Ask question

Ask question

All categories

3 years ago

8

A sample of Strontium-90 is found to have decayed to one-fourth of its original amount after 58.2 years. What is the half-life o

f strontium-90?

2 answers:

mafiozo [28]3 years ago

3 0

The half-life of strontium-90 is 28.8 years.

Igoryamba3 years ago

3 0

Answer: Half life of Sr-90 is 29.1 years.

Explanation: Half life is the time in which the substance remains halve of its original amount. Let's say originally we have 100 grams of Sr-90. From given information, it remains one-fourth of its original amount in 58.2 years.

One-fourth of 100 gram will be 25 gram. The equation used for solving this type of problems is:

$\frac{N}{N_0}=(0.5)^n$

where, n is the number of half lives, $N_0$ is the original amount of the radioactive substance and N is the remaining amount.

From above information:

$N_0$ = 100

N = 25

Let's plug in these values in the equation and find out the value of n.

$\frac{25}{100}=(0.5)^n$

$\frac{1}{4}=(0.5)^n$

$0.25=(0.5)^n$

$(0.5)^2=(0.5)^n$

n = 2

From above calculations, the value of n is 2 means there are two half lives.

From the given time and the value of n, we can calculate the half life of the substance using the equation:

half life = $\frac{T}{n}$

(Where T stands for time)

Let's plug in the values in the equation:

half life = $\frac{58.2years}{2}$

half life = 29.1 years

So, half life of Sr-90 is 29.1 years.

You might be interested in

Given a fixed amount of gas held at constant pressure, calculate the volume it would occupy if a 2.00 L sample were cooled from

aliina [53]

Answer:

1.82 L

Explanation:

We are given the following information;

Initial volume as 2.0 L
Initial temperature as 60.0°C
New volume as 30.0 °C

We are required to determine the new volume;

From Charles's law;

$\frac{V_1}{T_1}=\frac{V_2}{T_2}$

Where, $V_1 and V_2$ are initial and new volume respectively, while $T_1 and T_2$ are initial and new temperatures respectively;

$T_1= 333 K$

$T_2=303K$

$V_1 =2.0L$

Rearranging the formula;

$V_2=\frac{V_1T_2}{T_1}$

$= \frac{(2.0L)(303K)}{333K} \\=1.820 L$

Therefore, the new volume that would be occupied by the gas is 1.82 L

7 0

3 years ago

If 1.02 g of nickel reacted with 750. mL of 0.112 M hydrobromic acid, how much of each will be present at the end of the reactio

kati45 [8]

Answer:

35.1% is percent yield

Explanation:

<em>Full question: Assume no volume change. If you formed 0.0910 atm of gas, what is the percent yield?</em>

<em />

The reaction that is occurring is:

Ni + 3HBr → NiBr₃ + 3/2H₂(g)

First, we will determine moles of Ni and HBr to determine limiting reactant and theoretical yield

Using ideal gas law, we can determine the moles of hydrogen formed. Thus, we can find percent yield:

<em>Moles Ni (Molar mass: 58.69g/mol):</em>

1.02g * (1mol / 58.69g) = 0.01738moles Ni

<em>Moles HBr:</em>

0.750L * (0.112mol/L) = 0.084 moles of HBr.

For a complete reaction of the 0.084 moles of HBr you need:

0.084mol HBr * (1 mole Ni / 3 moles HBr) = 0.028 moles of Ni.

As there are just 0.01738 moles of Ni, the Ni is limiting reactant. Assuming a theoretical yield, moles of H₂ produced are:

0.01738moles Ni * (3/2 H₂ / 1 mol Ni) = 0.02607 moles H₂

Now, moles of H₂ produced are:

PV = nRT

PV/RT = n

<em>Where P is pressure (0.0910atm)</em>

<em>V is volume (2.50L)</em>

<em>R is gas constant (0.082atmL/molK)</em>

<em>T is absolute temperature in Kelvin (30°C + 273.15 = 303.15K)</em>

<em>And n are moles</em>

PV/RT = n

0.0910atm*2.50L/0.082atmL/molK*303.15K = n

0.00915 moles = n

<em />

And percent yield (Produced moles / Theoretical moles * 100) is:

0.00915 moles / 0.02607moles =

<h3>35.1% is percent yield</h3>

8 0

2 years ago

Which is a measurement of distance that is used around the world?

ivolga24 [154]

500 meters is the correct answer :)

8 0

3 years ago

Read 2 more answers

Sulfur dioxide and oxygen react to form sulfur trioxide during one of the key steps in sulfuric acid synthesis. An industrial ch

mamaluj [8]

Answer:

Explanation:

From the given information:

The equation for the reaction can be represented as:

$2SO_2 + O_2 \to 2SO_3$

The I.C.E table can be represented as:

2SO₂ O₂ 2SO₃

Initial: 14 2.6 0

Change: -2x -x +2x

Equilibrium: 14 - 2x 2.6 - x 2x

However, Since the amount of sulfur trioxide gas to be 1.6 mol.

SO₃ = 2x,

then x = 1.6/2

x = 0.8 mol

For 2SO₂; we have 14 - 2x

= 14 - 2(0.8)

= 14 - 1.6

= 12.4 mol

For O₂; we have 2.6 - x

= 2.6 - 1.6

= 1.0 mol

Thus;

[SO₂] = moles / volume = ( 12.4/50) = 0.248 M ,

[O₂] = 1/50 = 0.02 M ,

[SO₃] = 1.6/50 = 0.032 M

Kc = [SO₃]² / [SO₂]² [O₂]

= ( 0.032²) / ( 0.248² x 0.02)

= 0.8325

Recall that; the equilibrium constant for the reaction $2SO_2 + O_2 \to 2SO_3$ = 0.8325;

If we want to find:

$SO_2 + \dfrac{1}{2}O_2 \to SO_3$

Then:

$K_c = (0.8325)^{1/2}$

$\mathbf{K_c = 0.912}$

Since no temperature is given to use in the question, it will be impossible to find the final temperature of the mixture.

7 0

3 years ago

0.415 g of an unknown triprotic acid are used to make a 100.00 mL solution. Then 25.00 mL of this solution is transferred to an

Arturiano [62]

Answer:

Explanation:

Initial burette reading = 1.81 mL

final burette reading = 39.7 mL

volume of NaOH used = 39.7 - 1.81 = 37.89 mL .

37.89 mL of .1029 M NaOH is used to neutralise triprotic acid

No of moles contained by 37.89 mL of .1029 M NaOH

= .03789 x .1029 moles

= 3.89 x 10⁻³ moles

Since acid is triprotic , its equivalent weight = molecular weight / 3

No of moles of triprotic acid = 3.89 x 10⁻³ / 3

= 1.30 x 10⁻³ moles .

8 0

3 years ago