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deff fn [24]
1 year ago
13

The mass of radium-226 in a sample is found to have decreased from 45.00g to 5.625g in a period of 4800 years.From this informat

ion. Calculate the half life of radium-226
Chemistry
1 answer:
yan [13]1 year ago
4 0

Answer:

Half life = 1600 years

Explanation:

Given data:

Total mass of sample = 45.00 g

Mass remain = 5.625 g

Time period = 4800 years

Half life of radium-226 = ?

Solution:

First of all we will calculate the number of half lives passes,

At time zero 45.00 g

At first half life = 45.00 g/ 2= 22.5 g

At 2nd half life = 22.5 g/ 2 = 11.25 g

At 3rd half life = 11.25 g/ 2= 5.625 g

Half life:

Half life = Time elapsed / number of half lives

Half life = 4800 years / 3

Half life = 1600 years

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Sodium (chemical symbol Na) is atomic number 11 and is in the first column of the periodic table so it has one valence electron.
lisabon 2012 [21]
Na is cation so it loses electron to be positive and become stable losing one valence shells one electron so it's oxidation number is +1 ie A is correct
7 0
2 years ago
Read 2 more answers
A 5.00 liter gas sample is collected at a temperature and pressure of 27.0 degree C and 1.20 atm. It is desired to transfer the
kipiarov [429]

Answer:

The temperature of the gas in the 3.00 liter container, must be 150K

Explanation:

Let's apply the Ideal Gas Law, to find out the moles

P . V = n . R . T

1,20 atm . 5L = n . 0,082 L.atm/mol.K . 300 K

(1,20 atm . 5L) /  (0,082 mol.K/L.atm . 300 K) = n

6/24,6 mol = n = 0,244 moles

We have the moles now, so let's find the temperature in our new conditions.

P . V = n . R . T

1 atm . 3L = 0,244 moles . 0,082 L.atm/mol.K . T° in K

(1 atm . 3L / 0,244 moles . 0,082 mol.K/L.atm) = T° in K

3/20,008 K = T° in K = 150K

3 0
1 year ago
Small beads of iridium-192 are sealed in a plastic tube and inserted through a needle into breast tumors. If an Ir-192 sample ha
Stells [14]

Answer:

296.1 day.

Explanation:

  • The decay of radioactive elements obeys first-order kinetics.
  • For a first-order reaction: k = ln2/(t1/2) = 0.693/(t1/2).

Where, k is the rate constant of the reaction.

t1/2 is the half-life time of the reaction (t1/2 = 1620 years).

∴ k = ln2/(t1/2) = 0.693/(74.0 days) = 9.365 x 10⁻³ day⁻¹.

  • For first-order reaction: <em>kt = lna/(a-x).</em>

where, k is the rate constant of the reaction (k = 9.365 x 10⁻³ day⁻¹).

t is the time of the reaction (t = ??? day).

a is the initial concentration of Ir-192 (a = 560.0 dpm).

(a-x) is the remaining concentration of Ir-192 (a -x = 35.0 dpm).

<em>∴ kt = lna/(a-x)</em>

(9.365 x 10⁻³ day⁻¹)(t) = ln(560.0 dpm)/(35.0 dpm).

(9.365 x 10⁻³ day⁻¹)(t) = 2.773.

<em>∴ t </em>= (2.773)/(9.365 x 10⁻³ day⁻¹) =<em> 296.1 day.</em>

5 0
2 years ago
The density of liquid Z is 0.9237 g/mL. A student masses a cup
Ulleksa [173]

Answer:

37.65mL

Explanation:

Given parameters:

density of liquid  Z = 0.9237g/mL

Mass of liquidZ + mass of cup = 50.7g

Mass of cup= 15.92g

Volume of liquid Z in cup=?

Solution:

 Density is the mass per unit volume of a substance. It is mathematically expressed as shown below:

    Density = \frac{mass}{volume}

To find the volume of liquid Z, we know the density of the liquid but we dont know the mass yet.

Mass of liquidZ = 50.7g - mass of cup = 50.7g - 15.92g = 34.98g

Therefore:

 Volume of liquidZ = \frac{mass of liquidZ}{density of liquidZ}

                               =  \frac{34.98}{0.9237}

                               = 37.65mL

3 0
1 year ago
Now consider the reaction when 45.0 g NaOH have been added. What amount of NaOH is this, and what amount of FeCl3 can be consume
gayaneshka [121]

The correct answer is that 1.125 mol of NaOH is available, and 60.75 g of FeCl₃ can be consumed.

The mass of NaOH is 45 g

The molar mass of NaOH = 40 g/mol

The moles of NaOH = mass / molar mass

= 45 / 40

= 1.125

Thus, 1.125 mol NaOH is available

3 NaOH + FeCl₃ ⇒ Fe (OH)₃ + 3NaCl

3 mol of NaOH react with 1 mol of FeCl₃

1.125 moles of NaOH will react with x moles of FeCl₃

x = 1.125 / 3

x = 0.375 mol

0.375 mol FeCl₃ can take part in reaction

The molar mass of FeCl₃ is 162 g/mol

The mass of FeCl₃ = moles × mass

= 0.375 × 162

= 60.75 g

Thus, the amount of FeCl₃, which can be consumed is 60.75 g

6 0
2 years ago
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