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3 years ago

What is the name of the isotope produced when Californium-251 emits an alpha particle?

Chemistry

1 answer:

Charra [1.4K]3 years ago

3 0

<h3>Answer:</h3>

Curium-247 <em>i.e.</em> ²⁴⁷₉₆Cm

<h3>Explanation:</h3>

Alpha decay is given by following general equation,

ᵃₓA → ⁴₂He + ᵃ⁻⁴ₓ₋₂B

Where;

A = Parent Isotope

B = Daughter Isotope

ᵃ = Mass Number

ₓ = Atomic Number

Californium-251 is the parent isotope in our case and it has 98 protons (atomic number) and is given as,

²⁵¹₉₈Cf

The alpha decay reaction of Californium-251 will be as,

²⁵¹₉₈Cf → ⁴₂He + ²⁴⁷₉₆B

The symbol for B with atomic number 96 was found to be the atom of Curium (Cm) by inspecting periodic table. Hence, the final equation is as follow,

²⁵¹₉₈Cf → ⁴₂He + ²⁴⁷₉₆Cm

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How many half-lives are required for the concentration of reactant to decrease to 1.56% of its original value?4247.56.56

neonofarm [45]

Answer:

6 half-lives are required for the concentration of reactant to decrease to 1.56% of its original value.

Explanation:

Using integrated rate law for first order kinetics as:

$[A_t]=[A_0]e^{-kt}$

Where,

$[A_t]$ is the concentration at time t

$[A_0]$ is the initial concentration

Given:

Concentration is decreased to 1.56 % which means that 0.0156 of $[A_0]$ is decomposed. So,

$\frac {[A_t]}{[A_0]}$ = 0.0156

Thus,

$\frac {[A_t]}{[A_0]}=e^{-k\times t}$

$0.0156=e^{-k\times t}$

kt = 4.1604

The expression for the half life is:-

Half life = 15.0 hours

$t_{1/2}=\frac {ln\ 2}{k}$

Where, k is rate constant

So,

$k=\frac {ln\ 2}{t_{1/2}}$

$\frac{4.1604}{t}=\frac {ln\ 2}{t_{1/2}}$

$t = 6\times t_{1/2}$

<u>6 half-lives are required for the concentration of reactant to decrease to 1.56% of its original value.</u>

6 0

3 years ago

Starting with 2.50 mol of N2 gas (assumed to be ideal) in a cylinder at 1.00 atm and 20.0C, a chemist first heats the gas at con

arsen [322]

Answer:

a) $T_b=590.775k$

b) $W_t=1.08*10^4J$

d) $Q=3.778*10^4J$

d) $\triangle V=4.058*10^4J$

Explanation:

From the question we are told that:

Moles of N2 $n=2.50$

Atmospheric pressure $P=100atm$

Temperature $t=20 \textdegree C$

$t = 20+273$

$t = 293k$

Initial heat $Q=1.36 * 10^4 J$

Generally the equation for change in temperature is mathematically given by

$\triangle T=\frac{Q}{N*C_v}$

Where

$C_v=Heat\ Capacity \approx 20.76 J/mol/K$

$T_b-T_a=\frac{1.36 * 10^4 J}{2.5*20.76 }$

$T_b-293k=297.775$

$T_b=590.775k$

Generally the equation for ideal gas is mathematically given by

$PV=nRT$

For v double

$T_c=2*590.775k$

$T_c=1181.55k$

Therefore

$PV=Wbc$

$Wbc=(2.20)(8.314)(1181_590.778)$

$Wbc=10805.7J$

Total Work-done $W_t$

$W_t=Wab+Wbc$

$W_t=0+1.08*10^4$

$W_t=1.08*10^4J$

Generally the equation for amount of heat added is mathematically given by

$Q=nC_p\triangle T$

$Q=2.20*2907*(1181.55-590.775)\\$

$Q=3.778*10^4J$

Generally the equation for change in internal energy of the gas is mathematically given by

$\triangle V=nC_v \triangle T$

$\triangle V=2.20*20.76*(1181.55-293)k$

$\triangle V=4.058*10^4J$

3 0

2 years ago

What types of atoms release nuclear radiation

djyliett [7]

Isotopes of elements where the nucleas is unstable generally release nuclear radiation. So unstable atoms

4 0

3 years ago

Which is the correct equation for the reaction of magnesium with hydrochloric acid to produce hydrogen gas and magnesium chlorid

djyliett [7]

Answer:

Mg(s) + 2 HCl(aq) --> MgCl 2(aq) + H 2(g)

8 0

2 years ago

Calculate the mass of sodium azide required to decompose and produce 2.104 moles of nitrogen. Refer to the periodic table to get

Alexxx [7]

91 grams of sodium azide required to decompose and produce 2.104 moles of nitrogen.

Explanation:

2NaN3======2Na+3N2

This is the balanced equation for the decomposition and production of sodium azide required to produce nitrogen.

From the equation:

2 moles of NaNO3 will undergo decomposition to produce 3 moles of nitrogen.

In the question moles of nitrogen produced is given as 2.104 moles

so,

From the stoichiometry,

3N2/2NaN3=2.104/x

= 3/2=2.104/x

3x= 2*2.104

= 1.4 moles

So, 1.4 moles of sodium azide will be required to decompose to produce 2.104 moles of nitrogen.

From the formula

no of moles=mass/atomic mass

mass=no of moles*atomic mass

1.4*65

= 91 grams of sodium azide required to decompose and produce 2.104 moles of nitrogen.

4 0

3 years ago