aspirin, 5 gr." How many mg of aspirin are in a tablet containing 5 gr?

Answers

Answer:

325mg of Aspririn

Explanation:

First you should note the information that the problem gives you:

- The bottle of Aspirin has 5gr (grains)

- 1gr(grain) = 65mg (miligrams)

Also, the problem is asking about how many aspirin are in 5 gr (grains), so you should use a conversion factor, as follows:

-First you should put the quantity you need to convert:

$5grAspirin$

-Then you write the denominator of the conversion factor that must have the same units that you want to convert, in this case gr:

$5grAspirin*\frac{}{1grAspirin}$

-Then you write the numerator with the units that you want to obtain and the numerical equivalence between the units, in this case:

$5grAspirin*(65mgAspririn)/(1grAspirin)$

-Finally you multiply numerators and divide by denominators:

$5grAspirin*(65mgAspririn)/(1grAspirin)=325mgAspririn$

Calculate the pH of a buffer solution made by adding 15.0 g anhydrous sodium acetate (NaC2H3O2) to 100.0 mL of 0.200 M acetic acid. Assume there is no change in volume on adding the salt to the acid. (pKa for acetic acid is 4.74 or Ka is 1.8 x 10-5)3.

Answers

Answer:

pH of Buffer Solution 5.69

Explanation:

Mole of anhydrous sodium acetate = $(Given mass)/(Molecular mass)$

= $(15)/(82)$

= 0.18 mole

100 ml of 0.2 molar acetic acid means

= M x V

= 0.2 x 100

= 20 mmol

= 0.02 mole

Using Henderson equation to find pH of Buffer solution

pH = pKa + log $([Salt])/([Acid])$

= 4.74 + log $(0.18)/(0.02)$

= 4.74 + log 9

= 5.69

So pH of the Buffer solution = 5.69

A student uses paper towels to clean up a small chemical spill in the lab.Where should the dirty paper towels be placed?
A in the lab sink
B in the wastebasket
C in the chemical waste container
D in the hazardous waste container

Answers

Answer:

C. in the chemical wastebasket

C is the answer: chemical waste container

Consider the neutralization reaction 2HNO3(aq) + Ba(OH)2 ( aq ) ⟶ 2H2O ( l ) + Ba ( NO3)2 ( aq ). A 0.125 L sample of an unknown HNO 3 solution required 32.3 mL of 0.200 M Ba ( OH ) 2 for complete neutralization. What is the concentration of the HNO 3 solution?

Answers

Answer:

The concentration of the HNO3 solution is 0.103 M

Explanation:

Step 1: Data given

Volume of the unknow HNO3 sample = 0.125 L

Volume of 0.200 M Ba(OH)2 = 32.3 mL = 0.0323 L

Step 2: The balanced equation

2HNO3(aq) + Ba(OH)2 ( aq ) ⟶ 2H2O ( l ) + Ba( NO3)2 (aq)

Step 3:

n2*C1*V1 = n1*C2*V2

⇒ n2 = the number of moles of Ba(OH)2 = 1

⇒ C1 = the concentration of HNO3 = TO BE DETERMINED

⇒ V1 = the volume of the HNO3 solution = 0.125 L

⇒ n1 = the number of moles of HNO3 = 2

⇒ C2 = the concentration of Ba(OH)2 = 0.200 M

⇒ V2 = the volume of Ba(OH)2 = 0.0323 L

1*C1 * 0.125 L = 2*0.200M * 0.0323 L

C1 = (2*0.200*0.0323)/0.125

C1 = 0.103 M

The concentration of the HNO3 solution is 0.103 M

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Use the following structures of amino acids to answer the questions below. Note that the difference in the structures (the side chains) is highlighted by gray shading.A student performed chromatography of the four amino acids and theresults were shown in the chromatogram below. If an anion exchangecolumn (column is positively charged) was used in a neutral buffer,assign each amino acid to the corresponding peak in the chromatogram.

How many sulfide ions are in 15 dg of sodium sulfide?O 1.16 x 1022O 3.04 x 1028O 5.76 x 1029O 6.02 x 1023O 3.13