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Readme [11.4K]
3 years ago
15

Terbium-147 undergoes positron emission to become a stable atom. What is that stable atom?

Chemistry
1 answer:
ivanzaharov [21]3 years ago
7 0

We have to get the stable atom formed after positron emission from Terbium-147.

The stable atom is (D) ₆₄Gd¹⁴⁷.

Positron is radioactive decay. Positron is a type of beta particle β⁺.

Positron emission decreases proton number relative to neutron number, positron decay results in nuclear transmutation, changing an atom of one chemical element with an atomic number that is less by one.

Terbium on positron emission produces Gadolinium with one atomic number less than Terbium. So, the positron emission reaction is as shown below:

Tb¹⁴⁷→ ₆₄Gd¹⁴⁷ + ₁e⁰

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STatiana [176]

Explanation:

  • loss of electron or an increase in oxidation number
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DETERMINE THE MASS 1.366 mol of NH3
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Determine the mass of each of the following amounts. a. 1.366 mol of
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Calculate the unit cell edge length for an 85 wt% fe-15 wt% v alloy. All of the vanadium is in solid solution, and, at room temp
Lady bird [3.3K]

Answer is 0.289nm.

Explanation: The wt % of Fe and wt % of V is given for a Fe-V alloy.

wt % of Fe in Fe-V alloy = 85%

wt % of V in Fe-V alloy = 15%

We need to calculate edge length of the unit cell having bcc structure.

Using density formula,

\rho_{ave}=\frac{Z\times M_{ave}}{a^3\times N_A}

For calculating edge length,

a=(\frac{Z\times M_{ave}}{\rho_{ave}\times N_A})^{1/3}

For calculating M_{ave}, we use the formula

M_{ave}= \frac{100}{\frac{(wt\%)_{Fe}}{M_{Fe}}+\frac{(wt\%)_{V}}{M_V}}

Similarly for calculating (\rho)_{ave}, we use the formula

\rho_{ave}= \frac{100}{\frac{(wt\%)_{Fe}}{\rho_{Fe}}+\frac{(wt\%)_{V}}{\rho_V}}

From the periodic table, masses of the two elements can be written

M_{Fe}= 55.85g/mol

M_{V}=50.941g/mol

Specific density of both the elements are

(\rho)_{Fe}=7.874g/cm^3\\(\rho)_{V}=6.10g/cm^3

Putting  M_{ave} and \rho_{ave} formula's in edge length formula, we get

a=\left [\frac{Z\left (\frac{100}{\frac{(wt\%)_{Fe}}{M_{Fe}}+\frac{(wt\%)_{Fe}}{M_{Fe}}}  \right )}{N_A\left (\frac{100}{\frac{(wt\%)_V}{\rho_V}+\frac{(wt\%)_V}{\rho_V}}  \right )}  \right ]^{1/3}

a=\left [\frac{2atoms/\text{unit cell}\left (\frac{100}{\frac{85\%}{55.85g/mol}+\frac{15\%}{50.941g/mol}}  \right )}{(6.023\times10^{23}atoms/mol)\left (\frac{100}{\frac{85\%}{7.874g/cm^3}+\frac{15\%}{6.10g/cm^3}}  \right )}  \right ]^{1/3}

By calculating, we get

a=2.89\times10^{-8}cm=0.289nm

7 0
3 years ago
A piece of iron (mass = 25.0 g) at 398 k is placed in a styrofoam coffee cup containing 25.0 ml of water at 298 k. assuming that
Firlakuza [10]
The final temperature of the water is the equilibrium temperature, or the also the final temperature of the iron after a long period of time. Applying the conservation of energy:

m,iron*C,iron*ΔT = - m,water*C,water*ΔT

The density of water is 1000 g/mL.

(25 g)(0.449 J/g·°C)(T - 398 K) = - (25 mL)(1000 g/mL)(4.18 J/g·°C)(T - 298)
Solving for T,
<em>T = 298.01 K</em>
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The awnser is C because if it was a claim it would be proven with facts and if it were evidence it would also be proven with facts
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