Ask question

Ask question

All categories

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.

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