Write the chemical formula for disilicon hexabromide.

An isotope undergoes radioactive decay. The new isotope that forms has an atomic number that is 2 less than the original isotope

Mnenie [13.5K]

Alpha Decay is the answer.

3 0

3 years ago

Read 2 more answers

You start out with 800 moles of a radioactive substance. After 24 hours, 100 moles remain. What is the half-life of the substanc

Nookie1986 [14]

The half life of the substance is 8 hours , Option D is the correct answer.

What is Half Life ?

Half-life is the time required by an unstable element to reduce to half of its initial value.

The term is commonly used in nuclear physics.

The formula is

$\rm N(t) = \; N_{0} \;(\dfrac{1}{2} )^{\frac{t}{t_{1/2} } }$

N(t) = quantity of the substance remaining

N ₀ = initial quantity of the substance

t = time elapsed

$\rm t_{1/2}$ = half life of the substance

The data given in the question is N ₀ = 800 moles , N(t)= 100 , t= 24 hours

$\rm t_{1/2} =\; ?$

Substituting values in the equation

$\rm 100 = \; 800 \;(\dfrac{1}{2} )^{\frac{24}{t_{1/2} } }$

$\rm \dfrac{100}{800} = \;(\dfrac{1}{2} )^{\frac{24}{t_{1/2} } }\\\\\rm \dfrac{1}{8} = \;(\dfrac{1}{2} )^{\frac{24}{t_{1/2} } }\\\\\rm (\dfrac{1}{2})^{3} = \;(\dfrac{1}{2} )^{\frac{24}{t_{1/2} } }\\\\\rm 3=\dfrac{24}{t_{1/2} }\\\\t_{1/2}= \dfrac{24}{3}\\\\t_{1/2} = 8 hours$

Therefore half life of the substance is 8 hours , Option D is the correct answer.

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3 0

2 years ago

The standard reduction potentials for the Ag+|Ag(s) and Zn2+| Zn(s) half-cell reactions are +0.799 V and -0.762 V, respectively.

mihalych1998 [28]

Answer: The potential of the given cell is 1.551 V

Explanation:

The given chemical cell follows:

$Zn(s)|Zn^{2+}(0.125M)||Ag^{+}(0.240M)|Ag(s)$

Oxidation half reaction: $Zn(s)\rightarrow Zn^{2+}(0.125M)+2e^-;E^o_{Zn^{2+}/Zn}=-0.762V$

Reduction half reaction: $Ag^{+}(0.240M)+e^-\rightarrow Ag(s);E^o_{Ag^{+}/Ag}=0.799V$ ( × 2)

Net cell reaction: $Zn(s)+2Ag^{+}(0.240M)\rightarrow Zn^{2+}(0.125M)+2Ag(s)$

Oxidation reaction occurs at anode and reduction reaction occurs at cathode.

To calculate the $E^o_{cell}$ of the reaction, we use the equation:

$E^o_{cell}=E^o_{cathode}-E^o_{anode}$

Putting values in above equation, we get:

$E^o_{cell}=0.799-(-0.762)=1.561V$

To calculate the EMF of the cell, we use the Nernst equation, which is:

$E_{cell}=E^o_{cell}-\frac{0.059}{n}\log \frac{[Zn^{2+}]}{[Ag^{+}]^2}$

where,

$E_{cell}$ = electrode potential of the cell = ? V

$E^o_{cell}$ = standard electrode potential of the cell = +1.561 V

n = number of electrons exchanged = 2

$[Zn^{2+}]=0.125M$

$[Ag^{+}]=0.240M$

Putting values in above equation, we get:

$E_{cell}=1.561-\frac{0.059}{2}\times \log(\frac{(0.125)}{(0.240)^2})$

$E_{cell}=1.551V$

Hence, the potential of the given cell is 1.551 V

6 0

4 years ago

What are good life notes about science

Grace [21]

Answer:

science is pointless and you will never need it so just get thru the class and let it go.

Explanation:

have a good day in need to get rid of the adds so I'm answering as many questions ap

7 0

3 years ago

A volume of 25cm3 of a carbonate solution of concentration 0.2mol dm-3 was neutralized by 20 cm3 of acid of concentration 0.5 mo

Galina-37 [17]

Answer: 1 mol of carbonate to 2 mol of acid

Explanation:

To calculate the number of moles for given molarity, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}\times 1000}{\text{Volume of solution in ml}}$

a) $\text{Moles of carbonate}=\frac{0.2moldm^{-3}\times 25cm^3}{1000}=0.005mol$

b) $\text{Moles of acid}=\frac{0.5moldm^{-3}\times 20cm^3}{1000}=0.01mol$

Thus the mole ratio of carbonate to acid is = $\frac{0.005}{0.01}=\frac{1}{2}$

7 0

3 years ago