1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
Greeley [361]
3 years ago
6

Enter a balanced nuclear equation for the production of lead-212, used in the treatment of some cancers, from the beta decay of

thallium-212.
Chemistry
1 answer:
MariettaO [177]3 years ago
8 0

Answer:

\frac{212}{81}Tl ------>\beta  + \frac{212}{82}Pb

Explanation:

Beta emission involves the conversion of a neutron to a proton, an electron and a neutrino. This leads to a reduction in the neutron-proton ratio.

During a beta emission, the mass number of the parent and daughter nuclei remain the same. The atomic number of the daughter nucleus increases by one unit above that of the parent nucleus as we can see from the equation.

Hence, the daughter nucleus is found one place after its parent in the periodic table.

You might be interested in
Which element am I? Give my symbol ONLY! What a robber says as he's running away! ​
Olin [163]

Answer:

ummmmmmmmmmmmmmmm..mmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmm

Explanation:

ummmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmm.................... candyunicorns1999 has left the chat

5 0
3 years ago
Read 2 more answers
Lorelai discovers a compound that is 64.8 g C, 13.62 g H, and 21.58 g O. What is the empirical formula of Lorelai's compound
jenyasd209 [6]

The empirical formula of Lorelai's compound is C₄H₁₀O

<h3>Data obtained from the question</h3>
  • C = 64.8 g
  • H = 13.62 g
  • O = 21.58 g
  • Empirical formula =?

<h3>How to determine the empirical formula</h3>

Divide by their molar mass

C = 64.8 / 12 = 5.4

H = 13.62 / 1 = 13.62

O = 21.58 / 16 = 1.35

Divide by the smallest

C = 5.4 / 1.35 = 4

H = 13.62 / 1.35 = 10

O = 1.35 / 1.35 = 1

Thus, the empirical formula of the compound is C₄H₁₀O

Learn more about empirical formula:

brainly.com/question/24297883

#SPJ1

7 0
2 years ago
What would be an appropriate way to display the results of the experiment outlined?
miv72 [106K]
The best way (in my opinion) is to put it in a table or spreadsheet. 
6 0
3 years ago
Read 2 more answers
Metal carbonates decompose to the metal oxide and CO2 on heating according to this general equation. Mx(CO3)is) → My(s) + yCO2(g
Mrrafil [7]

Answer:

84.11 g/mol

Explanation:

A metal from group 2A will form the cation M²⁺, and the ion carbonate is CO₃²⁻, so the metal carbonate must be: MCO₃, and the reaction:

MCO₃(s) → MO(s) + CO₂(g)

For the stoichiometry of the reaction, 1 mol of MCO₃(s) will produce 1 mol of CO₂. Using the ideal gas law, it's possible to calculate the number of moles of CO₂:

PV = nRT , where P is the pressure, V is the volume(0.285 L), R is the gas constant (62.36 mmHg*L/mol*K), n is the number of moles, and T is the temperature (25 + 273 = 298 K).

69.8*0.285 = n*62.36*298

18583.28n = 19.893

n = 0.00107 mol

So, the number of moles of the metal carbonate is 0.00107. The molar mass is the mass divided by the number of moles:

0.0900/0.00107 = 84.11 g/mol

4 0
3 years ago
Please help me solve this!
yulyashka [42]

Answer : The image is attached below.

Explanation :

For O_3:

Molar mass, M = 48 g/mol

Mass, m = 24 g

Moles, n = \frac{m}{M}=\frac{24g}{48g/mol}=0.5mol

Number of particles, N = n\times 6.022\times 10^{23}=0.5\times 6.022\times 10^{23}=3.0\times 10^{23}

For NH_3:

Molar mass, M = 17 g/mol

Mass, m = 170 g

Moles, n = \frac{m}{M}=\frac{170g}{17g/mol}=10mol

Number of particles, N = n\times 6.022\times 10^{23}=10\times 6.022\times 10^{23}=6.0\times 10^{24}

For F_2:

Molar mass, M = 38 g/mol

Mass, m = 38 g

Moles, n = \frac{m}{M}=\frac{38g}{38g/mol}=1mol

Number of particles, N = n\times 6.022\times 10^{23}=1\times 6.022\times 10^{23}=6.0\times 10^{23}

For CO_2:

Molar mass, M = 44 g/mol

Moles, n = 0.10 mol

Mass, m = n\times M=0.10mol\times 44g/mol=4.4g

Number of particles, N = n\times 6.022\times 10^{23}=0.10\times 6.022\times 10^{23}=6.0\times 10^{22}

For NO_2:

Molar mass, M = 46 g/mol

Moles, n = 0.20 mol

Mass, m = n\times M=0.20mol\times 46g/mol=9.2g

Number of particles, N = n\times 6.022\times 10^{23}=0.20\times 6.022\times 10^{23}=1.2\times 10^{23}

For Ne:

Molar mass, M = 20 g/mol

Number of particles = 1.5\times 10^{23}

Moles, n = \frac{N}{6.022\times 10^{23}}=\frac{1.5\times 10^{23}}{6.022\times 10^{23}}=0.25mol

Mass, m = n\times M=0.25mol\times 20g/mol=5g

For N_2O:

Molar mass, M = 44 g/mol

Number of particles = 1.2\times 10^{24}

Moles, n = \frac{N}{6.022\times 10^{23}}=\frac{1.2\times 10^{24}}{6.022\times 10^{23}}=1.9mol

Mass, m = n\times M=1.9mol\times 44g/mol=83.6g

For unknown substance:

Number of particles = 3.0\times 10^{23}

Mass, m = 8.5 g

Moles, n = \frac{N}{6.022\times 10^{23}}=\frac{3.0\times 10^{23}}{6.022\times 10^{23}}=0.50mol

Molar mass, M = \frac{m}{n}=\frac{8.5g}{0.50mol}=17g/mol

The substance is NH_3.

3 0
3 years ago
Other questions:
  • Are there molecules of sodium chloride within a sodium chloride crystal?
    9·1 answer
  • Combination reactions always forms what?
    10·1 answer
  • What is the scientific notation of 38,802,000
    8·1 answer
  • A form of phosphorus called red phosphorus is used in match heads. When 0.062 g of red phosphorus burns in air, it forms 0.142 g
    6·1 answer
  • Determine the heat needed to warm 25.3 g of copper from 22 degrees celsius to 39 degrees celsius.
    11·1 answer
  • A can of soda holds 12fl oz. how many mL is this
    13·2 answers
  • Which statement is true of the motion of the ball?
    11·1 answer
  • PLEASE HURRY! What is the name of this alkane?
    9·1 answer
  • 1. You may "feel" cold when you touch certain kinds of matter because​
    14·1 answer
  • Write the symbol for every chemical element that has atomic number less than 13 and atomic mass greater than 17.2 u.
    7·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!