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2 years ago

Which isotope in each pair is more stable? Why?(c) ²⁸₁₂Mg or ²⁴₁₂Mg

Chemistry

1 answer:

3241004551 [841]2 years ago

7 0

Among ²⁸₁₂Mg ²⁴₁₂Mg, ²⁴₁₂Mg is more stable.

<h3>Briefly explained</h3>

We can identify which isotope in the pair is more stable. Let's review the criteria that are used to determine a stable isotope. A stable isotope will have more neutrons than protons with a neutron to proton ratio somewhere between one and 1.52 On the high atomic number, we're up to 152 at low atomic numbers were closer to one.

More stable isotopes have an even number of protons and neutrons. And stable isotopes have their proton or neutron number equal to one of these magic numbers.

We have ²⁸₁₂Mg with 16 neutrons and 12 protons with the neutron to proton ratio of 1.33. And we compare that to ²⁴₁₂Mg With 12 neutrons and 12 protons and the neutrons to protons ratio of one. So we do have a relatively small Z value. So the new trying to approach on ratio should be close to one. Wherefore ²⁸₁₂Mg , although both of these are even we do have a pretty high neutron to proton ratio for a low atomic number.

So ²⁸₁₂Mg is likely less stable While²⁴₁₂Mg is more stable.

Learn more about stable isotope

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The equilibrium constant, Kp, for the following reaction is 10.5 at 350 K: 2CH2Cl2(g) CH4(g) CCl4(g) Calculate the equilibrium p

Ostrovityanka [42]

The complete question is as follows: The equilibrium constant, Kp, for the following reaction is 10.5 at 350 K: 2CH2Cl2(g) CH4(g) + CCl4(g) Calculate the equilibrium partial pressures of all species when CH2Cl2(g) is introduced into an evacuated flask at a pressure of 0.968 atm at 350 K. PCH2Cl2 = atm PCH4 = atm PCCl4 = atm

Answer: The equilibrium partial pressures of all species, that is, $CH_{4}$ , $CCl_{4}$ and $CH_{2}Cl_{2}$ is 0.420 atm, 0.420 atm and 0.128 atm.

Explanation:

For the given reaction equation, the initial and equilibrium concentration of involved species is as follows.

$2CH_{2}Cl_{2}(g) \rightarrow CH_{4}(g) + CCl_{4}(g)\\$

Initial: 0.968 atm 0 0

Equilibrium: (0.968 - 2x) x x

Now, $K_{p}$ for this reaction is as follows.

$K_{p} = \frac{P_{CH_{4}}P_{CCl_{4}}}{P^{2}_{CH_{2}Cl_{2}}}\\10.5 = \frac{x \times x}{(0.968 - 2x)^{2}}\\x = 0.420$

$P_{CH_{4}} = x = 0.420 atm\\P_{CCl_{4}} = x = 0.420 atm\\P_{CH_{2}Cl_{2}} = (0.968 atm - 2x) = (0.968 atm - 2(0.420)) = 0.128 atm$

Thus, we can conclude that the equilibrium partial pressures of all species, that is, $CH_{4}$ , $CCl_{4}$ and $CH_{2}Cl_{2}$ is 0.420 atm, 0.420 atm and 0.128 atm.

3 0

3 years ago

5.22*10^-3 = ?*10^-2

Andrew [12]

x=0.522 is the answer

x times :10^-2=5.22 times 10^{-3}\

simplify 10^-3: 5.22/1000= x times 10^-2 = 5.22/1000

simplify 10^-2 1/100 = x 1/100=5.22/1000

Multiply both sides by 100 100x 1/100 = 5.22x100/1000

simplify 0.522 sorry I took so long I did the best i could to explain it

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3 years ago

The solid fuel in the booster stage of the space shuttle is a mix- ture of ammonium perchlorate and aluminum powder. Upon igni-

nataly862011 [7]

Answer:

For A: The mass of aluminium required will be 183 g

For B: The mass of alumina produced will be $6.63\times 10^6g$

Explanation:

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ .....(1)

For a:

Given mass of $NH_4ClO_4$ = 1.325 kg = 1325 g (Conversion factor: 1 kg = 1000 g)

Molar mass of $NH_4ClO_4$ = 117.50 g/mol

Putting values in equation 1, we get:

$\text{Moles of }NH_4ClO_4=\frac{1325g}{117.50g/mol}=11.28mol$

For the given chemical reaction:

$6NH_4ClO_4(s)+10Al(s)\rightarrow 5Al_2O_3(s)+3N_2(g)+6HCl(g)+9H_2O(g)$

By stoichiometry of the reaction:

6 moles of $NH_4ClO_4$ reacts with 10 moles of aluminium

So, 11.28 moles of $NH_4ClO_4$ will react with = $\frac{10}{6}\times 11.28=6.77mol$ of aluminium

Now, calculating the mass of aluminium by using equation 1, we get:

Molar mass of aluminium = 27.00 g/mol

Moles of aluminium = 6.77 moles

Putting values in equation 1, we get:

$6.77mol=\frac{\text{Mass of aluminium}}{27.00g/mol}\\\\\text{Mass of aluminium}=(6.77mol\times 27.00g/mol)=183g$

Hence, the mass of aluminium required will be 183 g

For b:

Given mass of aluminium = $3.500\times 10^3=3.5\times 10^6g$ (Conversion factor: 1 kg = 1000 g)

Molar mass of aluminium = 27.00 g/mol

Putting values in equation 1, we get:

$\text{Moles of aluminium}=\frac{3.5\times 10^6g}{27.00g/mol}=1.3\times 10^5mol$

For the given chemical reaction:

$6NH_4ClO_4(s)+10Al(s)\rightarrow 5Al_2O_3(s)+3N_2(g)+6HCl(g)+9H_2O(g)$

By stoichiometry of the reaction:

10 moles of aluminium produces 5 moles of alumina

So, $1.3\times 10^5mol$ of aluminium will react with = $\frac{5}{10}\times 1.3\times 10^5=6.5\times 10^4mol$ of alumina

Now, calculating the mass of alumina by using equation 1, we get:

Molar mass of alumina = 101.96 g/mol

Moles of alumina = $6.5\times 10^4mol$

Putting values in equation 1, we get:

$6.5\times 10^4mol=\frac{\text{Mass of alumina}}{101.96g/mol}\\\\\text{Mass of alumina}=(6.5\times 10^4mol\times 101.96g/mol)=6.63\times 10^6g$

Hence, the mass of alumina produced will be $6.63\times 10^6g$

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3 years ago

What features form when rocks bend?

bezimeni [28]

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