If a sample of Uranium-283 undergoes alpha decay what would be the complete chemical equation?

QAmmunity.org

My account
Edit my Profile
Private messages
My favorites

Ask a Question
Questions
Unanswered
Tags
Categories
Ask a Question

asked Apr 2, 2023 143k views

14 votes

If a sample of Uranium-283 undergoes alpha decay what would be the complete chemical equation?

Chemistry
high-school

Tiziano asked

by Tiziano

8.6k points

0 Comments

Please log in or register to add a comment.

Please log in or register to answer this question.

1 Answer

8 votes

Answer:

23892U→l23490Th+42He

Step-by-step explanation:

Humberto Pinheiro answered Apr 7, 2023

by Humberto Pinheiro

8.6k points

0 Comments

Please log in or register to add a comment.

← Prev Question Next Question →

Related questions

asked Apr 27, 2021 5.6k views

RACTICE Uranium (molar mass = 283.03 g/mol) has the largest molar mass of any element naturally found on Earth. What is the mass of 7.50 mol of uranium?

Andy Fowler asked Apr 27, 2021

by Andy Fowler

8.4k points

Physics
high-school

1 answer

4 votes

5.6k views

asked Nov 5, 2024 167k views

Identify the following substances involved in radioactive reactions, describing how you obtained the answer: a. An element which undergoes alpha decay to form uranium-236 b. The product of the alpha decay

Luis Averhoff asked Nov 5, 2024

by Luis Averhoff

9.2k points

Chemistry
high-school

1 answer

5 votes

167k views

asked Oct 28, 2024 108k views

When Uranium 238 undergoes alpha decay according to the equation below, what will be the mass number and atomic number of the resulting element? 238₉₂U→4₂ He+?

Carl Woodhouse asked Oct 28, 2024

by Carl Woodhouse

7.7k points

Chemistry
high-school

1 answer

1 vote

108k views

Ask a Question

Welcome to QAmmunity.org, where you can ask questions and receive answers from other members of our community.

9.5m questions

12.2m answers

0 Comments

Please log in or register to add a comment.

Please log in or register to answer this question.

1 Answer

0 Comments

Please log in or register to add a comment.

Related questions

Other Questions